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grigory [225]
3 years ago
12

SOMEONE HELP ASAP!!

Mathematics
2 answers:
Andreas93 [3]3 years ago
7 0

Answer:

7

Step-by-step explanation:

The circumference of a circle is given by

C = pi *d where d is the diameter

7 pi = pi *d

Divide each side by pi

7 = d

The diameter is 7

RideAnS [48]3 years ago
4 0
7 hope this helps :)
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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}}.

4 0
4 years ago
Factor the expression.
grigory [225]
Hey!

First, let's write the problem,
4r^2-64
Factor out the common term, which is 4,
=4\left(r^2-16\right)
Then, let's factor this part: r^2-16
We are going to factor it using the difference of squares rule.
=\left(r+4\right)\left(r-4\right)
Our final factored answer would be,
=4\left(r+4\right)\left(r-4\right)

Thanks!
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4 x 2x 4 being length and 2x being width
kow [346]

Answer:

1/4

Step-by-step explanation:

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4 years ago
Someone please help me with this problem! Will mark brainliest!
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Answer:

I need a question, then I will help

Step-by-step explanation:

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