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2 years ago

Explain how to find the distance between two integers using the difference.

Mathematics

2 answers:

rewona [7]2 years ago

6 0

Answer: Plot on a number line

Step-by-step explanation: The best way is to plot on a number line. That's the easiest way in my opinion.

ad-work [718]2 years ago

4 0

Answer:

Find the difference between two integers by adding the additive inverse. The distance is the absolute value of the difference. Check the distance by plotting the original two integers on a number line and counting the units between them.

Step-by-step explanation: Its the right answer

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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:

f is continuous on the closed interval [a, b].
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

3 years ago

Emily can write 1/6 pages for Ms.McCrimmon in 1/12 minutes. What was her unit rate to write full pages?

Vinil7 [7]

1/6:1/12 is 2:1

She can write 2 pages per minute

3 0

3 years ago

Simplify<br> ^3<br> V-27n^27

Sauron [17]

Apply the radical rule to separate terms:

Cubicroot(-27) and cubic root(n^27)

Cubicroot(-27) = -3

Now you have -3 cubicroot(n^27)

Using the exponent rule

N^27 can be rewritten as (n^9)^3

Now you have -3 cubicroot((n^9)^3)

The cubic root and the 3rd power cancel out to get the final answer of

-3n^9

8 0

3 years ago

If the scale factor of a scale drawing is greater than one, the scale drawing is larger than the actual object

galina1969 [7]

False, for example, if the scale factor is 2, the scale drawing would still be smaller because it would be twice as smaller than the actual object.

5 0

3 years ago

If f varies inversely as g, find f when g= -6.<br> F= 4 when g= -6<br> F=

Naddik [55]

Step-by-step explanation:

f=1/g

4=1/-6

=-24

find f

f=1/-6

4 0

3 years ago