Use the distributive property to expand the following expression.

Two drivers, Alison and Kevin, are participating in a drag race. Beginning from a standing start, they each proceed with a const

Vilka [71]

Answer:

Alison wins against Kevin by 0.93 s

Step-by-step explanation:

Alison covers the last 1/4 of the distance in 3 seconds, at a constant acceleration $a_a$ , we have the following equation of motion

$s/4 = a_at_a^2/2$

where s (m) is the total distance, ta = 3 s is the time

$s = 4a_a3^2/2 = 18a_a$

$a_a = s/18$

Similarly, Kevin overs the last 1/3 of the distance in 4 seconds, at a constant acceleration $a_k$ , we have the following equation of motion:

$s/3 = a_kt_k^2/2$

tk = 4 s is the time

$s = 3a_k4^2/2 = 24a_k$

$a_k = s/24$

Since $a_a = s/18$ we can conclude that $a_a > a_k$ , so Alison would win.

The time it takes for Alison to cover the entire track

$s = a_aT_a^2/2$

$T_a^2 = 2s/a_a = 2s/(s/18) = 36$

$T_a = \sqrt{36} = 6 s$

The time it takes for Kevin to cover the entire track

$s = a_kT_k^2/2$

$T_k^2 = 2s/a_k = 2s/(s/24) = 48$

$T_a = \sqrt{48} = 6.93 s$

So Alison wins against Kevin by 6.93 - 6 = 0.93 s

4 0

3 years ago

Find the volume of the sphere. Round your answer to the nearest tenth.

dusya [7]

Answer:

Step-by-step explanation:

Remark

All you need do is find the radius and put it into the formula V = (4/3) pi r^3.

Givens

d = 12 yards

r = d/2

Solution

r = 12 yds/2 = 6 yds

V = 4/3 pi r^3

V = 4/3 * 3.14 * 6*6*6

V = 904.32

Answer 904.4 yd^3

8 0

2 years ago

Can someone help me with this please it would be a big help

Anuta_ua [19.1K]

Answer in order

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answer
A. -4.6 - (0.4 + 7.2)

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answer
B. -10 + (-90) + (-40)

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answer

associate property of addition

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answer
Commutative <span>Property
</span>
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answer
-147

7 0

3 years ago

A college professor never finishes his lecture before the end of the hour and always finishes his lectures within 2 min after th

Citrus2011 [14]

Answer:

Value of k is $\displaystyle\frac{3}{8}$

Step-by-step explanation:

We are given the following information in the question:

$f(x) = kx^2, 0\leq x \leq 2\\~~~~~~~= 0, \text{ otherwise}$

where x is the time elapses between the end of the hour and the end of the lecture.

We have to find the values of k.

Since, f(x) is the pdf, then,

$\displaystyle\int^\infty_{-\infty} f(x) = 1\\\\\displaystyle\int^2_{0} f(x) = 1\\\\\displaystyle\int^2_{0} kx^2 = 1\\\\k\bigg[\frac{x^3}{3}\bigg]^2_0 = 1\\\\k\times \frac{8}{3} = 1\\\\k = \frac{3}{8}$