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2 years ago

Absolute value to find distance

Mathematics

1 answer:

qaws [65]2 years ago

6 0

Answer:

Step-by-step explanation:

The distance between two any points <em>a</em> i <em>b</em> is |a-b| as we don't know which of them is larger number.

So:

a=-2 and b=1 means distance |-2-1| = |-3| = 3

{If we know the value of given points we can subtract the smaller one from the larger. It also works: -2<1, so the distance would be 2-(-1)=2+1=3}

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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 $60 \:{\frac{mi}{h^2}}$ .

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3 years ago

If a is the set of all integers choose the set B that will make the following statement false

Tju [1.3M]

Answer:

Try the 2nd one

Step-by-step explanation:

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3 years ago

Read 2 more answers

What is the slope of f(x)=x^2+3 when x=-1 ?

irinina [24]

F(x)=(-1)^2+3
f(x)=1+3
f(x)=4

The slope is 0 because 4 is a constant.

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3 years ago

Which graph represents the function f(x)=

zhenek [66]

Step-by-step explanation:

you don't show us the choices.

anyway, the real graph must be similar to the one you are showing, as it goes to - and + infinity for x = 2.

because the denominator "x-2" will be 0 for x = 2.

but the horizontal limits are both y = +3 (and not 0).

because (3x-2)/(x-2) goes more and more to 3/1 the larger (or smaller in the - direction) x gets.

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1 year ago

working together, melissa and jing can mow a lawn in 5 hours. It would take melissa 8 hours to do the job alone. What is the val

lubasha [3.4K]

Total Time = (M x K) / (M + K)
5 = (8 x K) / (8 + K)
40 + 5 K = 8K
40 = 3K
King can do the job in 13.33 Hours

King could mow (1 / 13.33) of the lawn in one hour.

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2 years ago