All categories

3 years ago

What is the slope line of D?

Mathematics

1 answer:

aalyn [17]3 years ago

6 0

Answer:

2/5 (rise over run)

Step-by-step explanation:

You might be interested in

7/2x+1/2=10 1/2+9/2x what is the value of x

gogolik [260]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

P=a+b+c solve for b c=90x+35

Pavel [41]

Answer:

x=bc-35 over 90

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Find the measure of BCD

Lerok [7]

please mark me as brainliest!!

4 0

3 years ago

Someone please solve this. no where is an answer to this

serg [7]

Answer:

a=-8/5

Step-by-step explanation:

(-1/2a-5)=3a+1

To find the opposite of 1/2 a−5, find the opposite of each term.

-1/2a-(-5)= -3a+1

The opposite of −5 is 5.

-1/2a+5=-3a+1

add 3a to both sides

-1/2a+5+3a=1

Combine-1/2and 3a to get 5/2a

5/2a+5=1

Subtract 5 from both side

5/2a=1-5

Subtract 5 from 1 to get −4.

5/2a=-4

Multiply both sides by 2/5 =0.4, the reciprocal of 5/2=2.5.

a=-4x(2/5)

express -4x(2/5)=1.6 as single fraction.

a= -4x2/5

Multiply −4 and 2 to get −8.

a=-8/5

Fraction −8/5 ≈−1.6 can be rewritten as −8/5 =−1.6 by extracting the negative sign.

a=-8/5

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

4 years ago

Read 2 more answers