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2 years ago

How can I find the slope and Y- intercept for each equation

1 answer:

7 0

Answer:

You plug in numbers into the y=mx+b equation and solve for which ever one you are finding for.
m=slope
x=x intercept
b=y intercept
apply the numbers to this equation and solve or x

You might be interested in

What is the solution to the equation 5x = 40?

Dima020 [189]

To find X, we can divide 5 into 40.
40 divided by 5 is 8.
Now we can do 5 x 8. That is 40.

X=40

OR we can do it another way.
40 divided by 5 is 8.
Cross out 5 divided by 5.

Now we have 8. Still.
X is 8

7 0

2 years ago

Read 2 more answers

Add the following polynomials (2x + 5) + (-1x -4) Answer options: 3x + 9, 3x + 1, 1x + 1, 1x + 9

Leni [432]

Answer:

1x+1

Step-by-step explanation:

just open the parentheses

3 0

2 years ago

What is 1/4(x+16)-x equal too

schepotkina [342]

<span>1/4(x + 16) - x =

First, distribute the 1/4.

= 1/4x + 1/4 * 16 - x

= 1/4x + 4 - x

= 1/4x + 4 - 4/4x

= 1/4x - 4/4x + 4

= -3/4x + 4

$= -\dfrac{3}{4}x + 4$ </span>

8 0

2 years ago

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

2 years ago

Which is it? please help me and i will give brainliest

Cerrena [4.2K]

Answer: A, 1/2

Step-by-step explanation:

he is simply using 1 cup of bs for each cup of ws so when you do it its i cup of ws then its hald of 1 bs

3 0

2 years ago

Read 2 more answers