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gregori [183]
3 years ago
12

A circle with radius of 4cm sits inside a circle with radius of 11cm.

Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
6 0

Answer:

Area of shaded region = 329.70 cm²

Step-by-step explanation:

Area of a circle = πr²

Area of bigger circle = 3.14 × 11²

= 3.14 × 11 × 11

= 379.94 cm²

Area of smaller circle = 3.14 * 4²

= 3.14 × 4 × 4

= 50.24 cm²

Area of shaded region

= Area of bigger circle - Area of smaller circle

= (379.94 - 50.24) cm²

= 329.70 cm²

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A pedestal in a craft store is in the shape
Angelina_Jolie [31]
Given:
side lengths of right triangles = 12 cm ; 16 cm ; 20 cm
lateral area = 192 cm²

lateral area of a triangular prism = perimeter * height
192 cm² = (12 cm + 16 cm + 20 cm) * height
192 cm² = 48 cm * height
192 cm² / 48 cm = height
4 cm = height

The height of the pedestal in the shape of a triangular prism is 4 cm.

4 0
3 years ago
Read 2 more answers
I don't really get it ​
blsea [12.9K]

I can't see the whole pic...

what is written on the side?? I can see part of it and I think it is telling you what to do. If not what is thew question?? All you are showing is the lines.

3 0
3 years ago
Chatty made this square picture. She wants to glue yarn around the border to make a frame. How much yarn does she need?
Romashka [77]

Answer:

60 inches

Step-by-step explanation:

s = 15 in

Perimeter of a Square = 4 s

                                     = 4 × 15

                                     = 60 in

∴Chatty needs 60 inches of yarn.

Hope that helps!!

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7 0
3 years ago
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5x-y=1/4 intercept form
MAVERICK [17]
5x - y = 1/4; this equation is written in slope - intercept form

Slope - Intercept form formula: y = mx + b

-Move '5x' to the left side of the equation
-Move 'y' to the right side of the equation

y = 5x - 1/4
3 0
3 years ago
Can I get some help, it is due in 15 minutes and I need help. Thanks, any help appreciated
koban [17]

Well, I'm way past the 15 min mark, but here's how to do the question.


With this, you will need to use the distance formula, \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}, on XY, YZ, and ZX.



XY: \sqrt{(3-1)^2+(1-6)^2}


Firstly, solve inside the parentheses: \sqrt{(2)^2+(-5)^2}


Next, solve the exponents: \sqrt{4+25}


Next, solve the addition, and XY's distance will be √29



(The process is the same with the other 2 sides, so I'll go through them real quickly)


YZ:

\sqrt{(6-3)^2+(3-1)^2}\\ \sqrt{(3)^2+(2)^2}\\ \sqrt{9+4}\\ \sqrt{13}



ZX:

\sqrt{(1-6)^2+(6-3)^2}\\ \sqrt{(-5)^2+(3)^2}\\ \sqrt{25+9}\\ \sqrt{34}



Now that we got the 3 sides, we can add them up: \sqrt{29}+\sqrt{13} +\sqrt{34} =14.8


In short, your answer is 14.8, or the second option.

8 0
3 years ago
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