If jasmine spent $280 on 20 pairs of earrings,what was the cost of 1 pair?

Mathematics

1 answer:

strojnjashka [21]1 year ago

7 0

Answer:

Step-by-step explanation:

280 divided by 20 is 14 when you divide by those 2 numbers you get how much they would cost. also 14x20 is 280

You might be interested in

Find the slope of the line that passes through (10, 8) and (5, 4).

GarryVolchara [31]

Answer:

M=8-4 / 10-5

M= 4/5

Step-by-step explanation:

3 0

2 years ago

Darin skateboarded to the west at 8 m/s. In 7 seconds, he slowed down to 3

MakcuM [25]

Answer:

C) -0.71m/s²

Step-by-step explanation:

V = V0 + at

V = final veloctiy

V0 = Initial velocity

a = acceleration

t = time

3 = 9 + a7

-5 = a7

$-\frac{5}{7}$ = a

a ≅ -0,71

3 0

2 years ago

WHAT IS THE EQUATION OF THE LINE IN SLOPE - INTERCEPT FORM

Maksim231197 [3]

Y=10x-40
i hope that helps!!

5 0

2 years ago

Solve. 1,032 divided by 24 Enter the answer in the box.<br> Response area with 1 text input box

AlexFokin [52]

Answer:

$258\quad \mathrm{Remainder}\quad \:0$

Step-by-step explanation:

$\frac{1032}{4} \\\\\begin{matrix}4\overline{|\smallspace1032}\space\space\space\space\space\space\space\space\space\space\space\space\end{matrix}\\\\\mathrm{Divide}\:10\:\mathrm{by}\:4\:\mathrm{to\:get}\:2\\\begin{matrix}\space\space\space\space\space\space\emptyspace2\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\\ 4\overline{|\smallspace1032}\space\space\space\space\space\space\space\space\space\space\space\space\end{matrix}\\\\$

$\mathrm{Multiply\:the\:quotient\:digit}\:\left(2\right)\:\mathrm{by\:the\:divisor}\:4\\\begin{matrix}\space\space\space\space\space\space\emptyspace2\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\\4\overline{|\smallspace1032}\space\space\space\space\space\space\space\space\space\space\space\space\\ \space\space\space\space\space\space\underline{\emptyspace8}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\end$