How do you get 189 as the answer? Please help

A car rental charge is $100 plus $5.00 per mile traveled. Create an equation for the situation above.

Elis [28]

Answer:
X = 100$ + 5a$
Where X is the price of a car rent
And a is the amount of mile travel

3 0

2 years ago

In 2015 there were 759 laptops at lightning high school starting a 2016 the school box 75 more laptops at the end of each year t

nikdorinn [45]

Answer:

Step-by-step explanation:

Given the following :

Number of laptops in 2015 = 759

t(x) = 75x + 759 ; model to determine number of laptops at the school. X years after 2015.

Number of laptops in the school in 2020:

4 0

3 years ago

Rewrite in simplest terms: 4(-5u +3) – 4(8u - 1)

viktelen [127]

Answer:

You distribute then combine

your answer is -52u+16

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

I have trouble with these two questions, even i did my equation but i don’t get the answers

Sophie [7]

Answer:

$318.17, out of state college is $161.12 cheaper per month

Step-by-step explanation:

72*24 = 1728 + 410*4 = 1728 + 1640 = 3368 + 450 = 18/12 = $318.17

8700+650*9 = 8700 + 5850 = 14550/9 = 1616.66

8500+400*9 = 8500 + 3600 = 12100 + 1000 = 13100/9 = 1455.56

1616.67 - 1455.56= $161.11 (use $161.12)

3 0

2 years ago

If x = a cosθ and y = b sinθ , find second derivative

Olin [163]

I'm guessing the second derivative is for y with respect to x, i.e.

$\dfrac{\mathrm d^2y}{\mathrm dx^2}$

Compute the first derivative. By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta$

$x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta$

Now compute the second derivative. Notice that $\frac{\mathrm dy}{\mathrm dx}$ is a function of $\theta$ ; so denote it by $f(\theta)$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}$

By the chain rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$f=-\dfrac ba\cot\theta\implies\dfrac{\mathrm df}{\mathrm d\theta}=\dfrac ba\csc^2\theta$

and so the second derivative is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac ba\csc^2\theta}{-a\sin\theta}=-\dfrac b{a^2}\csc^3\theta$

4 0

2 years ago