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3 years ago

Find the slope of the line going through the points (0,6) and (-3,9)

Mathematics

1 answer:

andre [41]3 years ago

7 0

Answer:

The slope of the line is 5.

The point slope form of the line is y=5x+6.

Step-by-step explanation:

We can find the slope using this formula: $\frac{y_{2} }{x_{2} } -\frac{y_{1} }{x_{1} }$

$\frac{6}{0} -\frac{9}{-3} =$

$\frac{6}{0} -(-\frac{9}{3} )=$

$\frac{6}{0} +\frac{9}{3} =$

$\frac{15}{3} =$

The slope of the line is 5.

To find point slope form, we use the equation y=mx+b where m is the slope and b is the y-intercept.

The y-intercept of this slope is 6 and the slope is 5 so we get, y=5x+6

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2 years ago

Help help area and perimeter

natita [175]

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2 years ago

A survey was conducted to determine the amount of time, on average, during a given week SCAD students spend outside of class on

fiasKO [112]

Answer:

Standard Deviation = 5.928

Step-by-step explanation:

a) Data:

Days Hours spent (Mean - Hour)²

1 5 61.356

2 7 34.024

3 11 3.360

4 14 1.362

5 18 26.698

6 22 84.034

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mean

77/6 = 12.833 and 210.83/6 = 35.139

Therefore, the square root of 35.139 = 5.928

b) The standard deviation of 5.928 shows how the hours students spend outside of class on class work varies from the mean of the total hours they spend outside of class on class work.

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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