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Semmy [17]
3 years ago
6

Katlyn sent 34,008 text messages last

Mathematics
2 answers:
Over [174]3 years ago
8 0

Answer:

2,834 texts

Step-by-step explanation:

There are 12 months in a year so divided the total number of texts by 12 then that will give you the monthly number of texts sent

Rudiy273 years ago
7 0
2834
If you divide the number of text messages (34,008) by the number of months in a year (12) you get 2834.
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HELP ME PLEASE I WILL MARK YOU AS A BRAINLIEST!!!!!!
telo118 [61]

Answer:

Statement C is correct.

4 0
3 years ago
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At 2:00 PM a car's speedometer reads 30 mi/h. At 2:20 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:20 the acce
Bad White [126]

Answer:

Let v(t) be the velocity of the car t hours after 2:00 PM. Then \frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }.  By the Mean Value Theorem, there is a number c such that 0 < c with v'(c)=60 \:{\frac{mi}{h^2}}. Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly 60 \:{\frac{mi}{h^2}}.

Step-by-step explanation:

The Mean Value Theorem says,

Let be a function that satisfies the following hypotheses:

  1. f is continuous on the closed interval [a, b].
  2. f is differentiable on the open interval (a, b).

Then there is a number c in (a, b) such that

f'(c)=\frac{f(b)-f(a)}{b-a}

Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.

By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.

Let v(t) be the velocity of the car t hours after 2:00 PM. Then v(0 \:h) = 30 \:{\frac{mi}{h} } and v( \frac{1}{3} \:h) = 50 \:{\frac{mi}{h} } (note that 20 minutes is 20/60=1/3 of an hour), so the average rate of change of v on the interval [0 \:h, \frac{1}{3} \:h] is

\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }

We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in (0 \:h, \frac{1}{3} \:h) at which v'(c)=60 \:{\frac{mi}{h^2}}.

c is a time time between 2:00 and 2:20 at which the acceleration is 60 \:{\frac{mi}{h^2}}.

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3 years ago
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Tanya [424]
Bag D; there are 12 counters in total and 2 blue counters, therefore the probability is 2/12, which is equal to 1/6.
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2 years ago
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Katyanochek1 [597]

Plug in the x value.

3(1-2)2=?
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3 years ago
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lesya [120]

Answer:

the answer is 10,000

Step-by-step explanation:

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