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Firdavs [7]
3 years ago
13

Surface area hw help pls

Mathematics
1 answer:
Natali [406]3 years ago
8 0

Answer:

Find the perimeter first then the are

Step-by-step explanation:

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Find the area of the shape
Tamiku [17]

Answer:

1296 sorry if im wrong

Step-by-step explanation:

:):):):):):):):):):):):):):):):):):):):):):):)

6 0
3 years ago
Solve the inequality 2x+10<-28
Otrada [13]

Answer:

x < -19

Step-by-step explanation:

2x+10<-28

Subtract 10 from each side

2x+10-10<-28-10

2x< -38

Divide each side by 2

2x/2 <-38/2

x < -19

8 0
4 years ago
What is the equation of the line that passes through the points (15, 9) and (Negative 2, 9)? y =
Alja [10]

Answer:

Y=9

Step-by-step explanation:

Because it passes through two points in a straight line, it must have no coefficient for x, just a Y-Intercept equal to the Y of the two points.

7 0
3 years ago
Lucia draws a square and plots the center of the square. (image)
ss7ja [257]

Answer:

C

Step-by-step explanation:

Lucia's claim is correct since any rotation that is a multiple of 45° carries a square onto itself

8 0
3 years ago
Graph the image of the figure after a dilation with a scale factor of 1/3 centered at (1, −2) . Use the polygon tool to graph th
brilliants [131]

Answer:

New image will have (2,1) , (-2,0) and (-2,-2) as new vertex.

Below is the graph.

Step-by-step explanation:

Given the triangle with points (4,7) , (-8,4) and (-8,-2) having dilation center at (1,-2) .

We have to dilate using a factor of 1/3 and graph the new image.

Let us find all the three points,

Formula,

New point = [old point - dilation center]* scale factor + dilation center

So,

For (4,7)

[(4,7)- (1,-2)]*\frac{1}{3} + (1,-2)

[3,9]*\frac{1}{3} +(1,-2)

[1,3]+(1,-2)= (2,1)

For (-8,4)

[(-8,4)- (1,-2)]*\frac{1}{3} + (1,-2)

[-9,6]*\frac{1}{3} +(1,-2)

[-3,2]+(1,-2)= (-2,0)

For (-8,-2)

[(-8,-2)- (1,-2)]*\frac{1}{3} + (1,-2)

[-9,0]*\frac{1}{3} +(1,-2)

[-3,0]+(1,-2)= (-2,-2)

Therefore these (2,1) , (-2,0) and (-2,-2) are the new vertices after dilation.

4 0
4 years ago
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