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3 years ago

What is the length of the missing leg?

Mathematics

1 answer:

Vaselesa [24]3 years ago

4 0

Answer:

15 ft

Step-by-step explanation:

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Which rule matches the function graphed in the image?

Roman55 [17]

How do you find the correct equation? Follow these steps.

First, plug 0 in for X and solve for Y . Does the y value match what is on the graph? 0 in this case?

a. 0 = -2(0) True - This could be an answer
b. 0 = 0 + 4 False - This cannot be an answer
c. 0 = 2(0) True - This could be an answer
d. 0 = 0 - 4 False - This cannot be an answer

Now we are down to 2 possible answers A and C. Now take another point on the line, say x=-2 and repeat, does the answer match what is on the graph? (4).

a. 4 = -2(-2) True
c. 4 = 2(-2) False

As only A matches, A must be the answer. With graphing problems, when you have to find the correct equation, always choose X=0 as the first point (assuming it has a Y value) and eliminate answers. I hope this helps.

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4 years ago

Determine whether each event is impossible, unlikely, as likely as not, likely, or certain

chubhunter [2.5K]

Answer:

likely

Step-by-step explanation:

because you can tell its in the 1-12

6 0

3 years ago

Find the exact area of the surface obtained by rotating the curve about the x-axis. y = 1 + ex , 0 ≤ x ≤ 9

tekilochka [14]

The surface area is given by

$\displaystyle2\pi\int_0^9(1+e^x)\sqrt{1+e^{2x}}\,\mathrm dx$

since $y=1+e^x\implies y'=e^x$ . To compute the integral, first let

$u=e^x\implies x=\ln u$

so that $\mathrm dx=\frac{\mathrm du}u$ , and the integral becomes

$\displaystyle2\pi\int_1^{e^9}\frac{(1+u)\sqrt{1+u^2}}u\,\mathrm du$

$=\displaystyle2\pi\int_1^{e^9}\left(\frac{\sqrt{1+u^2}}u+\sqrt{1+u^2}\right)\,\mathrm du$

Next, let

$u=\tan t\implies t=\tan^{-1}u$

so that $\mathrm du=\sec^2t\,\mathrm dt$ . Then

$1+u^2=1+\tan^2t=\sec^2t\implies\sqrt{1+u^2}=\sec t$

so the integral becomes

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\left(\frac{\sec t}{\tan t}+\sec t\right)\sec^2t\,\mathrm dt$

$=\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\left(\frac{\sec^3t}{\tan t}+\sec^3 t\right)\,\mathrm dt$

Rewrite the integrand with

$\dfrac{\sec^3t}{\tan t}=\dfrac{\sec t\tan t\sec^2t}{\sec^2t-1}$

so that integrating the first term boils down to

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\frac{\sec t\tan t\sec^2t}{\sec^2t-1}\,\mathrm dt=2\pi\int_{\sqrt2}^{\sqrt{1+e^{18}}}\frac{s^2}{s^2-1}\,\mathrm ds$

where we substitute $s=\sec t\implies\mathrm ds=\sec t\tan t\,\mathrm dt$ . Since

$\dfrac{s^2}{s^2-1}=1+\dfrac12\left(\dfrac1{s-1}-\dfrac1{s+1}\right)$

the first term in this integral contributes

$\displaystyle2\pi\int_{\sqrt2}^{\sqrt{1+e^{18}}}\left(1+\frac12\left(\frac1{s-1}-\frac1{s+1}\right)\right)\,\mathrm ds=2\pi\left(s+\frac12\ln\left|\frac{s-1}{s+1}\right|\right)\bigg|_{\sqrt2}^{\sqrt{1+e^{18}}}$

$=2\pi\sqrt{1+e^{18}}+\pi\ln\dfrac{\sqrt{1+e^{18}}-1}{1+\sqrt{1+e^{18}}}$

The second term of the integral contributes

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\sec^3t\,\mathrm dt$

The antiderivative of $\sec^3t$ is well-known (enough that I won't derive it here myself):

$\displaystyle\int\sec^3t\,\mathrm dt=\frac12\sec t\tan t+\frac12\ln|\sec t+\tan t|+C$

so this latter integral's contribution is

$\pi\left(\sec t\tan t+\ln|\sec t+\tan t|\right)\bigg|_{\pi/4}^{\tan^{-1}(e^9)}=\pi\left(e^9\sqrt{1+e^{18}}+\ln(e^9+\sqrt{1+e^{18}})-\sqrt2-\ln(1+\sqrt2)\right)$

Then the surface area is

$2\pi\sqrt{1+e^{18}}+\pi\ln\dfrac{\sqrt{1+e^{18}}-1}{1+\sqrt{1+e^{18}}}+\pi\left(e^9\sqrt{1+e^{18}}+\ln(e^9+\sqrt{1+e^{18}})-\sqrt2-\ln(1+\sqrt2)\right)$

$=\boxed{\left((2+e^9)\sqrt{1+e^{18}}-\sqrt2+\ln\dfrac{(e^9+\sqrt{1+e^{18}})(\sqrt{1+e^{18}}-1)}{(1+\sqrt2)(1+\sqrt{1+e^{18}})}\right)\pi}$

4 0

4 years ago

I'll give u brainliest!!!! Discuss the locations of the angle of depression and angle of elevation of a right triangle. Which on

barxatty [35]

The angle of elevation of an object as seen by an observer is the angle between the horizontal and the line from the object to the observer's eye (the line of sight). This angle is seen outside of the triangle. If the object is below the level of the observer, then the angle between the horizontal and the observer's line of sight is called the angle of depression. This angle is seen inside the triangle. The angle of depression equals the angle of elevation because they from alternate interior angles or congruent angles located on different parallel lines cut by a transversal line.

3 0

3 years ago

Which of the following expressions is

AURORKA [14]

Answer:

Option F

Step-by-step explanation:

(y+3)-(-3y-5)

y+3+3y+5

4y+8

5 0

3 years ago

What is the length of the missing leg?​

What is the length of the missing leg?