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3 years ago

PLEASE HELP I CANT FAIL

Mathematics

1 answer:

myrzilka [38]3 years ago

5 0

Answer:

A) Miguel: d=25t+10

gabby: d=30t

B) gabby will catch up with Miguel in 2 hours at a distance of 60 miles

Step-by-step explanation:

Lets start of with the equation:

There is 2 possible equations we can use for this:

A) Miguel: d=25t+10 gabby: d=30t OR

D) Miguel: d=10t+25 gabby: d= 30t

The resonable response for this would be A because Miguel starts off with 10 miles so you would add 10 to your equation so response D is incorrect because you are not multiplying 10t, but rather 25t.

Now let's do the actual answer

There is 2 possible answer for this:

B) gabby will catch up with Miguel in 2 hours at a distance of 60 miles
C) gabby will catch up with Miguel in 1.25 at a distance of 37.5 miles

The response that would make the most sense would be B. Let me explain

We are going to be using response C for now. You would have to multiply Miguel: 25(1.25) + 10

31.25 + 10

=41.25

Gabby: 30(1.25)

= 37.50

Miguel is still ahead of Gabby so Answer C is incorrect

We are going to be using response B now. You would have to multiply

Miguel: 25(2) + 10

50 + 10

=60

Gabby: 30(2)

=60

Now they are both the same mile so this would mean Answer B is correct

plz give brainliest :'D

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Determine whether the integral converges.

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You have one mistake which occurs when you integrate $\dfrac1{1-p^2}$ . The antiderivative of this is not in terms of $\tan^{-1}p$ . Instead, letting $p=\sin r$ (or $\cos r$ , if you want to bother with more signs) gives $\mathrm dp=\cos r\,\mathrm dr$ , making the indefinite integral equality

$\displaystyle-\frac12\int\frac{\mathrm dp}{1-p^2}=-\frac12\int\frac{\cos r}{1-\sin^2r}\,\mathrm dr=-\frac12\int\sec r\,\mathrm dr=\ln|\sec r+\tan r|+C$

and then compute the definite integral from there.

$-\dfrac12\ln|\sec r+\tan r|\stackrel{r=\sin^{-1}p}=-\dfrac12\ln\left|\dfrac{1+p}{\sqrt{1-p^2}}=\ln\left|\sqrt{\dfrac{1+p}{1-p}}\right|$
$\stackrel{p=u/2}=-\dfrac12\ln\left|\sqrt{\dfrac{1+\frac u2}{1-\frac u2}}\right|=-\dfrac12\ln\left|\sqrt{\dfrac{2+u}{2-u}}\right|$
$\stackrel{u=x+1}=-\dfrac12\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|$
$\implies-\dfrac12\displaystyle\lim_{t\to\infty}\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|\bigg|_{x=2}^{x=t}=-\frac12\left(\ln|-1|-\ln\left|\sqrt{\frac5{-1}}\right|\right)=\dfrac{\ln\sqrt5}2=\dfrac{\ln5}4$

Or, starting from the beginning, you could also have found it slightly more convenient to combine the substitutions in one fell swoop by letting $x+1=2\sec y$ . Then $\mathrm dx=2\sec y\tan y\,\mathrm dy$ , and the integral becomes

$\displaystyle\int_2^\infty\frac{\mathrm dx}{(x+1)^2-4}=\int_{\sec^{-1}(3/2)}^{\pi/2}\frac{2\sec y\tan y}{4\sec^2y-4}\,\mathrm dy$
$\displaystyle=\frac12\int_{\sec^{-1}(3/2)}^{\pi/2}\csc y\,\mathrm dy$
$\displaystyle=-\frac12\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2}}^{y=\pi/2}$
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Another way to do this is to notice that the integrand's denominator can be factorized.

$x^2+2x-3=(x+3)(x-1)$

So,

$\dfrac1{x^2+2x-3}=\dfrac1{(x+3)(x-1)}=\dfrac14\left(\dfrac1{x-1}-\dfrac1{x+3}\right)$

There are no discontinuities to worry about since you're integrate over $[2,\infty)$ , so you can proceed with integrating straightaway.

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$=\displaystyle\frac14\lim_{t\to\infty}\ln\left|\frac{x-1}{x+3}\right|\bigg|_{x=2}^{x=t}$
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Just goes to show there's often more than one way to skin a cat...

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We can isolate y in the linear equation to get:

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<h3>How to isolate the variable?</h3>

Here we have the linear equation:

8x - 2y = 16

Now, remember that we can perform the same operation in both sides of the equation, and the equation don't changes.

To isolate y, we can start by subtracting 8x in both sides, so we get:

-2y = 16 - 8x

Now we should divide both sides by -2:

y = (16 - 8x)/(-2) = -8 + 4x

y = 4x - 8

If you want to learn more about linear equations:

brainly.com/question/1884491

#SPJ1

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