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3 years ago

8

A toy has been marked up by 75%. The tag lists the price as $63. Find the wholesale price.

1 answer:

Lady bird [3.3K]3 years ago

3 0

Answer:

110.25

Step-by-step explanation:

You do 1.75 because it is a mark up and ur paying 175 percent

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SOLVE EQUATION BY TAKING SQUARE ROOTS.<br> -100+81a^2=0

IceJOKER [234]

Answer:

a = ±10/9

Step-by-step explanation:

$-100+81a^2=0\\81a^2=100\\a^2=\frac{100}{81}\\\sqrt{a^2}=\sqrt{\frac{100}{81} }\\a=10/9\\a=-10/9$

7 0

3 years ago

Read 2 more answers

Find the integral, using techniques from this or the previous chapter.<br> ∫x(8-x)3/2 dx

Soloha48 [4]

Answer:

$\int x(8-x)^{3/2}dx= -\frac{16}{5} (8-x)^{\frac{5}{2}} +\frac{2}{7} (8-x)^{\frac{7}{2}} +C$

Step-by-step explanation:

For this case we need to find the following integral:

$\int x(8-x)^{3/2}dx$

And for this case we can use the substitution $u = 8-x$ from here we see that $du = -dx$ , and if we solve for x we got $x = 8-u$ , so then we can rewrite the integral like this:

$\int x(8-x)^{3/2}dx= \int (8-u) u^{3/2} (-du)$

And if we distribute the exponents we have this:

$\int x(8-x)^{3/2}dx= - \int 8 u^{3/2} + \int u^{5/2} du$

Now we can do the integrals one by one:

$\int x(8-x)^{3/2}dx= -8 \frac{u^{5/2}}{\frac{5}{2}} + \frac{u^{7/2}}{\frac{7}{2}} +C$

And reordering the terms we have"

$\int x(8-x)^{3/2}dx= -\frac{16}{5} u^{\frac{5}{2}} +\frac{2}{7} u^{\frac{7}{2}} +C$

And rewriting in terms of x we got:

$\int x(8-x)^{3/2}dx= -\frac{16}{5} (8-x)^{\frac{5}{2}} +\frac{2}{7} (8-x)^{\frac{7}{2}} +C$

And that would be our final answer.

8 0

3 years ago

angelo earns 2080 each month his total deductions are 30% of his pay how much is deducted from his pay each month

Elden [556K]

624 is how much is deducted from his pay every month

5 0

3 years ago

If twice a number, added to 3 is divided by the number plus 7, the result is three halves. Find the number.

morpeh [17]

The answer to the question

6 0

4 years ago

Find the product of all real values of r for which 1/2x=r-x/7

Dahasolnce [82]

Answer:

$r = \±\sqrt{14$

$Product = -14$

Step-by-step explanation:

Given

$\frac{1}{2x} = \frac{r - x}{7}$

Required

Find all product of real values that satisfy the equation

$\frac{1}{2x} = \frac{r - x}{7}$

Cross multiply:

$2x(r - x) = 7 * 1$

$2xr - 2x^2 = 7$

Subtract 7 from both sides

$2xr - 2x^2 -7= 7 -7$

$2xr - 2x^2 -7= 0$

Reorder

$- 2x^2+ 2xr -7= 0$

Multiply through by -1

$2x^2 - 2xr +7= 0$

The above represents a quadratic equation and as such could take either of the following conditions.

(1) No real roots:

This possibility does not apply in this case as such, would not be considered.

(2) One real root

This is true if

$b^2 - 4ac = 0$

For a quadratic equation

$ax^2 + bx + c = 0$

By comparison with $2x^2 - 2xr +7= 0$

$a = 2$

$b = -2r$

$c =7$

Substitute these values in $b^2 - 4ac = 0$

$(-2r)^2 - 4 * 2 * 7 = 0$

$4r^2 - 56 = 0$

Add 56 to both sides

$4r^2 - 56 + 56= 0 + 56$

$4r^2 = 56$

Divide through by 4

$r^2 = 14$

Take square roots

$\sqrt{r^2} = \±\sqrt{14$

$r = \±\sqrt{14$

Hence, the possible values of r are:

$\sqrt{14$ or $-\sqrt{14$

and the product is:

$Product = \sqrt{14} * -\sqrt{14}$

$Product = -14$

8 0

3 years ago