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Inessa [10]
3 years ago
8

Area of solid shapes ​

Mathematics
1 answer:
daser333 [38]3 years ago
5 0

Answer:

Surface-area. If a solid is composed of flat surfaces, such as the cube on the right, the surface area is simply the sum of the areas of the flat surfaces (called faces). So, for example, if a each edge of a cube has a length s, the area of one face is s2 since it is a square.

Step-by-step explanation:

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Use the given numbers above the table in multiplying integers and to fill in the blanks inside the box.​
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Step-by-step explanation:

1 × 5 = 5

(- 2) × (- 3) = 6

4 × (- 6) = - 24

I hope I've helped you.

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Write 230% in simplest form
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Lin runs 5 laps around a track in ten minutes how many laps per minute
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Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (1, 5), F (3, 5), a
hram777 [196]

Answer:

(1,1)

Step-by-step explanation:

Given: E, F, G, H denote the three coordinates of the area fenced

To find: coordinates of point H

Solution:

According to distance formula,

length of side joining points (x_1,y_1)\,,\,(x_2,y_2) is equal to \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

So,

EF=\sqrt{(3-1)^2+(5-5)^2}=2\,\,units\\FG=\sqrt{(6-3)^2+(1-5)^2}=\sqrt{9+16}=5\,\,units\\GH=\sqrt{(x-6)^2+(y-1)^2}\\EH=\sqrt{(x-1)^2+(y-5)^2}

Perimeter of a figure is the length of its outline.

EF+FG+GH+EH=16\\2+5+\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16-2-5\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9

Put (x,y)=(1,1)

\sqrt{(1-6)^2+(1-1)^2}+\sqrt{(1-1)^2+(1-5)^2}=9\\\sqrt{25}+\sqrt{16}=9\\5+4=9\\9=9

This is true.

So, the point (1,1) satisfies the equation \sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9

So, point H is (1,1).

7 0
3 years ago
Line A has a slope of -
Yuki888 [10]

Answer:

Graphing

Step-by-step explanation:

Refer to the image below!

3 0
3 years ago
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