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mojhsa [17]
3 years ago
7

I don’t understand anything someone please help me :/

Mathematics
1 answer:
Artist 52 [7]3 years ago
8 0

Answer:

answer choice c=6

Step-by-step explanation:

since the question says that round to the nearest

mile then you would round 5.7 to get you 6 so

your answer is 6

Hope this help you :))

 please give me branliest.                                                              

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Can you help me with math?
GaryK [48]

Answer:

n = 143

Step-by-step explanation:

To find the value of n, you multiply 13 by 10 because that is the inverse of dividing, and you get 130. Since there is a greater than sign, the answer is 143 because the value of n has to be greater than 130.

5 0
3 years ago
Read 2 more answers
Tamara has a cell phone plan that charges $0.07 per minute plus a monthly fee of $19.00. She budgets $29.50 per month for total
Dominik [7]

Answer:

The Answer is A: 150

Step-by-step explanation:

$29.50 - $19.00 = $10.5

$10.5 / $0.07 = 150 Minutes.

5 0
3 years ago
Angle A and B are supplementary angles. The measure of angle B is twice the measure of angle A minus 24. Find the measure of the
ivann1987 [24]

Answer:

A = 58

B = 112

Step-by-step explanation:

A + B = 180                           Supplementary angles

B = 2A - 24                           Given

Substitute for B in the first equation

A + 2A - 24  = 180

3A - 24 = 180                         Add 24 to both sides

3A = 180 + 24                        Combine the right

3A = 204                               Divide both sides by 3

A = 204/3

A = 68

B = 2A - 24

B = 2*68 - 24

B = 136 - 24

B = 112

5 0
3 years ago
Determine whether the integral is convergent or divergent. [infinity] 7 e−1/x x2 dx convergent divergent Changed: Your submitted
inessss [21]

I suppose the integral could be

\displaystyle\int_7^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx

In that case, since -\frac1x\to0 as x\to\infty, we know e^{-1/x}\to1. We also have \left(e^{-1/x}\right)'=\frac{e^{-1/x}}{x^2}>0, so the integral is approach +1 from below. This tells us that, by comparison,

\displaystyle\frac{e^{-1/x}}{x^2}\le\frac1{x^2}\implies\int_7^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx\le\int_7^\infty\frac{\mathrm dx}{x^2}

and the latter integral is convergent, so this integral must converge.

To find its value, let u=-\frac1x, so that \mathrm du=\frac{\mathrm dx}{x^2}. Then the integral is equal to

\displaystyle\int_{-1/7}^0e^u\,\mathrm du=e^0-e^{-1/7}=1-\frac1{\sqrt[7]{e}}

4 0
3 years ago
.....................
Oksana_A [137]

Answer: hi .................

3 0
3 years ago
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