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3 years ago

Write the equation of a line passing through (5,-3) with a slope of 3/8

Mathematics

1 answer:

ANTONII [103]3 years ago

6 0

Answer:

y=3/8x-3

Step-by-step explanation:

the equation is in the form of y intercept so the 3/8x is your slope which would be your mx in y=mx+b the b is the the point where it passes through the y axis when x is zero and since the 3 is negative you could either put + -3 or just -3 like i did. then you get the equation y=3/8x-3

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Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

3 years ago

WILL MARK BRAINLIEST IF CORRECT! A mini-cone has a diameter of 2 inches and holds 12.5 in3 of ice cream. A mega-cone holds 133.1

Arada [10]

I think the answer is B

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3 years ago

Find the area of the triangle.

madreJ [45]

Start by finding the area of each smaller right triangle.
Right triangle area = (1/2)hb -> h = height & b = base
b = 4 & h = 3
(1/2)(4)(3) = 6
6*2 = 12 (there are two right triangles)
The area of the entire triangle is 12yd^2

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4 years ago

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Agata [3.3K]

Answer:

C sorry if its wrong

Step-by-step explanation:

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