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stealth61 [152]

3 years ago

13

Nate has 5 green marbles and 3 red marbles. Select the ratios that compares the number of red marbles to the total number of mar

bles. Mark all that apply.
3 to 8
5 to 8
3:8
5:3
5/8
3/8

1 answer:

gtnhenbr [62]3 years ago

6 0

Answer:

5 to 8 & 5/8

Step-by-step explanation:

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I’m confused please help

Law Incorporation [45]

So x equals 30 because u did 180-150 and y is also 30 since you found x you then add 90 plus 30 and it will give u 120 then 180 minus 120 will be 60 then 90 - 60 because u are trying to find y, and u will get 30.

3 0

3 years ago

John has $6 to buy school supplies. He buys 6 notebooks that are $0.52 each. He spends the remaining money buying erasers that a

Anuta_ua [19.1K]

John bought 18 erasers.

Explanation: He has $6.00

He spends $3.12 on notebooks
Then you do $6.00 - $3.12 and you get $2.88

If it’s ¢16 per eraser and he has no money left over after buying erasers than you would divide $2.88 by $0.16 which turns into 288 divided by 16 which comes out to 18.

Therefor John bought 18 erasers.

5 0

3 years ago

Greatest to least of 2, -1, -2, 0, 1

Wittaler [7]

Answer:

2,1,0,-1,-2

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

HELP BEEN STUCH FOR 30 MIN

Jobisdone [24]

Start by plugging 0 into the equation and see which ones give you 350 back. Those equations are the right ones,

6 0

3 years ago

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. Verify y

mariarad [96]

Answer:

the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis is;

$\frac{\pi }{2} [e^2 - 1 ]$ or 10.036

Step-by-step explanation:

Given the data in the question;

y = $y = e^{(x - 1 )$ , y = 0, x = 1, x = 2.

Now, using the integration capabilities of a graphing utility

y = $y = e_2}^{(x - 1 )_$ , y = 0

Volume = $\pi \int\limits^2_1 ( e^{x-1)^2} - (0)^2 dx$

Volume = $\pi \int\limits^2_1 ( e^{x-1)^2 dx$

Volume = $\pi \int\limits^2_1 e^{2x-2}dx$

Volume = $\frac{\pi }{e^2} \int\limits^2_1 e^{2x}dx$

Volume = $\frac{\pi }{e^2} [\frac{e^{2x}}{2}]^2_1$

Volume = $\frac{\pi }{2e^2} [e^4 - e^2 ]$

Volume = $\frac{\pi }{2} [e^2 - 1 ]$ or 10.036

Therefore, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis is;

$\frac{\pi }{2} [e^2 - 1 ]$ or 10.036

3 0

3 years ago