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2 years ago

Practice Solving Distance Problems

Mathematics

1 answer:

Radda [10]2 years ago

5 0

$\Huge\green{\tt{✏Answer✏}}$

The problem asked to find how many inches Tonya grows during the year

 $\Huge\blue{\tt{✏Explanation✏}}$ 

<h2>here my friend</h2>

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<h2>hit that heart and dont forget to follow hehe thx<3</h2>

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Slope intercept from y-4=-3(x-3) step by step please

Maurinko [17]

Okay first you need to do distribute the -3 to remove the parentheses:

y - 4 = -3x + 9

Then you need to move -4 to the right by adding on both sides

y = -3x + 9 + 4

The answer is —> y = -3x + 13

7 0

2 years ago

B = -1-7ab=−1−7a I need help with this question

Elan Coil [88]

b = -1-7ab=−1−7a

Can be rewritten to be easier to understand.

$b = -1-7ab\\-1 -7a = -1-7ab$

In addition you can say the following equation is valid due to the way your problem was given to you.

$b = -1-7a$

Of course, you can simplify equation 2

$-1-7a = -1-7ab\\-7a = -7ab\\a = ab\\1 = b\\b = 1\\$

This ends up giving you b so you don't end up having to do any substitution in order to find b. Now substitute the value for b into either equation. I'm choosing the third because it's easiest.

$b = -1-7a$

$1 = -1 -7a\\2 = -7a\\-\frac{2}{7} = a\\a = -\frac{2}{7}$

And there you go :)

8 0

3 years ago

How do you solve this limit of a function math problem?

hram777 [196]

If you know that

$e=\displaystyle\lim_{x\to\pm\infty}\left(1+\frac1x\right)^x$

then it's possible to rewrite the given limit so that it resembles the one above. Then the limit itself would be some expression involving $e$ .

For starters, we have

$\dfrac{3x-1}{3x+3}=\dfrac{3x+3-4}{3x+3}=1-\dfrac4{3x+3}=1-\dfrac1{\frac34(x+1)}$

Let $y=\dfrac34(x+1)$ . Then as $x\to\infty$ , we also have $y\to\infty$ , and

$2x-1=2\left(\dfrac43y-1\right)=\dfrac83y-2$

So in terms of $y$ , the limit is equivalent to

$\displaystyle\lim_{y\to\infty}\left(1-\frac1y\right)^{\frac83y-2}$

Now use some of the properties of limits: the above is the same as

$\displaystyle\left(\lim_{y\to\infty}\left(1-\frac1y\right)^{-2}\right)\left(\lim_{y\to\infty}\left(1-\frac1y\right)^y\right)^{8/3}$

The first limit is trivial; $\dfrac1y\to0$ , so its value is 1. The second limit comes out to

$\displaystyle\lim_{y\to\infty}\left(1-\frac1y\right)^y=e^{-1}$

To see why this is the case, replace $y=-z$ , so that $z\to-\infty$ as $y\to\infty$ , and

$\displaystyle\lim_{z\to-\infty}\left(1+\frac1z\right)^{-z}=\frac1{\lim\limits_{z\to-\infty}\left(1+\frac1z\right)^z}=\frac1e$

Then the limit we're talking about has a value of

$\left(e^{-1}\right)^{8/3}=\boxed{e^{-8/3}}$

# # #

Another way to do this without knowing the definition of $e$ as given above is to take apply exponentials and logarithms, but you need to know about L'Hopital's rule. In particular, write

$\left(\dfrac{3x-1}{3x+3}\right)^{2x-1}=\exp\left(\ln\left(\frac{3x-1}{3x+3}\right)^{2x-1}\right)=\exp\left((2x-1)\ln\frac{3x-1}{3x+3}\right)$

(where the notation means $\exp(x)=e^x$ , just to get everything on one line).

Recall that

$\displaystyle\lim_{x\to c}f(g(x))=f\left(\lim_{x\to c}g(x)\right)$

if $f$ is continuous at $x=c$ . $\exp(x)$ is continuous everywhere, so we have

$\displaystyle\lim_{x\to\infty}\left(\frac{3x-1}{3x+3}\right)^{2x-1}=\exp\left(\lim_{x\to\infty}(2x-1)\ln\frac{3x-1}{3x+3}\right)$

For the remaining limit, write

$\displaystyle\lim_{x\to\infty}(2x-1)\ln\frac{3x-1}{3x+3}=\lim_{x\to\infty}\frac{\ln\frac{3x-1}{3x+3}}{\frac1{2x-1}}$

Now as $x\to\infty$ , both the numerator and denominator approach 0, so we can try L'Hopital's rule. If the limit exists, it's equal to

$\displaystyle\lim_{x\to\infty}\frac{\frac{\mathrm d}{\mathrm dx}\left[\ln\frac{3x-1}{3x+3}\right]}{\frac{\mathrm d}{\mathrm dx}\left[\frac1{2x-1}\right]}=\lim_{x\to\infty}\frac{\frac4{(x+1)(3x-1)}}{-\frac2{(2x-1)^2}}=-2\lim_{x\to\infty}\frac{(2x-1)^2}{(x+1)(3x-1)}=-\frac83$