Ask question

Ask question

All categories

2 years ago

10

The cost of 9 scarves is $87.75. What is the unit price?

1 answer:

Reil [10]2 years ago

7 0

Answer:

the answer is 9.75 for one

Step-by-step explanation:

divide 87.75/9

hope this helped

You might be interested in

I need help to answer this

Anna [14]

Subtract 59 from both sides, you get x=-179

6 0

3 years ago

Which function is greater for<br> x < 2 (for all x values less than 2) ?

kodGreya [7K]

Answer:

Function 1

Step-by-step explanation:

8 0

2 years ago

The sum of two numbers is twenty-four. The difference of the two numbers is four.

RSB [31]

Answer:

$x + y = 24$

$x - y = 4$

Now, add these two equations.

You get,

$2x = 28$

$x = \frac{28}{2}$

$x = 14$

Given,

$x + y = 24$

$14 + y = 24$

$y = 24 - 14$

$y = 10$

You can test this to the other equation as well.

$x - y = 4$

$14 - 10 = 4$

Hence, the two numbers are 14 and 10.

3 0

2 years ago

Complete the square to transform the quadratic equation into the form (x – p)2 = q. x2 - 12x - 5 = 7 A) (x - 36)2 = 9 B) (x - 6)

iren2701 [21]

x2 - 12x - 12 = 0

(x - 6)2 - 48 = 0

(x - 6)2 = 48

Hence, the answer is (B).

6 0

3 years ago

Read 2 more answers

What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove

wolverine [178]

Answer:

The solutions are:

$b=0,\:b=4$

Step-by-step explanation:

Considering the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

Solving the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

$\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2$

$5b^3-10=6b^3-4b^2-10$

$\mathrm{Switch\:sides}$

$6b^3-4b^2-10=5b^3-10$

$6b^3-4b^2-10+10=5b^3-10+10$

$6b^3-4b^2=5b^3$

$\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}$

$6b^3-4b^2-5b^3=5b^3-5b^3$

$b^3-4b^2=0$

$Using\:the\:Zero\:Factor\:Principle:$ $if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

So,

$b=0,b-4=0$

$b=0,b=4$

Therefore, the solutions are:

$b=0,\:b=4$

4 0

3 years ago

Read 2 more answers