A salesperson sells $2,500 in merchandise and earns a commission of $300. What is the commission rate as a percentage?

A particle moves on a straight line and has acceleration a(t)=24t+2. Its position at time t=0 is s(0)=3 and its velocity at time

user100 [1]

Answer:

It's position at time t = 5 is 593.

Step-by-step explanation:

The velocity v(t) is the integral of the acceleration a(t)

The position s(t) is the integral of the velocity v(t)

We have that:

The acceleration is:

$a(t) = 24t + 2$

Velocity:

$v(t) = \int {a(t)} \, dt = \int {24t + 2} \, dt = 12t^{2} + 2t + K$

K is the initial velocity, that is v(0). Since V(0) = 13, K = 13

Then

$v(t) = 12t^{2} + 2t + 13$

Position:

$s(t) = \int {s(t)} \, dt = \int {12t^{2} + 2t + 13} \, dt = 4t^{3} + t^{2} + 13t + K$

Since s(0) = 3

$s(t) = 4t^{3} + t^{2} + 13t + 3$

What is its position at time t=5?

This is s(5).

$s(t) = 4t^{3} + t^{2} + 13t + 3$

$s(5) = 4*5^{3} + 5^{2} + 13*5 + 3$

$s(5) = 593$

It's position at time t = 5 is 593.

3 0

3 years ago

Add and Subtract Rational Expressions with a Common Denominator

Sladkaya [172]

Answer:

$\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{q^2+9}{q^2+6q+5}$

Step-by-step explanation:

<u>Simplifying Rational Expressions</u>

If two or more rational expressions have the same denominator, the add and subtract operations are done only with the numerator. The final denominator will be the common of both.

The expression is:

$\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}$

Operating on the numerators:

$\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{4q^2-q+3-(3q^2-q-6)}{q^2+6q+5}$

Removing parentheses:

$\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{4q^2-q+3-3q^2+q+6}{q^2+6q+5}$

Simplifying:

$\boxed{\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{q^2+9}{q^2+6q+5}}$

The expression cannot be further simplified.

7 0

2 years ago

Why does it take 3 copies of 1/6 to show the same amount as 1 copy of 1/2?

antiseptic1488 [7]

You can solve this by putting the number of copies in the top part of the fraction. If you replace the one in the top part of 1/6 with 3, you get the fraction 3/6. Half of six is three and half of two is one, so both fractions are equal. You can test this be dividing the 3 in the top by 3 and the 6 in the bottom by three. Your result will be 1/2. Although there are three times ad many copies, each copy is three times as small, so it balances out.

6 0

3 years ago

Round 1231.50432 to the nearest tenth

Verizon [17]

Answer:

1231.5

Step-by-step explanation:

I hope this help.

5 0

3 years ago

I need help please me!!!

Romashka [77]

Answer: a

Step-by-step explanation:

8 0

3 years ago

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