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3 years ago

What is the Tan J? URGENT HELP PLS

Mathematics

1 answer:

BARSIC [14]3 years ago

8 0

Answer:

4th Option, $\frac{12}{5}$

Hope this helps!

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Joe takes out a loan for $19,000. He will pay the loan back monthly. The interest rate is 7.6% per year and the length of the lo

Y_Kistochka [10]

I believe it should be 8,000

6 0

2 years ago

When 23 is subtracted from 5 times a certain number the result is 87

allsm [11]

5x-23=87
add 23 to both sides
5x=110
divide both sides by 5
x=22

To check your answer...
5(22)-23=87
110-23=87
87=87

8 0

3 years ago

one-third of the people from country A claim that they are from country B, and the rest admit they are from country A. One-fourt

In-s [12.5K]

Answer: 3 : 2

Step-by-step explanation:

Let A represents the total population of country A and B represents the total population of country B.

According to the question,

$\text{The population of country A that admit they are from B} = \frac{1}{3}\text{ of }A$

⇒ $\text{ The population of A that admit they are from country A }= A - \frac{1}{3} \text{ of } A$

$= \frac{3-1}{3} A$

$= \frac{2}{3} A$

$\text{The population of country B that admit they are from A} = \frac{1}{4}\text{ of }B$

⇒ $\text{ The total population that claims that they are from A }= \frac{2}{3} A +\frac{1}{4} B$

But, Again according to the question,

The total population that claims that they are from A = one half of the total population of A and B.

⇒ $\frac{2}{3} A + \frac{1}{4} B= \frac{1}{2}(A+B)$

⇒ $\frac{2}{3} A + \frac{1}{4} B= \frac{1}{2}A+\frac{1}{2}B$

⇒ $\frac{2}{3} A - \frac{1}{2}A= \frac{1}{2}B-\frac{1}{4} B$

⇒ $\frac{4}{6} A - \frac{3}{6}A= \frac{2}{4}B-\frac{1}{4} B$

⇒ $\frac{1}{6} A = \frac{1}{4} B$

⇒ $A =\frac{6}{4}B$

⇒ $\frac{A}{B} =\frac{3}{2}$

8 0

3 years ago

99 POINT QUESTION, PLUS BRAINLIEST!!!

VladimirAG [237]

First, we have to convert our function (of x) into a function of y (we revolve the curve around the y-axis). So:

$y=100-x^2\\\\x^2=100-y\qquad\bold{(1)}\\\\\boxed{x=\sqrt{100-y}}\qquad\bold{(2)} \\\\\\0\leq x\leq10\\\\y=100-0^2=100\qquad\wedge\qquad y=100-10^2=100-100=0\\\\\boxed{0\leq y\leq100}$

And the derivative of x:

$x'=\left(\sqrt{100-y}\right)'=\Big((100-y)^\frac{1}{2}\Big)'=\dfrac{1}{2}(100-y)^{-\frac{1}{2}}\cdot(100-y)'=\\\\\\=\dfrac{1}{2\sqrt{100-y}}\cdot(-1)=\boxed{-\dfrac{1}{2\sqrt{100-y}}}\qquad\bold{(3)}$

Now, we can calculate the area of the surface:

$A=2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\left(-\dfrac{1}{2\sqrt{100-y}}\right)^2}\,\,dy=\\\\\\= 2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=(\star)$

We could calculate this integral (not very hard, but long), or use (1), (2) and (3) to get:

$(\star)=2\pi\int\limits_0^{100}1\cdot\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\left|\begin{array}{c}1=\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\end{array}\right|= \\\\\\= 2\pi\int\limits_0^{100}\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\\\\\\ 2\pi\int\limits_0^{100}-2\sqrt{100-y}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\dfrac{dy}{-2\sqrt{100-y}}=\\\\\\$

$=2\pi\int\limits_0^{100}-2\big(100-y\big)\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\left(-\dfrac{1}{2\sqrt{100-y}}\, dy\right)\stackrel{\bold{(1)}\bold{(2)}\bold{(3)}}{=}\\\\\\= \left|\begin{array}{c}x=\sqrt{100-y}\\\\x^2=100-y\\\\dx=-\dfrac{1}{2\sqrt{100-y}}\, \,dy\\\\a=0\implies a'=\sqrt{100-0}=10\\\\b=100\implies b'=\sqrt{100-100}=0\end{array}\right|=\\\\\\= 2\pi\int\limits_{10}^0-2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=(\text{swap limits})=\\\\\\$

$=2\pi\int\limits_0^{10}2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4}\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^4}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^2}{4}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{\dfrac{x^2}{4}\left(4x^2+1\right)}\,\,dx= 4\pi\int\limits_0^{10}\dfrac{x}{2}\sqrt{4x^2+1}\,\,dx=\\\\\\=\boxed{2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx}$

Calculate indefinite integral:

$\int x\sqrt{4x^2+1}\,dx=\int\sqrt{4x^2+1}\cdot x\,dx=\left|\begin{array}{c}t=4x^2+1\\\\dt=8x\,dx\\\\\dfrac{dt}{8}=x\,dx\end{array}\right|=\int\sqrt{t}\cdot\dfrac{dt}{8}=\\\\\\=\dfrac{1}{8}\int t^\frac{1}{2}\,dt=\dfrac{1}{8}\cdot\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}=\dfrac{1}{8}\cdot\dfrac{t^\frac{3}{2}}{\frac{3}{2}}=\dfrac{2}{8\cdot3}\cdot t^\frac{3}{2}=\boxed{\dfrac{1}{12}\left(4x^2+1\right)^\frac{3}{2}}$

And the area:

$A=2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx=2\pi\cdot\dfrac{1}{12}\bigg[\left(4x^2+1\right)^\frac{3}{2}\bigg]_0^{10}=\\\\\\= \dfrac{\pi}{6}\left[\big(4\cdot10^2+1\big)^\frac{3}{2}-\big(4\cdot0^2+1\big)^\frac{3}{2}\right]=\dfrac{\pi}{6}\Big(\big401^\frac{3}{2}-1^\frac{3}{2}\Big)=\boxed{\dfrac{401^\frac{3}{2}-1}{6}\pi}$

Answer D.

6 0

3 years ago

Read 2 more answers

Janelle has 5 hours to spend training for an upcoming race. She completes her training by running full speed the distance of the

VikaD [51]

Answer:

The distance cover by him walking back is 11.25 miles

Step-by-step explanation:

Given as :

Time for Janelle to spend on training = 5 hours

The running speed of Janelle = 9 mph

The walking back with 3 mph