Brainliest available

check if ur capable of me: )

u to everyone who

C, D

Explanation:

Length: L = x + 3

width: x ≥ y + 2   ⇒  y ≤ x - 2

wheels: y

0.50(L * x) + 4(2.25y) ≤ 50

0.50(x + 3)(x) + 4(2.25y) ≤ 50

0.50(x² + 3x) + 9y ≤ 50

                        9y ≤ 50 - 0.50(x² + 3x)

                        9y ≤ 50 - 0.50x² - 1.5x

                        ÷9               ÷9

                          y ≤

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15 per person is the rate

22 is the constant

y = 15x + 22

divide that by the number of people (x) to get:

y =

C for the first problem

D for the second One

Step-by-step explanation:

First One

This problem is a monster. There are so many letters flying around that you hardly know what is going on. I'll try and break it down a little at a time.

Width of the car = x

radius of the car = y

The width of the car must be at least 2 inches greater than radius

x ≥ y + 2 or

x - 2 ≥ y or

y ≤ x - 2

Area: L * w

Area of the base = x*(x + 3)

Cost of Area = x*(x + 3)*0.5

Cost of Area = (x^2 + 3x)*0.5

Cost of Area = 1/2 x^2 + 1.5x              We'll add fractions later.

The radius = y which is more commonly

For every inch of the radius, there is a cost of 2.25

For every inch of the radius, the cost = 2.25 * y

There are 4 wheels which cost 2.25*4*y= 9y

Total cost 9y + 1/2x^2 + 1.5x ≤ 50

Now here's where the problem get's really hairy. You have to solve for y

9y ≤ 50 - 1/2x^2  - 1.5x    50 dollars - base still has to be more than 9y

Now divide by 9. This too can be a little ugly.

9y / 9 ≤ 50 - (1/2 // 9/1) x^2 - 1.5/9 x

y ≤ 50/9 - 1/2 * 1/9*x^2  - 1/6 x

y ≤ 50/9 - 1/18 x^2 - 1/6 x

I'll leave you to find the correct letter.

Second Problem

The real snake in this problem is the 22 dollars. What to do with it. It says flat rate and I interpret that to mean there is no x with it. That eliminates B.

I think the only equation you can use is

C(x) = (15x + 22) / x    All the people there have to share in paying the 22 dollars is what that statement means.

The answer is D

4.08%

step-by-step explanation: