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2 years ago

Graph the function f(x)=(1/3)^x+2-6

Mathematics

1 answer:

Hitman42 [59]2 years ago

3 0

I got this picture from photomath, hope it helps

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Please help me I need to get a good grade on this

Julli [10]

1 4/5 - 2/5 = r you minus the total she has already walked or ran from the total she plans on running to get “r”

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3 years ago

3x-7>5 Find the solution, please show your work. Thank You!1

miskamm [114]

To find x just change the > to a =
3x-7=5
add 7
3x=12
then divide by 3
x=4

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3 years ago

Read 2 more answers

Identify the terms of the expression. 3r^2+4r+8

Andreas93 [3]

Answer:

The terms

Step-by-step explanation:

The terms in this expression are coefficients, constants, variables, and exponents. 3 and 4 would be the coefficients. 8 would be the constant. r is the variable and 2 is the exponent.

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3 years ago

Read 2 more answers

The question is on the image.

irakobra [83]

2 x 2 x 5 = 20

4 x 2 x 2 = 16

20 + 16 = 36

Ans: 36 cubic feet

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3 years ago

The curves y = √x and y=(2-x) and the Cartesian axes form two distinct regions in the first quadrant. Find the volumes of rotati

makkiz [27]

Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

$y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\$

And the second one would be

$y = 2-x \,\,\,\,\,, 1 \leq x \leq 2$ .

If you rotate the first region around the "y" axis you get that

$A_1 = 2\pi \int\limits_{0}^{1} x\sqrt{x} dx = \frac{4\pi}{5} = 2.51$

And if you rotate the second region around the "y" axis you get that

$A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188$

And the sum would be 2.51+4.188 = 6.698

If you revolve just the outer curve you get

If you rotate the first region around the x axis you get that

$A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708$

And if you rotate the second region around the x axis you get that

$A_2 = \pi \int\limits_{1}^{2} (2-x)^2 dx = \frac{\pi}{3} = 1.0472$

And the sum would be 1.5708+1.0472 = 2.618

7 0

3 years ago