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3 years ago

6

Karana's mom bought her two items using a credit card. The first item cost $235, and the second item cost $369. Later, she retur

ned both items to the store. What was the change in the amount owed to the credit card company from returning both items?
A.-$134
B.$134
C.$604
D.-$604

1 answer:

mamaluj [8]3 years ago

8 0

Answer:

C. 604

Step-by-step explanation:

Karana returned the items so you are adding the money back to the credit card which would be 604. 369 + 235 = 604

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a circle pizza with a diameter of 10 inches. The pizza is in a square box with side lengths of 14 inches. In square inches, how

julsineya [31]

Answer:

117.45 in^2

Step-by-step explanation:

Given data

Diameter of pizza= 10in

Radius of pizza=10/2= 5in

Lenght of pizza box= 14in

Area of pizza= πr^2

A= 3.142*5^2

A= 3.142*25

A= 78.55 in^2

Area of pizza box= L^2

A= 14^2

A= 196 in^2

Hence the empty space is

=196-78.55

=117.45 in^2

5 0

4 years ago

Can you please help me

Ket [755]

Answer:

The scale factor for the dilation of rectangle ABCD to rectangle MNOP is $\frac{3}{4}$ ⇒ B

Step-by-step explanation:

In similar rectangles, their corresponding dimensions are proportional, which means $\frac{L_{1} }{L_{2}}$ = $\frac{W_{1} }{W_{2}}$ , where L is the length and W is the width

∵ Rectangle ABCD is similar to rectangle MNOP

∴ Their dimensions are proportional

∵ The dimensions of rectangle ABCD are 6 m, 14 m

∴ L $_{1}$ = 14 and W $_{1}$ = 6

∵ The dimensions of rectangle MNOP are 4.5 m, 10.5 m

∴ L $_{2}$ = 10.5 and W $_{2}$ = 4.5

∵ The rectangle MNOP is the image of rectangle ABCD after dilation

→ To find the scale factor of dilation find the ratio between the

corresponding dimensions in the two rectangles (image/pre-image)

∵ $\frac{L_{2} }{L_{1}}$ = $\frac{10.5}{14}$ = $\frac{3}{4}$

∵ $\frac{W_{2} }{W_{1}}$ = $\frac{4.5}{6}$ = $\frac{3}{4}$

∴ The scale factor for the dilation is $\frac{3}{4}$

The scale factor for the dilation of rectangle ABCD to rectangle MNOP is $\frac{3}{4}$

6 0

3 years ago

Aneli [31]

$\frac{8x}{7}$ +9=30
$\frac{8x}{7}$ =21 {subtract 9 from both sides}
8x=147 {multiply both sides by 7 to cancel out the division (fraction)}
x=18.375 {divide both sides by 8}
The answer is A) 18 3/8

5 0

4 years ago

A rectangular school yard has a perimeter of 114 meters and an area of 702 square meters. What are the dimensions of the school

Lostsunrise [7]

18 by 39.

Reason:
2x+2y=114 (perimeter)
XY=702 (area)
X=18
Y=39

6 0

3 years ago

Perimeter the coordinates of the vertices of a quadrilateral are R(-1,3), S(3,3), T(5,-1), and U(-2,-1)

Crank

Answer:

The perimeter is equal to $P=(11+2\sqrt{5}+\sqrt{17})\ units$ or $P=19.59\ units$

Step-by-step explanation:

we have

The coordinates of the vertices are

R(-1,3), S(3,3), T(5,-1), and U(-2,-1)

plot the figure to better understand the problem

see the attached figure

we know that

The perimeter of a quadrilateral is the sum of its four length sides

so

$P=RS+ST+TU+UR$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the distance RS

$R(-1,3)\\S(3,3)$

substitute the values

$d=\sqrt{(3-3)^{2}+(3+1)^{2}}$

$d=\sqrt{(0)^{2}+(4)^{2}}$

$RS=4\ units$

step 2

Find the distance ST

$S(3,3)\\T(5,-1)$

substitute the values

$d=\sqrt{(-1-3)^{2}+(5-3)^{2}}$

$d=\sqrt{(-4)^{2}+(2)^{2}}$

$ST=2\sqrt{5}\ units$

step 3

Find the distance TU

$T(5,-1)\\U(-2,-1)$

substitute the values

$d=\sqrt{(-1+1)^{2}+(-2-5)^{2}}$

$d=\sqrt{(0)^{2}+(-7)^{2}}$

$TU=7\ units$

step 4

Find the distance UR

$U(-2,-1)\\R(-1,3)$

substitute the values

$d=\sqrt{(3+1)^{2}+(-1+2)^{2}}$

$d=\sqrt{(4)^{2}+(1)^{2}}$

$UR=\sqrt{17}\ units$

step 5

Find the perimeter

$P=RS+ST+TU+UR$

substitute the values

$P=4+2\sqrt{5}+7+\sqrt{17}$

$P=(11+2\sqrt{5}+\sqrt{17})\ units$ -----> exact value

$P=(11+4.47+4.12)=19.59\ units$ -----> approximate value

5 0

3 years ago