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3 years ago

Can someone please help me with this question!ASAP

Mathematics

1 answer:

Marina86 [1]3 years ago

6 0

Answer:

I can only say it is not the first one

Step-by-step explanation:

I don't really know the rest of it but I know that the first one is not correct

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3 years ago

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What is the area of a trapezoid with side lengths of 13,13,20,30 and a height of 12 cm

Dennis_Churaev [7]

Answer:

Area = $156cm^{2}$

Step-by-step explanation:

A=a+b

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I hope this helps you :)

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2 years ago

Which of the following is a binomial experiment

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23/50 equals what percent

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3 years ago

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The German autobahns are the nationally coordinated motorway system in Germany and have no general speed limit, but the advisory

melamori03 [73]

Answer:

The required equation is: $s=0.0508v^2$

The stopping distance of your car if you were moving at 95 mph is 458.47 feet.

Step-by-step explanation:

Consider the provided information.

Part (A)

Let us consider "s" is the distance "v" is the velocity and "k" is the constant of variation.

It is given that the stopping distance of a car varies directly as the square of the velocity of car when the brakes are applied

This can be written as:

$s=kv^2$ ......(1)

It is given that A car moving at 85 mph can stop in 367 feet.

Substitute v=85 and s=367 in above formula and solve for k.

$367=k(85)^2$

$367=7225k$

$\frac{367}{7225}=k$

$0.0508=k$

Now to find the equation that relates the stopping distance to the velocity substitute the value of k in equation 1.

$s=0.0508v^2$

Hence, the required equation is: $s=0.0508v^2$

Part (B)

Calculate the stopping distance of your car if you were moving at 95 mph.

Substitute the value of v=95 in $s=0.0508v^2$

$s=0.0508(95)^2$

$s=0.0508\times 9025$

$s=458.47$

Hence, the stopping distance of your car if you were moving at 95 mph is 458.47 feet.

4 0

3 years ago