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2 years ago

What is. the fraction1/5 of 80?

Mathematics

2 answers:

Ksivusya [100]2 years ago

8 0

$\dfrac{1}{5}\cdot 80=16$

Yuliya22 [10]2 years ago

5 0

$\frac{1}{5} \times 80=\frac{80}{5}=\frac{16 \times 5}{5}=16$

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For how many positive integers n is it possible to have a triangle with side lengths five, 12, and n

Andre45 [30]

Step-by-step explanation:

you know, a single side cannot be longer than the other 2 sides combined.

otherwise, the triangle cannot "close".

so, it starts with 12 cannot be longer than 5 + n.

therefore, n must be at least 7.

and n cannot be longer than 5+12 = 17

7 and 17 I would normally rule out a well, because in these cases the triangle would just be a flat line, when the 3rd side is as long as the other 2 combined.

so, in reality for a real, visible triangle, the range of valid values is 8 .. 16.

that is 9 positive integer values.

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Which if any of the points is a whole of F (X)?

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Answer:

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4 months ago

Find the measure of the indicated angle. Need help please.<br> I need explanation

Zolol [24]

Answer:

BC =21.03540021

Step-by-step explanation:

We know the measure of angle B since the sum of the angles of a triangle add to 180

A + B+ C = 180

61+ B + 12 =180

B = 180 - 12 -61

B =107

Then we can use the law of sines to find BC

sin B sin A

-------- = -------------

AC BC

sin 107 sin 61

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23 BC

Using cross products

BC sin 107 = 23 sin 61

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2 years ago

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Please help with this math problem, lost in the sauce over here.

lilavasa [31]

Answer:

the answer i got was 501600

Step-by-step explanation:

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2 years ago

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