MathsGee Answers - Recent questions and answers in Mathematics
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Powered by Question2AnswerWhy do we study lots of mathematics when we mostly end up relying on spreadsheets for basic calculations on the job?
https://mathsgee.com/30230/study-mathematics-mostly-relying-spreadsheets-calculations
<p>Why do we study lots of mathematics when we mostly end up relying on spreadsheets for basic calculations on the job?</p>
<p><img alt="" src="https://mathsgee.com/?qa=blob&qa_blobid=9679235276023771423" style="height:600px; width:600px"></p>
<p> </p>Mathematicshttps://mathsgee.com/30230/study-mathematics-mostly-relying-spreadsheets-calculationsSun, 20 Jun 2021 14:38:54 +0000Answered: For the following function state the following: its domain, any discontinuities and their types, what values should redefine the function to remove any removable discontinuities (give the extended function).
https://mathsgee.com/30228/following-following-discontinuities-removable-discontinuities?show=30229#a30229
Polynomials are continuous on their entire domain of all real numbers. So, rational functions like $f$ can only be discontinuous when the denominator is equal to $0 .$ This happens in two places:<br />
$x=3$ and $x=-3$. We'll check the limits from each side at each of these points to determine the type of discontinuity. For $x=3$,<br />
<br />
$$<br />
\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{-}} f(x)<br />
<br />
$$<br />
<br />
<br />
<br />
$$<br />
<br />
=\lim _{x \rightarrow 3} f(x)=\lim _{x \rightarrow 3} \frac{(x+2)(x+1)(x-3)}{(x+3)(x-3)}<br />
<br />
$$<br />
<br />
<br />
<br />
$$<br />
<br />
=\lim _{x \rightarrow 3} \frac{(x+2)(x+1)}{(x+3)}=\frac{20}{6}=\frac{10}{3}<br />
$$<br />
<br />
So, $f$ has a removable discontinuity at $x=3$ because the left and right limits are the same. For $x=3$<br />
<br />
$$<br />
\lim _{x \rightarrow-3^{+}} f(x)=-\infty \text { and } \lim _{x \rightarrow-3^{+}} f(x)=\infty<br />
$$<br />
<br />
So, $f$ has an infinite discontinuity at $x=-3$ because both of the left and right limits go to $\pm \infty$. The value we got from the limits at $x=3$ gives us the value we need to redefine $f$ as to remove the discontinuity. The extended function is therefore<br />
<br />
$$<br />
f_{e}(x)=\left\{\begin{array}{ll}<br />
f(x) & x \neq 3 \\<br />
\frac{10}{3} & x=3<br />
\end{array}\right.<br />
$$Mathematicshttps://mathsgee.com/30228/following-following-discontinuities-removable-discontinuities?show=30229#a30229Sun, 20 Jun 2021 03:43:45 +0000Answered: What are the different types of discontinuities in calculus?
https://mathsgee.com/30226/what-are-the-different-types-of-discontinuities-in-calculus?show=30227#a30227
<p>There are four major types of discontinuity.</p>
<p>
<br>
<strong>Removable:</strong></p>
<p>If $f$ is discontinuous at $c$ but we can remove the discontinuity by setting $f$ equal to its limit at $c$, then $f$ has a removable discontinuity at $c$.</p>
<p><strong>Jump:</strong></p>
<p>If $f$ is discontinuous at $c$, and both of the one-sided limits exist but are different, then $f$ has a jump discontinuity at $c$.</p>
<p><strong>Infinite:</strong></p>
<p>If $f$ has a vertical asymptote at $c$, meaning one or both sides go to $\pm \infty$, then $f$ has an infinite discontinuity at $c$.</p>
<p><strong>Oscillating:</strong></p>
<p>If $f$ oscillates without limit at $c$, then $f$ has an oscillating discontinuity at $c$. An example of such a function would be $\sin \frac{1}{x}$ at $x=0$.</p>
<p>It might seem strange that $\sin \frac{1}{x}$ has an oscillating discontinuity at $x=0$ because we were able to find the limit as $x$ approaches of 0 of $x \sin \frac{1}{x}$, a very similar function. However, remembering how we applied the Ham Sandwich Theorem to find this limit, we see that the $x$ term bounds the amplitude of the oscillations, allowing the limit to be 0 .</p>Mathematicshttps://mathsgee.com/30226/what-are-the-different-types-of-discontinuities-in-calculus?show=30227#a30227Sun, 20 Jun 2021 03:41:29 +0000Answered: Find the points of continuity and discontinuity of the following functions $$ \text { 1. } f(x)=\frac{1}{x^{2}+1} \quad \text { 2. } g(x)=e^{1 / x} $$
https://mathsgee.com/30224/find-points-continuity-discontinuity-following-functions?show=30225#a30225
1. There are no points where $f(x)$ or its limit are undefined. Further, there are no points where $f$ and its limit at that point are different. So, $f$ is continuous on $(-\infty, \infty)$ and discontinuous on $\emptyset$.<br />
<br />
2. Since $1 / x$ is undefined at $x=0, g(x)$ is also undefined at $x=0$. At every other point, $g$ and its limit are defined and are equal. So, $g$ is continuous on $(\infty, 0) \cup(0, \infty)$ and discontinuous on $[0]$.Mathematicshttps://mathsgee.com/30224/find-points-continuity-discontinuity-following-functions?show=30225#a30225Sun, 20 Jun 2021 03:38:34 +0000Answered: When is a real-valued function continuous?
https://mathsgee.com/30222/when-is-a-real-valued-function-continuous?show=30223#a30223
<p>When we were looking at limits, we noticed that we can't always substitute to find the limit, even if the function is defined there. In the example given to show that substitution and the limit can give different results, we saw a special type of function that seemed to have a "hole" at the point we were interested in finding the limit of. This function is said to be discontinuous at this point, and in this section we'll define when a function is or isn't continuous at a point based on this idea of the limit and substitution giving different values.</p>
<p>Definition. Let $f(x)$ be a real-valued function defined over $D \subseteq \mathbb{R} . f(x)$ is continuous at some point $x=c$ if all of the following hold.</p>
<ol>
<li>$\lim _{x \rightarrow c} f(x)$ exists</li>
<li>$f(c)$ is defined</li>
<li>$\lim _{x \rightarrow c} f(x)=f(c)$ (substitution works)</li>
</ol>
<p>Otherwise, $f(x)$ is discontinuous at $c .{ }^{1}$
<br>
We say that a function is continuous on an interval if it's continuous on every point in that interval. </p>Mathematicshttps://mathsgee.com/30222/when-is-a-real-valued-function-continuous?show=30223#a30223Sun, 20 Jun 2021 03:36:16 +0000Answered: Find the limit $\lim _{x \rightarrow \infty} x+e^{-x}$
https://mathsgee.com/30220/find-the-limit-lim-x-rightarrow-infty-x-e-x?show=30221#a30221
Looking at the two terms, we can see that as $x$ gets large, $e^{-x}$ gets very small, contributing less and less to the overall value. So, we can say that this function as a right end behavior model of $x$, so the limit is $\infty$.Mathematicshttps://mathsgee.com/30220/find-the-limit-lim-x-rightarrow-infty-x-e-x?show=30221#a30221Sun, 20 Jun 2021 03:33:29 +0000Answered: Find the limit $\lim _{x \rightarrow \infty} \frac{x^{3}-6 x+1}{x^{2}+2 x-3}$
https://mathsgee.com/30212/find-the-limit-lim-x-rightarrow-infty-frac-x-3-6-x-1-x-2-2-x-3?show=30219#a30219
Since the numerator degree is bigger than the denominator degree, we'll need to use the end behavior model. The end behavior model tells us that the numerator term dominates and has positive values, so the limit evaluates to $\infty$.Mathematicshttps://mathsgee.com/30212/find-the-limit-lim-x-rightarrow-infty-frac-x-3-6-x-1-x-2-2-x-3?show=30219#a30219Sun, 20 Jun 2021 03:32:07 +0000Answered: Find the limit $\lim _{x \rightarrow-\infty} \frac{x-9}{2 x-x^{2}}$
https://mathsgee.com/30213/find-the-limit-lim-x-rightarrow-infty-frac-x-9-2-x-x-2?show=30218#a30218
Since the denominator has higher degree than the numerator, there is a HA at $y=0$, so the limit evaluates to 0 .Mathematicshttps://mathsgee.com/30213/find-the-limit-lim-x-rightarrow-infty-frac-x-9-2-x-x-2?show=30218#a30218Sun, 20 Jun 2021 03:31:16 +0000Answered: Find the limit $\lim _{x \rightarrow \infty} \frac{6 x^{2}-4 x^{5}+7 x-1}{12 x^{5}-3 x^{2}+2}$
https://mathsgee.com/30214/find-the-limit-lim-rightarrow-infty-frac-x-2-4-x-5-7-x-1-12-x-5-3-x-2-2?show=30217#a30217
Since the numerator and denominator have the same degree, the limit is the ratio of the highest-degree coefficients, $\dfrac{-1}{3}$.Mathematicshttps://mathsgee.com/30214/find-the-limit-lim-rightarrow-infty-frac-x-2-4-x-5-7-x-1-12-x-5-3-x-2-2?show=30217#a30217Sun, 20 Jun 2021 03:30:14 +0000Answered: Find the limit $\lim _{x \rightarrow \infty} \frac{3 x+1}{|x|+2}$
https://mathsgee.com/30215/find-the-limit-lim-x-rightarrow-infty-frac-3-x-1-x-2?show=30216#a30216
The numerator and denominator have the same degree. For $x>0,|x|+2=x+2$, so the limit is the ratio of highest-degree coefficients, $3 .$Mathematicshttps://mathsgee.com/30215/find-the-limit-lim-x-rightarrow-infty-frac-3-x-1-x-2?show=30216#a30216Sun, 20 Jun 2021 03:29:23 +0000Answered: What are horizontal asymptotes when dealing with limits in calculus?
https://mathsgee.com/30210/what-horizontal-asymptotes-when-dealing-with-limits-calculus?show=30211#a30211
Horizontal Asymptotes are a special type of end-behavior model.<br />
<br />
Definition. The line $y=b$ is a horizontal asymptote of $y=f(x)$ if $\lim _{x \rightarrow \infty} f(x)=b$ or $\lim _{x \rightarrow-\infty} f(x)=$ $b$.<br />
We can determine horizontal asymptotes for rational functions (usually quotient of polynomials). There are a few cases to consider<br />
1. If the numerator is a higher degree than the denominator, there is no horizontal asymptote, so we'll need a different method to calculate what happens at $\pm \infty$.<br />
2. If the denominator is a higher degree than the numerator, then there is a horizontal asymptote at $y=0$.<br />
3. If the numerator and denominator have the same degree, there is a horizontal asymptote at $y=k$ where $\mathrm{k}$ is the ratio of the highest degree terms.Mathematicshttps://mathsgee.com/30210/what-horizontal-asymptotes-when-dealing-with-limits-calculus?show=30211#a30211Sun, 20 Jun 2021 03:25:16 +0000Answered: What is the End Behavior Model?
https://mathsgee.com/30208/what-is-the-end-behavior-model?show=30209#a30209
<p>When $x$ is numerically large, we can often model the behavior of a complicated function with a simplier one that behaves roughly the same for numerically large input values and is the same in the limit. There are a few rules that these follow.</p>
<ol>
<li>For a polynomial, the end-behavior is highest-degree term.</li>
<li>For a rational function, like a ratio of polynomials, the end behavior is the ratio of the highest degree terms.</li>
<li>For more complicated functions, we may need to use some reasoning about the graph of the function and limit properties to determine end-behavior.</li>
</ol>Mathematicshttps://mathsgee.com/30208/what-is-the-end-behavior-model?show=30209#a30209Sun, 20 Jun 2021 03:23:39 +0000Answered: How do you define infinite limits in calculus?
https://mathsgee.com/30206/how-do-you-define-infinite-limits-in-calculus?show=30207#a30207
Although our limit definition works for finite values of $c$, it's also useful to think about what happens as $c$ goes to $\pm \infty$. We'll need to add to our limit definition to incorporate infinite values, since it doesn't make sense to talk about neighborhoods at infinity.<br />
<br />
Definition. Let $f$ be a real-valued function defined on some subset $D \subseteq \mathbb{R}$ that contains arbitrarily large values.<br />
$$<br />
\lim _{x \rightarrow \infty} f(x)=L<br />
$$<br />
if for every real $\epsilon>0$, there is a real number $N>0$ such that for all $x \in D$,<br />
$$<br />
x>N \Longrightarrow|f(x)-L|<\epsilon<br />
$$<br />
All the same properties that we described for finite limits, like the Sum and Difference Rule, still hold for infinite limits.Mathematicshttps://mathsgee.com/30206/how-do-you-define-infinite-limits-in-calculus?show=30207#a30207Sun, 20 Jun 2021 03:21:41 +0000Answered: Evaluate the following limit $$ \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta} $$
https://mathsgee.com/30204/evaluate-following-limit-theta-rightarrow-frac-theta-theta?show=30205#a30205
<p>We’ll need to use some geometric ideas to solve this limit. Consider the following on a unit circle.</p>
<p><img alt="" src="https://mathsgee.com/?qa=blob&qa_blobid=8265065801070537833" style="height:232px; width:436px"></p>
<p>We can see that the area of the swept arc is between the two triangle with base of length 1 and heights of $\sin \theta$ and $\tan \theta$. So, we can write the following inequality.
<br>
$$
<br>
\begin{aligned}
<br>
\frac{1}{2} & \sin \theta \leq \frac{1}{2} \theta \leq \frac{1}{2} \frac{\sin \theta}{\cos \theta} \\
<br>
\sin \theta \leq \theta & \leq \frac{\sin \theta}{\cos \theta}
<br>
\end{aligned}
<br>
$$</p>
<p>Taking the reciprocal of each part and multiplying by $\sin \theta$,
<br>
$$
<br>
1 \geq \frac{\sin \theta}{\theta} \geq \cos \theta
<br>
$$
<br>
Taking the limit of as $\theta$ approaches 0 of each term,
<br>
$$
<br>
\begin{aligned}
<br>
&1 \geq \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta} \geq \lim _{\theta \rightarrow 0} \cos \theta \\
<br>
&1 \geq \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta} \geq 1
<br>
\end{aligned}
<br>
$$
<br>
So, by the Sandwich Theorem,
<br>
$$
<br>
\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1
<br>
$$</p>Mathematicshttps://mathsgee.com/30204/evaluate-following-limit-theta-rightarrow-frac-theta-theta?show=30205#a30205Sun, 20 Jun 2021 03:19:48 +0000Answered: How does the sandwich/squeeze theorem for limits work?
https://mathsgee.com/30202/how-does-the-sandwich-squeeze-theorem-for-limits-work?show=30203#a30203
We can use the Sandwich Theorem (also called the Squeeze Theorem) to indirectly find limits by "sandwiching" the function in question between two functions we do know the limit of. If these two sandwiching functions go to the same value in the limit, then so to must the function in question.<br />
<br />
Theorem (The Sandwich Theorem). If $g(x) \leq f(x) \leq h(x)$ and $\lim _{x \rightarrow c} g(x)=\lim _{x \rightarrow c} h(x)=L$,<br />
then $\lim _{x \rightarrow c} f(x)=L$.Mathematicshttps://mathsgee.com/30202/how-does-the-sandwich-squeeze-theorem-for-limits-work?show=30203#a30203Sun, 20 Jun 2021 03:17:04 +0000Answered: How do you explain left and right hand limits in calculus?
https://mathsgee.com/30200/how-do-you-explain-left-and-right-hand-limits-in-calculus?show=30201#a30201
Our definition of the limit requires that the function get arbitrarily close to the limit value when approaching from both the left and right hand sides. However, we can evaluate limits by specifying that we only approach from one side. We usually notate this with a superscript $+$ or $-$ next to the $x$ limit value. So,<br />
$$<br />
\lim _{x \rightarrow 0^{+}} f(x)<br />
$$<br />
would mean "the limit of $f(x)$ as $x$ approaches 0 from the right", while<br />
$$<br />
\lim _{x \rightarrow 0^{-}} f(x)<br />
$$<br />
would mean "the limit of $f(x)$ as $x$ approaches 0 from the left."<br />
<br />
Our definition of the limit from both sides requires the left and right sides to be the same. If they are different, the the limit does not exist.<br />
$$<br />
\lim _{x \rightarrow c^{+}} f(x)=\lim _{x \rightarrow c^{-}} f(x) \Longrightarrow \lim _{x \rightarrow c^{+}} f(x)=\lim _{x \rightarrow c^{-}} f(x)=\lim _{x \rightarrow c} f(x)<br />
$$<br />
$\lim _{x \rightarrow c^{+}} f(x) \neq \lim _{x \rightarrow c^{-}} f(x) \Longrightarrow \lim _{x \rightarrow c} f(x)$ does not exist (DNE).Mathematicshttps://mathsgee.com/30200/how-do-you-explain-left-and-right-hand-limits-in-calculus?show=30201#a30201Sun, 20 Jun 2021 03:15:34 +0000Answered: Find the limit of $f(x)$ as $x$ approaches 2 for the following function. $$ f(x)=\left\{\begin{array}{ll} x^{2} & x \neq 2 \\ 0 & x=2 \end{array}\right. $$
https://mathsgee.com/30198/limit-approaches-following-function-begin-array-array-right?show=30199#a30199
Although it may seem obvious from our idea that limits describe behavior at a point that if $f(x)$ is defined at $x=c$, then $\lim _{x \rightarrow c} f(x)=f(c)$. However, this is not always the case. Remember that our definition of a limit required these $\epsilon$ and $\delta$ neighborhoods around the limit point. If $f(x)$ is defined at $x=c$, but $(c, f(c))$ is not a point in these neighborhoods for any $\epsilon>0$, then the limit will not evaluate to $f(c)$.<br />
<br />
We can clearly see that $f(2)=0$, but for $\epsilon=0.1$, for example, there is no $\delta$ that can satisfy our definition, as points like $\left(2-\delta, 4-2 \delta+\delta^{2}\right)$ would outside the neighborhood around $(2,0)$. In fact, the correct limit value is 4, the same as if $f(x)=x^{2}$ for all $x .$ There are some more nuances we'll need to describe before we can say when it's OK to substitute to evaluate a limit.Mathematicshttps://mathsgee.com/30198/limit-approaches-following-function-begin-array-array-right?show=30199#a30199Sun, 20 Jun 2021 03:13:56 +0000Answered: What are some of the properties of limits in calculus?
https://mathsgee.com/30196/what-are-some-of-the-properties-of-limits-in-calculus?show=30197#a30197
Limit have many nice properties all allow us to make useful simplifications when evaluating a limit. Let<br />
$$<br />
\lim _{x \rightarrow c} f(x)=L \text { and } \lim _{x \rightarrow c} g(x)=M<br />
$$<br />
<br />
Sum and Difference Rule: $\lim _{x \rightarrow c}(f(x) \pm g(x))=L \pm M$<br />
Product Rule: $\lim _{x \rightarrow c}(f(x) g(x))=L M$<br />
Constant Multiple Rule: $\lim _{x \rightarrow c} k \cdot f(x)=k \lim _{x \rightarrow c} f(x)=k L$<br />
Quotient Rule: $\lim _{x \rightarrow c} \frac{f(x)}{g(x)}=\frac{\lim _{x \rightarrow x} f(x)}{\lim _{x \rightarrow c} g(x)}=\frac{L}{M}$, if $M \neq 0$<br />
Power Rule: If $n \neq \in \mathbb{R}, \lim _{x \rightarrow c}(f(x))^{n}=\left(\lim _{x \rightarrow c} f(x)\right)^{n}=L^{n}$Mathematicshttps://mathsgee.com/30196/what-are-some-of-the-properties-of-limits-in-calculus?show=30197#a30197Sun, 20 Jun 2021 03:11:50 +0000Answered: In calculus, how are limits defined?
https://mathsgee.com/30194/in-calculus-how-are-limits-defined?show=30195#a30195
<p>Limits are a way of describing what happens to a function $f(x)$ as $x$ gets arbitrarily close to a value from some direction (positive or negative). This allows us not only to deal with "holes" in some functions but describe some of the building blocks of calculus, namely the derivative.</p>
<p>Definition. Let $f: D \subseteq \mathbb{R} \rightarrow \mathbb{R}$. Let $c \in R$ be a limit point (ie $c \in D$ or $c$ is on the boundary of $D$ ). $f$ has a limit $L$ as $x$ approaches $c$ if for any given positive real number $\epsilon$, there is a positive real number $\delta$ such that for all $x \in D$,
<br>
$$
<br>
0<|x-c|<\delta \Longrightarrow|f(x)-L|<\epsilon
<br>
$$
<br>
We write this as
<br>
$$
<br>
\lim _{x \rightarrow c} f(x)=L .
<br>
$$</p>
<p> </p>
<p><img alt="" src="https://mathsgee.com/?qa=blob&qa_blobid=14085047355738251891" style="height:418px; width:495px"></p>
<p>Visually, what this means is that for any "error bound" of $y$ values $\epsilon$, I can give you a corresponding error bound of $x$ values $\delta$ such that all values of $f(z)$ for $z \in(c-\delta, c+\delta)$ bound are between $L-\epsilon$ and $L+\epsilon$. We don't use this definition of the limit very often because it's a bit cumbersome. However, it's important to know that when we use the limit, this is the formal definition making things work.
<br>
Example. Use the $(\epsilon, \delta)$ definition of the limit to show that
<br>
$$
<br>
\lim _{x \rightarrow 0} x \sin \frac{1}{x}=0
<br>
$$
<br>
Letting $\epsilon>0$, we need to find corresponding $\delta>0$ that satisfies the definition for $L=0$. Knowing that sin is bounded between $-1$ and 1 ,
<br>
$$
<br>
\left|x \sin \frac{1}{x}-0\right|=\left|x \sin \frac{1}{x}\right|=|x|\left|\sin \frac{1}{x}\right| \leq|x| \text { . }
<br>
$$
<br>
Letting $\delta=\epsilon$, if $0<|x-0|<\delta$, then $\left|x \sin \frac{1}{x}-0\right| \leq|x|<\epsilon$, as required by the definition.</p>
<p> </p>Mathematicshttps://mathsgee.com/30194/in-calculus-how-are-limits-defined?show=30195#a30195Sun, 20 Jun 2021 03:09:57 +0000Answered: What are teaching and learning strategies?
https://mathsgee.com/30134/what-are-teaching-and-learning-strategies?show=30135#a30135
In simple terms, teaching strategies refer to "everything teachers do or should do in order to help their learners learn". Also called teaching practices, they can include everything from planning and organising lessons, classes, resources and assessments, to the individual actions and activities that teachers engage in during their classroom teaching.<br />
<br />
Learning strategies are the behaviours and thoughts students use as they attempt to complete various tasks associated with the process of learning a new concept or acquiring, storing, retrieving and using information.Mathematicshttps://mathsgee.com/30134/what-are-teaching-and-learning-strategies?show=30135#a30135Sat, 19 Jun 2021 23:47:32 +0000Answered: What is the multiplicative inverse of 1/16 ÷ 1/81 +-1/8?
https://mathsgee.com/30125/what-is-the-multiplicative-inverse-of-1-16-1-81-1-8?show=30126#a30126
<p><strong>Answer</strong></p>
<p>Multiplicative inverse $=16 / 79$</p>
<p> </p>
<p><strong>Explanation</strong></p>
<p>
<br>
a multiplicative inverse is basically a reciprocal
<br>
Multiplicative inverse of number is the number which if multiplied by original number result in 1
<br>
$1 / 16 \div 1 / 81+-1 / 8$
<br>
$1 / 16 \div 1 / 81=81 / 16$
<br>
$+-1 / 8=-1 / 8$</p>
<p>81/16 - $1 / 8$
<br>
$=81 / 16-2 / 16$
<br>
$=(81-2) / 16$
<br>
$=79 / 16$
<br>
Multiplicative inverse $=16 / 79$</p>Mathematicshttps://mathsgee.com/30125/what-is-the-multiplicative-inverse-of-1-16-1-81-1-8?show=30126#a30126Sat, 19 Jun 2021 07:18:20 +0000A bag has balls of four colours red, blue, pink,green. Half the total number of balls are pink. One -fourth of the number of green balls equals one third of the number of blue balls.
https://mathsgee.com/30124/balls-colours-number-balls-fourth-number-equals-number-balls
A bag has balls of four colours red, blue, pink,green. Half the total number of balls are pink. One -fourth of the number of green balls equals one third of the number of blue balls. The number of pink balls is 4 less than twice the total number of green and blue balls. The number of green balls is 22 less than the total number of blue and pink balls. Find the total number of balls in the bag.<br />
<br />
(A) 48<br />
<br />
(B) 36<br />
<br />
(C) 40<br />
<br />
(D) 54Mathematicshttps://mathsgee.com/30124/balls-colours-number-balls-fourth-number-equals-number-ballsSat, 19 Jun 2021 07:13:31 +0000Find a and $b$ if $3 \sqrt{7} / 3-\sqrt{7}+3=\sqrt{7} / 3+\sqrt{7}=a+b \sqrt{7}$
https://mathsgee.com/30123/find-a-and-b-if-3-sqrt-7-3-sqrt-7-3-sqrt-7-3-sqrt-7-a-b-sqrt-7
Find a and $b$ if $3 \sqrt{7} / 3-\sqrt{7}+3=\sqrt{7} / 3+\sqrt{7}=a+b \sqrt{7}$Mathematicshttps://mathsgee.com/30123/find-a-and-b-if-3-sqrt-7-3-sqrt-7-3-sqrt-7-3-sqrt-7-a-b-sqrt-7Sat, 19 Jun 2021 07:11:02 +0000$\left(-9^{*} 24\right)$ divided by $[-15-(23)]$
https://mathsgee.com/30122/left-9-24-right-divided-by-15-23
$\left(-9^{*} 24\right)$ divided by $[-15-(23)]$Mathematicshttps://mathsgee.com/30122/left-9-24-right-divided-by-15-23Sat, 19 Jun 2021 07:09:55 +0000if a×b ={(3,2) (3,5) (5,2) (5,4)} then find a and b
https://mathsgee.com/30121/if-a-b-3-2-3-5-5-2-5-4-then-find-a-and-b
if a&times;b ={(3,2) (3,5) (5,2) (5,4)} then find a and bMathematicshttps://mathsgee.com/30121/if-a-b-3-2-3-5-5-2-5-4-then-find-a-and-bSat, 19 Jun 2021 07:05:49 +0000If a, 4a and 6a+5 are in an arithmetic progression. What is the value of a?
https://mathsgee.com/30120/if-and-6a-are-in-an-arithmetic-progression-what-is-the-value-of
If a, 4a and 6a+5 are in an arithmetic progression. What is the value of a?Mathematicshttps://mathsgee.com/30120/if-and-6a-are-in-an-arithmetic-progression-what-is-the-value-ofSat, 19 Jun 2021 07:04:16 +0000Simplify $(2-\sqrt{\frac{5}{2}}+\sqrt{5})^2 - (2+\sqrt{\frac{5}{2}})^2$
https://mathsgee.com/30119/simplify-2-sqrt-frac-5-2-sqrt-5-2-2-sqrt-frac-5-2-2
Simplify $(2-\sqrt{\frac{5}{2}}+\sqrt{5})^2 - (2+\sqrt{\frac{5}{2}})^2$Mathematicshttps://mathsgee.com/30119/simplify-2-sqrt-frac-5-2-sqrt-5-2-2-sqrt-frac-5-2-2Sat, 19 Jun 2021 07:02:56 +0000P, Q, R, S, T and U are six students studying in a class. Each of them has a different height and weight. The tallest is not the heaviest .
https://mathsgee.com/30118/students-studying-different-height-weight-tallest-heaviest
P, Q, R, S, T and U are six students studying in a class. Each of them has a different height and weight. The tallest is not the heaviest .T is taller than only P but lighter than R, Q is taller than S and P and heavier than only T and U. P is lighter than only S, T is heavier than U. S is taller than U and Q is not the tallest. Who among them is the tallest? a) U b) P c) T d)RMathematicshttps://mathsgee.com/30118/students-studying-different-height-weight-tallest-heaviestSat, 19 Jun 2021 06:59:19 +0000Answered: Solve the simultaneous equations: $\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2$ and $\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1$
https://mathsgee.com/30116/solve-simultaneous-equations-frac-sqrt-frac-sqrt-frac-sqrt?show=30117#a30117
<p><strong>Answers</strong></p>
<p>$x = \dfrac{4}{9}$</p>
<p>$y = \dfrac{9}{25}$</p>
<p> </p>
<p><strong>Explanation</strong></p>
<p>Let $\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2$ be equation 1 and $\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1$ be equation 2.</p>
<p>multiply equation 1 by 2 so we can eliminate $x$</p>
<p>Equation 1 becomes:</p>
<p>$\frac{4}{\sqrt{x}}+\frac{6}{\sqrt{y}}=4$ </p>
<p>Now subtract equation 1 from equation 2 to get:</p>
<p>$\frac{6}{\sqrt{y}}-\frac{9}{\sqrt{y}}=4$ </p>
<p>$\frac{-3}{\sqrt{y}}=-5$ </p>
<p>multiply both side by $\sqrt{y}$ to get $-3=-5\sqrt{y}$ </p>
<p>$y = \dfrac{9}{25}$</p>
<p>To get $x$ we substitute $y$ in equation 1 to get:</p>
<p>$\frac{2}{\sqrt{x}}=2-5$</p>
<p>so</p>
<p>$\dfrac{2}{3}=\sqrt{x}$</p>
<p>$\therefore x = \dfrac{4}{9}$</p>Mathematicshttps://mathsgee.com/30116/solve-simultaneous-equations-frac-sqrt-frac-sqrt-frac-sqrt?show=30117#a30117Sat, 19 Jun 2021 06:56:38 +0000Answered: Find the partial fraction decomposition of the following expression: $$ \frac{x^{3}+3}{x^{2}-2 x-3} $$
https://mathsgee.com/30112/find-partial-fraction-decomposition-following-expression?show=30113#a30113
First we do polynomial long division to find that<br />
$$<br />
\frac{x^{3}+3}{x^{2}-2 x-3}=x+2+\frac{7 x+9}{x^{2}-2 x-3}<br />
$$<br />
Now that the numerator is of a lesser degree than the denominator, we can decompose it normally.<br />
$$<br />
x^{2}-2 x-3=(x-3)(x+1)<br />
$$<br />
<br />
So,<br />
$$<br />
\frac{7 x+9}{x^{2}-2 x-3}=\frac{A_{1}}{x-3}+\frac{A_{2}}{x+1}<br />
$$<br />
Multiplying each side by the denominator,<br />
$$<br />
7 x+9=A_{1}(x+1)+A_{2}(x-3) \text { . }<br />
$$<br />
At $x=-1$,<br />
$$<br />
2=-4 A_{2} \Longrightarrow A_{2}=\frac{-1}{2}<br />
$$<br />
At $x=3$,<br />
$$<br />
30=4 A_{1} \Longrightarrow A_{1}=\frac{15}{2}<br />
$$<br />
So,<br />
$$<br />
\frac{x^{3}+3}{x^{2}-2 x-3}=x+2+\frac{15 / 2}{x-3}+\frac{-1 / 2}{x+1}<br />
$$Mathematicshttps://mathsgee.com/30112/find-partial-fraction-decomposition-following-expression?show=30113#a30113Sat, 19 Jun 2021 02:22:47 +0000Answered: What are improper fractions?
https://mathsgee.com/30110/what-are-improper-fractions?show=30111#a30111
If the degree of the numerator is greater than or equal to the degree of the denominator, we have a case of improper fractions. In this case, we have to do polynomial long division to get a quotient and remainder and then decompose the remainder if necessary. So,<br />
$$<br />
\frac{P(x)}{Q(x)}=R(x)+\frac{S(x)}{Q(x)}<br />
$$Mathematicshttps://mathsgee.com/30110/what-are-improper-fractions?show=30111#a30111Sat, 19 Jun 2021 02:18:58 +0000Answered: Find the partial fraction decomposition of $\frac{3 x^{4}-2 x^{3}+6 x^{2}-3 x+3}{x^{5}+3 x^{4}+4 x^{3}+12 x^{2}+4 x+12}$.
https://mathsgee.com/30108/find-the-partial-fraction-decomposition-of-frac-x-4-4-x-12-2-12?show=30109#a30109
Factoring,<br />
$$<br />
x^{5}+3 x^{4}+4 x^{3}+12 x^{2}+4 x+12=(x+3)\left(x^{2}+2\right)^{2} .<br />
$$<br />
So,<br />
$$<br />
\frac{3 x^{4}-2 x^{3}+6 x^{2}-3 x+3}{x^{5}+3 x^{4}+4 x^{3}+12 x^{2}+4 x+12}=\frac{A_{1}}{x+3}+\frac{A_{2} x+B_{2}}{x^{2}+2}+\frac{A_{3} x+B_{3}}{\left(x^{2}+2\right)^{2}}<br />
$$<br />
Multiplying each side by the denominator,<br />
$$<br />
3 x^{4}-2 x^{3}+6 x^{2}-3 x+3=A_{1}\left(x^{2}+2\right)^{2}+\left(A_{2} x+B_{2}\right)\left(x^{2}+2\right)(x+3)+\left(A_{3} x+B_{3}\right)(x+3)<br />
$$<br />
At $x=-3$,<br />
$$<br />
363=121 A_{1} \Longrightarrow A_{1}=3<br />
$$<br />
Now, we'll use our result for $A_{1}$ and pick a value for $x$ that minimizes the number of things we need to solve for. We'll have to solve a linear system with 4 unknowns, so we'll need up to 4 values. At $x=0$<br />
$$<br />
3=3(2)^{2}+B_{2}(2)(3)+B_{3}(3) \Longrightarrow 2 B_{2}+B_{3}=-3<br />
$$<br />
<br />
At $x=1$,<br />
$$<br />
7=3(3)^{2}+\left(A_{2}+B_{2}\right)(3)(4)+\left(A_{3}+B_{3}\right)(4) \Longrightarrow 3 A_{2}+A_{3}+3 B_{2}+B_{3}=-5<br />
$$<br />
At $x=-1$,<br />
$$<br />
17=3(3)^{2}+\left(-A_{2}+B_{2}\right)(3)(2)+\left(-A_{3}+B_{3}\right)(2) \Longrightarrow-3 A_{2}-A_{3}+3 B_{2}+B_{3}=-5<br />
$$<br />
At $x=2$,<br />
$$<br />
53=3(6)^{2}+\left(2 A_{2}+B_{2}\right)(6)(5)+\left(2 A_{3}+B_{3}\right)(5) \Longrightarrow 12 A_{2}+2 A_{3}+6 B_{2}+B_{3}=-11 .<br />
$$<br />
Now we have the following system of equations:<br />
$$<br />
\left\{\begin{array}{ll}<br />
0 A_{2}+0 A_{3}+2 B_{2}+B_{3} & =-3 \\<br />
3 A_{2}+A_{3}+3 B_{2}+B_{3} & =-5 \\<br />
-3 A_{2}-A_{3}+3 B_{2}+B_{3} & =-5 \\<br />
12 A_{2}+2 A_{3}+6 B_{2}+B_{3} & =-11<br />
\end{array}\right.<br />
$$<br />
Solving,<br />
$$<br />
A_{2}=0, A_{3}=0, B_{2}=-2, \text { and } B_{3}=1 \text { . }<br />
$$<br />
So,<br />
$$<br />
\frac{3 x^{4}-2 x^{3}+6 x^{2}-3 x+3}{x^{5}+3 x^{4}+4 x^{3}+12 x^{2}+4 x+12}=\frac{3}{x+3}-\frac{2}{x^{2}+2}+\frac{1}{\left(x^{2}+2\right)^{2}}<br />
$$Mathematicshttps://mathsgee.com/30108/find-the-partial-fraction-decomposition-of-frac-x-4-4-x-12-2-12?show=30109#a30109Sat, 19 Jun 2021 02:17:05 +0000Answered: What are repeated quadratic factors?
https://mathsgee.com/30106/what-are-repeated-quadratic-factors?show=30107#a30107
If a quadratic factor that can't be broken into linear factors is repeated, then we can write $Q(x)=$ $R(x)\left(a x^{2}+b x+c\right)^{k}, k \geq 0$, and $R(x)$ is not divisible by $\left(a x^{2}+b x+c\right)^{k}$. Now we have to do a combination of what we did for repeated linear factors and quadratic factors. We say<br />
$$<br />
\frac{P(x)}{R(x)\left(a x^{2}+b x+c\right)^{k}}=(\text { Decomposition of } R(x))+\frac{A_{1} x+B_{1}}{a x^{2}+b x+c}+\ldots+\frac{A_{k} x+B_{k}}{\left(a x^{2}+b x+c\right)^{k}}<br />
$$<br />
We then solve for the coefficients in the numerator.Mathematicshttps://mathsgee.com/30106/what-are-repeated-quadratic-factors?show=30107#a30107Sat, 19 Jun 2021 02:14:30 +0000Answered: Find the partial fraction decomposition of the following expression: $$ \frac{6 x^{2}+21 x+11}{x^{3}+5 x^{2}+3 x+15} $$
https://mathsgee.com/30104/find-partial-fraction-decomposition-following-expression?show=30105#a30105
Factoring,<br />
$$<br />
x^{2}+5 x^{2}+3 x+15=(x+5)\left(x^{2}+3\right) \text { . }<br />
$$<br />
So,<br />
$$<br />
\frac{6 x^{2}+21 x+11}{x^{3}+5 x^{2}+3 x+15}=\frac{A_{1}}{x+5}+\frac{A_{2} x+B_{2}}{x^{2}+3}<br />
$$<br />
<br />
Multiplying each side by the denominator,<br />
$$<br />
6 x^{2}+21 x+11=A_{1}\left(x^{2}+3\right)+\left(A_{2} x+B_{2}\right)(x+5) \text { . }<br />
$$<br />
At $x=-5$,<br />
$$<br />
56=28 A_{1} \Longrightarrow A_{1}=2<br />
$$<br />
Now we'll use the previous result and another value for $x .$ We can use $x=0$ to not have to worry about the $A_{2}$ term. At $x=0$,<br />
$$<br />
11=2(3)+\left(B_{2}\right)(5) \Longrightarrow B_{2}=1 .<br />
$$<br />
Now we'll use the previous 2 results to find $A_{2} . x=1$ is a good choice to keep the numbers small. At $x=1$,<br />
$$<br />
38=2(1+3)+\left(A_{2}+1\right)(6) \Longrightarrow A_{2}=4 .<br />
$$<br />
So,<br />
$$<br />
\frac{6 x^{2}+21 x+11}{x^{3}+5 x^{2}+3 x+15}=\frac{2}{x+5}+\frac{4 x+1}{x^{2}+3}<br />
$$Mathematicshttps://mathsgee.com/30104/find-partial-fraction-decomposition-following-expression?show=30105#a30105Sat, 19 Jun 2021 02:12:40 +0000Answered: What are linear factors?
https://mathsgee.com/30102/what-are-linear-factors?show=30103#a30103
This is the the most basic type where the degree of the numerator is less than the degree of the denominator and the denominator factors into all linear factors with no repeated roots. In this case we can write<br />
$$<br />
\frac{P(x)}{Q(x)}=\frac{A_{1}}{\left(x-a_{1}\right)}+\ldots+\frac{A_{n}}{\left(x-a_{n}\right)}<br />
$$<br />
Multiplying each side by $Q(x)$<br />
$$<br />
P(x)=A_{1}\left(x-a_{2}\right) \ldots\left(x-a_{n}\right)+\ldots+A_{n}\left(x-a_{1}\right) \ldots\left(x-a_{n-1}\right)<br />
$$<br />
We can then find each $A_{i}$ by evaluating both sides at $x=a_{i}$, since every term except the ith has an $\left(x-a_{i}\right)$ factor that will go to $0 .$ So,<br />
$$<br />
A_{i}=\frac{P\left(a_{i}\right)}{\left(x-a_{i}\right) \ldots\left(x-a_{i-1}\right)\left(x-a_{i+1}\right) \ldots\left(x-a_{n}\right)} .<br />
$$Mathematicshttps://mathsgee.com/30102/what-are-linear-factors?show=30103#a30103Sat, 19 Jun 2021 02:09:50 +0000Answered: What are repeated linear factors?
https://mathsgee.com/30100/what-are-repeated-linear-factors?show=30101#a30101
If $Q(x)$ has repeated roots, it factors into<br />
$$<br />
Q(x)=R(x)(x-a)^{k}, k \geq 2 \text { and } R(a) \neq 0<br />
$$<br />
When making the common denominator for each repeated root of multiplicity $k$, we do<br />
$$<br />
\frac{P(x)}{R(x)(x-a)^{k}}=(\text { Decomposition of } R(x))+\frac{A_{1}}{x-a}+\ldots+\frac{A_{k}}{(x-a)^{k}} .<br />
$$<br />
You would then multiply each side by the denominator like in the linear factors case and solve for the coefficients. The only additional difficulty is that you might have to use previous results or solve a system of linear equations to get some of the constants.Mathematicshttps://mathsgee.com/30100/what-are-repeated-linear-factors?show=30101#a30101Sat, 19 Jun 2021 02:08:39 +0000Answered: What are quadratic factors?
https://mathsgee.com/30098/what-are-quadratic-factors?show=30099#a30099
If a quadratic doesn't have real roots, then we have a quadratic factor. Here, we'll assume that the quadratic factor isn't repeated. So, $Q(x)=R(x)\left(a x^{2}+b x+c\right), b^{2}-4 a c<0$, and $R(x)$ is not evenly divisible by $a x^{2}+b x+c .$ In this case, we say<br />
$$<br />
\frac{P(x)}{R(x)\left(a x^{2}+b x+c\right)}=(\text { Decomposition of } R(x))+\frac{A_{1} x+B_{1}}{a x^{2}+b x+c} .<br />
$$<br />
We then solve for the constants in the numerator, possibly having to solve a system of equations or. using previous results and less convenient values for $x$.Mathematicshttps://mathsgee.com/30098/what-are-quadratic-factors?show=30099#a30099Sat, 19 Jun 2021 02:07:15 +0000Answered: Find the partial fraction of the following expression: $$ \frac{x^{2}+5 x-6}{x^{3}-7 x^{2}+16 x-12} $$
https://mathsgee.com/30096/find-the-partial-fraction-the-following-expression-frac-12?show=30097#a30097
Factoring,<br />
$$<br />
x^{3}-7 x^{2}+16 x-12=(x-3)(x-2)^{2} \text { . }<br />
$$<br />
So,<br />
$$<br />
\frac{x^{2}+5 x-6}{x^{3}-7 x^{2}+16 x-12}=\frac{A_{1}}{x-3}+\frac{A_{2}}{x-2}+\frac{A_{3}}{(x-2)^{2}} .<br />
$$<br />
Multiplying each side by the denominator,<br />
$$<br />
x^{2}+5 x-6=A_{1}(x-2)^{2}+A_{2}(x-2)(x-3)+A_{3}(x-3) .<br />
$$<br />
At $x=2$,<br />
$$<br />
8=A_{3}(2-3) \Longrightarrow A_{3}=-8<br />
$$<br />
At $x=3$,<br />
$$<br />
18=A_{1}(3-2)^{2} \Longrightarrow A_{1}=18<br />
$$<br />
Now we'll use our results for $A_{1}$ and $A_{3}$ to find $A_{2}$ using a value for $x$ that isn't 2 or 3 so the $A_{2}$ term doesn't become $0 .$ A good choice is $x=0$. At $x=0$,<br />
$$<br />
-6=18(0-2)^{2}+A_{2}(0-2)(0-3)+-8(0-3) \Longrightarrow A_{2}=-17<br />
$$<br />
So,<br />
$$<br />
\frac{x^{2}+5 x-6}{x^{3}-7 x^{2}+16 x-12}=\frac{18}{x-3}-\frac{17}{x-2}-\frac{8}{(x-2)^{2}} .<br />
$$Mathematicshttps://mathsgee.com/30096/find-the-partial-fraction-the-following-expression-frac-12?show=30097#a30097Sat, 19 Jun 2021 02:05:21 +0000Answered: Find the partial fraction decomposition of the following expression: $$ \frac{2 x-1}{x^{3}-6 x^{2}+11 x-6} $$
https://mathsgee.com/30094/find-partial-fraction-decomposition-following-expression?show=30095#a30095
Factoring,<br />
$$<br />
x^{3}-6 x^{2}+11 x-6=(x-1)(x-2)(x-3) \text { . }<br />
$$<br />
So,<br />
$$<br />
\frac{2 x-1}{x^{3}-6 x^{2}+11 x-6}=\frac{A_{1}}{x-1}+\frac{A_{2}}{x-2}+\frac{A_{3}}{x-3} .<br />
$$<br />
Multiplying each side by the denominator,<br />
$$<br />
2 x-1=A_{1}(x-2)(x-3)+A_{2}(x-1)(x-3)+A_{3}(x-1)(x-2) .<br />
$$<br />
At $x=1$,<br />
$$<br />
1=A_{1}(1-2)(1-3) \Longrightarrow A_{1}=\frac{1}{2}<br />
$$<br />
At $x=2$,<br />
$$<br />
3=A_{2}(2-1)(2-3) \Longrightarrow A_{2}=-3<br />
$$<br />
At $x=3$,<br />
$$<br />
5=A_{3}(3-1)(3-2) \Longrightarrow A_{3}=\frac{5}{2}<br />
$$<br />
So,<br />
$$<br />
\frac{2 x-1}{x^{3}-6 x^{2}+11 x-6}=\frac{1 / 2}{x-1}+\frac{-3}{x-2}+\frac{5 / 2}{x-3}<br />
$$Mathematicshttps://mathsgee.com/30094/find-partial-fraction-decomposition-following-expression?show=30095#a30095Sat, 19 Jun 2021 02:02:56 +0000Answered: What are partial fractions of functions?
https://mathsgee.com/30092/what-are-partial-fractions-of-functions?show=30093#a30093
If we have a function of two polynomials $f(x)=\frac{P(x)}{Q(x)}$, it's often easier to break this quotient into a sum of parts where the denominator is a linear or quadratic factor and the numerator is always a smaller degree than the denominator.Mathematicshttps://mathsgee.com/30092/what-are-partial-fractions-of-functions?show=30093#a30093Sat, 19 Jun 2021 01:59:29 +0000Answered: What is the definition of $e$, the base of the natural logarithm?
https://mathsgee.com/30090/what-is-the-definition-of-e-the-base-the-natural-logarithm?show=30091#a30091
Definition. e is the base of the natural logarithm. It's defined by the limit<br />
$$<br />
e=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n}<br />
$$<br />
$\exp x=e^{x}$ and $\ln x$ are inverse functions of each other such that<br />
$$<br />
e^{\ln x}=x \text { and } \ln e^{x}=x \text { . }<br />
$$<br />
Just like other exponentials, the normal rules for adding, subtracting, and multiplying exponents apply:<br />
$$<br />
e^{x} e^{y}=e^{x+y}, \frac{e^{x}}{e^{y}}=e^{x-y}, \text { and }\left(e^{x}\right)^{k}=e^{x k} .<br />
$$<br />
Similar rules apply for logarithms:<br />
$$<br />
\ln x+\ln y=\ln x y, \ln x-\ln y=\ln \left(\frac{x}{y}\right) \text { , and } \ln \left(a^{b}\right)=b \ln a<br />
$$<br />
<br />
We can also write a logarithm of any base using natural logarithms:<br />
$$<br />
\log _{b} a=\frac{\ln a}{\ln b} \text { . }<br />
$$<br />
$e$ is also unique in that it is the only real number $a$ satisfying the equation<br />
$$<br />
\frac{\mathrm{d}}{\mathrm{d} x} a^{x}=a^{x}<br />
$$<br />
meaning $e^{x}$ is its own derivative.Mathematicshttps://mathsgee.com/30090/what-is-the-definition-of-e-the-base-the-natural-logarithm?show=30091#a30091Sat, 19 Jun 2021 01:57:31 +0000Answered: Using a right-angled triangle how can we derive the trigonometric identity $1+\cot ^{2} \theta=\csc ^{2} \theta$
https://mathsgee.com/30088/angled-triangle-derive-trigonometric-identity-theta-theta?show=30089#a30089
$\sin$ and $\cos$ form a right triangle with hypotenuse. So, using the Pythagorean Theorem,<br />
$$<br />
\sin ^{2} \theta+\cos ^{2} \theta=1<br />
$$<br />
<br />
By dividing by $\sin ^{2}$ or $\cos ^{2}$, we can also get<br />
$$<br />
1+\cot ^{2} \theta=\csc ^{2} \theta \text { and } \tan ^{2} \theta+1=\sec ^{2} \theta<br />
$$<br />
Together, these 3 identities are called the Pythagorean Identities.<br />
We can also relate functions and co-functions.<br />
$$<br />
\operatorname{xxx}(\theta)=\operatorname{coxxx}\left(\frac{\pi}{2}-\theta\right) .<br />
$$<br />
Some of the most useful and used identities are the sum and difference.<br />
$$<br />
\begin{aligned}<br />
\sin (\alpha \pm \beta) &=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\<br />
\cos (\alpha \pm \beta) &=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta \\<br />
\tan (\alpha \pm \beta) &=\frac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta} \\<br />
\sin \alpha \pm \sin \beta &=2 \sin \left(\frac{\alpha \pm \beta}{2}\right) \cos \left(\frac{\alpha \mp \beta}{2}\right) \\<br />
\cos \alpha+\cos \beta &=2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right) \\<br />
\cos \alpha-\cos \beta &=-2 \sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)<br />
\end{aligned}<br />
$$Mathematicshttps://mathsgee.com/30088/angled-triangle-derive-trigonometric-identity-theta-theta?show=30089#a30089Sat, 19 Jun 2021 01:55:11 +0000Answered: How is the unit circle and trig functions?
https://mathsgee.com/30086/how-is-the-unit-circle-and-trig-functions?show=30087#a30087
<p>Imagine a circle of radius 1 centered at the origin that we'll call the unit circle. The $x$ and $y$ coordinates of a point on the unit circle are completely determined by the angle $\theta$ in radians between the $x$ -axis and a line from the origin to the point.</p>
<p>The function $\cos \theta$ tells us $x$ -coordinate of the point, while $\sin \theta$ tells us the $y$ -coordinate of the point. The function $\tan \theta=\frac{\sin \theta}{\cos \theta}$ tells us the slope of the line from the origin to the point.</p>
<p>Most of the trig functions have geometric interpretations as shown below. The most used ones are $\sin , \cos , \tan =\frac{\sin }{\cos }$, $\cot =\frac{\cos }{\sin }, \csc =\frac{1}{\sin }$, and $\sec =\frac{1}{\cos }$.</p>
<p><img alt="" src="https://mathsgee.com/?qa=blob&qa_blobid=6515950759760939032" style="height:318px; width:556px"></p>
<p>We can also think about the inverses of these trig functions. These are either notated with a $-1$ exponent on the function, or the prefix arc in front of the function name. Many of these functions are only defined on a part of the domain $[0,2 \pi]$. Below is a table of the inverse trig functions and their domains.</p>
<p>
<br>
<img alt="" src="https://mathsgee.com/?qa=blob&qa_blobid=4091314170366737076" style="height:171px; width:289px"></p>Mathematicshttps://mathsgee.com/30086/how-is-the-unit-circle-and-trig-functions?show=30087#a30087Sat, 19 Jun 2021 01:50:10 +0000Answered: Factor the polynomial $x^{5}+x^{4}-2 x^{3}+4 x^{2}-24 x$.
https://mathsgee.com/30084/factor-the-polynomial-x-5-x-4-2-x-3-4-x-2-24-x?show=30085#a30085
<p>We can immediately see that there is no constant term, so $x=0$ is a root. Now we need to work on factoring $x^{4}+x^{3}-2 x^{2}+4 x-24$.
<br>
The factors of $-24$ are: $-24,-12,-8,-6,-4,-3,-2,-1,1,2,3,4,6,8,12$, and 24 . Starting from roots close to 0 and working outwards, we find that $x=2$ is a root. So, we synthetic divide like so</p>
<p>
<br>
<img alt="" src="https://mathsgee.com/?qa=blob&qa_blobid=13124437632561929140" style="height:78px; width:288px"></p>
<p>
<br>
to see that now we need to work on factoring $x^{3}+3 x^{2}+4 x+12 . x^{3}+3 x^{2}+4 x+12=x^{2}(x+$
<br>
3) $+4(x+3)=(x+3)\left(x^{2}+4\right)$, so $x=-3$ is a root, and we need to work on factoring $x^{2}+4$. $x^{2}+4$ has two complex roots $\pm 2 i$, so we'll leave it as a quadratic.
<br>
$$
<br>
x^{5}+x^{4}-2 x^{3}+4 x^{2}-24 x=x(x-2)(x-3)\left(x^{2}+4\right)
<br>
$$</p>Mathematicshttps://mathsgee.com/30084/factor-the-polynomial-x-5-x-4-2-x-3-4-x-2-24-x?show=30085#a30085Sat, 19 Jun 2021 01:46:38 +0000Answered: What are the steps and methods for factoring polynomials?
https://mathsgee.com/30082/what-are-the-steps-and-methods-for-factoring-polynomials?show=30083#a30083
We want to break up a polynomial like $f(x)=a_{0}+a_{1} x^{1}+\ldots a_{n} x^{n}$ into linear factors so that $f(x)=c\left(x-b_{1}\right) \cdot \ldots \cdot\left(x-b_{n}\right)$. This form makes it simple to see that the roots of $f$, solutions to $f(x)=0$, are $x=b_{1} \ldots b_{n}$.<br />
For quadratics, $f(x)=a x^{2}+b x+c$, there exists a simple formula that will give us both roots, the quadratic formula<br />
$$<br />
x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}<br />
$$<br />
We can see that when $b^{2}-4 a c<0$, like for $f(x)=x^{2}+5 x+10$, we will get complex roots $\alpha \pm \beta$ i. For any polynomial, these roots come in pairs, so if $\alpha+\beta i$ is a root, then so is $\alpha-\beta i$. This means that every conjugate pair $\alpha \pm \beta i$ has a quadratic equation with those roots. Sometimes we will not <br />
<br />
factor quadratics with complex roots into linear terms.<br />
Although there do exist explicit formulas for finding roots for cubic (degree 3) and quartic (degree 4) equations, they are too long and not useful enough to memorize. When working by hand, we instead use other tricks to find roots.<br />
<br />
There are a few useful tricks that can help. If the polynomial doesn't have a constant term, then 0 is a root. If all the coefficients sum to 0 , then 1 is a root. For certain polynomials with an even number of terms, like all cubics of the form $a x^{3}+b x^{2}+\operatorname{cax}+c b$ we can factor out a term from the first two and last two terms to get $x^{2}(a x+b)+c(a x+b)=(a x+b)\left(x^{2}+c\right)$. For other polynomials, we might just try guessing and checking values. However, we need a more efficient way that works in general.<br />
<br />
Since we are looking to find linear factors $f(x)=\left(x-b_{1}\right) \cdot \ldots \cdot\left(x-b_{n}\right)$, we can see that the constant term in the polynomial is the product of the roots $b_{1} \ldots b_{n} .$ In fact, since the coefficients of polynomials are completely determined by the roots and the leading coefficient, all the coefficients are sums and products of roots. You might remember when factoring quadratics that the coefficient of $x$ term is the sum of the two roots. These rules are called Vieta's formulas.<br />
<br />
So, if we have the constant term, we can check all of its integer factors to see if any are roots. For each root, we can divide, using a technique like synthetic division, to continue finding the rest of the roots. This method is especially useful on tests because the roots tend to be integers.Mathematicshttps://mathsgee.com/30082/what-are-the-steps-and-methods-for-factoring-polynomials?show=30083#a30083Sat, 19 Jun 2021 01:43:56 +0000Answered: What are complex conjugates?
https://mathsgee.com/30080/what-are-complex-conjugates?show=30081#a30081
A common operation on complex numbers is the complex conjugate. The complex conjugate of $z=\alpha+\beta i$ is $\bar{z}=\alpha-\beta i . z$ and $\bar{z}$ are called a conjugate pair.<br />
Conjugate pairs have the following properties. Let $z, w \in \mathbb{C}$.<br />
$$<br />
\begin{aligned}<br />
\overline{z \pm w} &=\bar{z} \pm \bar{w} \\<br />
\overline{z w} &=\overline{z w} \\<br />
\bar{z} &=z \Leftrightarrow z \in \mathbb{R} \\<br />
z \bar{z} &=|z|^{2}=|\bar{z}|^{2} \\<br />
\overline{\bar{z}} &=z \\<br />
\bar{z}^{n} &=\overline{z^{n}} \\<br />
z^{-1} &=\frac{\bar{z}}{|z|^{2}}<br />
\end{aligned}<br />
$$Mathematicshttps://mathsgee.com/30080/what-are-complex-conjugates?show=30081#a30081Sat, 19 Jun 2021 01:41:18 +0000Answered: How are complex numbers related to real numbers?
https://mathsgee.com/30078/how-are-complex-numbers-related-to-real-numbers?show=30079#a30079
Definition. $i$ is called the imaginary unit. It's defined by $i^{2}=-1$.<br />
<br />
The set of complex numbers ( $\mathbb{C}$ ) is an extension of the real numbers. Complex numbers have the form $z=\alpha+\beta i$, where $\alpha$ and $\beta$ are real numbers. The $\alpha$ part of $z$ is called the real part, so $\Re(z)=\alpha$. The $\beta$ part of $z$ is called the imaginary part, so $\Im(z)=\beta i$.<br />
<br />
Often, complex numbers are visualized as points or vectors in a 2D plane, called the complex plane, where $\alpha$ is the $x$ -component, and $\beta$ is the y-component. Thinking of complex numbers like points helps us define the magnitude of complex numbers and compare them. Since a point $(x, y)$ has a distance $\sqrt{x^{2}+y^{2}}$ from the origin, we can say the magnitude of $z,|z|$ is $\sqrt{\alpha^{2}+\beta^{2}}$. Thinking of complex numbers like vectors helps us understand adding two complex numbers, since you just add the components like vectors.Mathematicshttps://mathsgee.com/30078/how-are-complex-numbers-related-to-real-numbers?show=30079#a30079Sat, 19 Jun 2021 01:39:58 +0000Answered: Is $f(x)=2 x-x^{2}$ even, odd, or neither?
https://mathsgee.com/30076/is-f-x-2-x-x-2-even-odd-or-neither?show=30077#a30077
$$<br />
f(-x)=2(-x)-(-x)^{2}=-2 x-x^{2}<br />
$$<br />
Since $f(-x) \neq f(x)$ and $f(-x) \neq-f(x)$, the function is neither even nor odd.Mathematicshttps://mathsgee.com/30076/is-f-x-2-x-x-2-even-odd-or-neither?show=30077#a30077Sat, 19 Jun 2021 01:37:57 +0000Answered: When is a mathematical function classified as even?
https://mathsgee.com/30074/when-is-a-mathematical-function-classified-as-even?show=30075#a30075
We say that a function $f$ is even if it satisfies $f(-x)=f(x)$ for all $x \in D .$ Likewise, we say that a function $f$ is odd if it satisfies $f(-x)=-f(x)$ for all $x \in D .$ Geometrically, we can see that the graph of an even function is symmetric with respect to the $y$ -axis, while the graph of an odd function is symmetric with respect to the origin.Mathematicshttps://mathsgee.com/30074/when-is-a-mathematical-function-classified-as-even?show=30075#a30075Sat, 19 Jun 2021 01:36:52 +0000Answered: Find the inverse function of $$ f(x)=\frac{5 x+2}{4 x-3} $$
https://mathsgee.com/30072/find-the-inverse-function-of-f-x-frac-5-x-2-4-x-3?show=30073#a30073
We first make the substitutions to set up the algorithm:<br />
$$<br />
y=\frac{5 x+2}{4 x-3} \text { followed by } x=\frac{5 y+2}{4 y-3}<br />
$$<br />
After multiplying both sides by the denominator and simplifying, we have<br />
$$<br />
\Longrightarrow 4 x y-3 x=-5 y-2 \Longrightarrow y=f^{-1}(x)=\frac{3 x-2}{4 x+5} \text { . }<br />
$$Mathematicshttps://mathsgee.com/30072/find-the-inverse-function-of-f-x-frac-5-x-2-4-x-3?show=30073#a30073Sat, 19 Jun 2021 01:35:42 +0000