All categories

2 years ago

A frame is marked down by 22 percent from an original price of $14.50. What is the new price?

Mathematics

2 answers:

Valentin [98]2 years ago

7 0

22% = 22/100 22/100 * 14.50 = 3.19 (this is the amount discounted off original price) 14.50 - 3.19 = $11.31

Marat540 [252]2 years ago

7 0

Answer:11.31

Step-by-step explanation:

jus no

You might be interested in

23. Assuming the green is flat, what is the radius of the green?

tensa zangetsu [6.8K]

Answer:

Step-by-step explanation:

$r^{2} +32^{2} = (r+8)^{2}$

$r^{2} +1024=r^{2} +16r + 64$

$960 = 16r$

$60=r$

4 0

2 years ago

What is part of a line, has one endpoint, and continues in one direction

Lina20 [59]

The answer is a ray.

6 0

2 years ago

Read 2 more answers

What is the greatest common factor of 44 and 47?

Veseljchak [2.6K]

Answer:

Step-by-step explanation:

To get the GCF of 44 and 47, factor each value,then we choose copies of factors and multiply them.

4 0

2 years ago

Please help!!! Find the value of the variable and the measures of the angles.

e-lub [12.9K]

Answer:

x= 38°

<P = 57°

<R = 38

<Q = 85°

5 0

7 months ago

The total cost (in hundreds of dollars) to produce x units of a product is c(x) = (3x-2) / (8x+1), find the average cost for eac

olya-2409 [2.1K]

Answer:

a) $\frac{74}{10025}$

b) $\frac{3x-2}{x(8x+1)}$

c) $\frac{-24x^2+32x-2}{(8x^2+x)^2}$

Step-by-step explanation:

For total cost function $c(x)$ , average cost is given by $\frac{c(x)}{x}$ i.e., total cost divided by number of units produced.

Marginal average cost function refers to derivative of the average cost function i.e., $\left ( \frac{c(x)}{x} \right )'$

Given: $c(x)=\frac{3x-2}{8x+1}$

Average cost = $\frac{c(x)}{x}=\frac{3x-2}{x(8x+1)}$

At x = 50 units,

$\frac{c(50)}{50}=\frac{150-2}{50(400+1)}=\frac{148}{50(401)}=\frac{74}{10025}$

Average cost = $\frac{c(x)}{x}=\frac{3x-2}{x(8x+1)}$

Marginal average cost:

Differentiate average cost with respect to $x$

Take $f=3x-2\,,\,g=8x^2+x$

using quotient rule, $\left ( \frac{f}{g} \right )'=\frac{f'g-fg'}{g^2}$

Therefore,

$\left ( \frac{c(x)}{x} \right )'=\left ( \frac{3x-2}{8x^2+x} \right )'\\=\left ( \frac{3(8x^2+x)-(16x+1)(3x-2)}{(8x^2+x)^2} \right )\\=\frac{24x^2+3x-48x^2-3x+32x+2}{(8x^2+x)^2}\\=\frac{-24x^2+32x-2}{(8x^2+x)^2}$

3 0

2 years ago