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Andrew [12]
2 years ago
12

Which of the following could be the ratio between the lengths of the two legs

Mathematics
1 answer:
aliya0001 [1]2 years ago
7 0
B c and f because each can go into each one
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A square piece of paper 10 cm on a side is rolled to form the lateral surface area of a cylinder and then a top and bottom are a
Anika [276]

Answer:

The surface = pi*10cm2

Step-by-step explanation:

The area of any rectangle is Base *  Height,

We know that the square originally has 10cm High, when is rolled to form a cylinder, it's new shape has a bottom side that is equals to a circle.

The length of a circle is 2*pi*Radio= pi*Diameter.

Since the square has the same length in each side, we know that pi*Diameter= 10cm*pi

The surface (area) of the cylinder is= Base * Height= 10cm*10cm*pi

 

6 0
2 years ago
How do you find the area of a circle equation easy? (Also to find points on circle using the equation)
baherus [9]

the area of a circle is A = πr², where r = radius of the circle.

how to find points in the circle? depends on what the scenario is, if you know circle's center and its radius, simply use the distance formula to get any of the points, if you have only the equation of it in standard form, you can use the x,y coordinates with substitution, if you have an sketch, you can get them off the grid, so depends on what you have as components.

8 0
3 years ago
Help me please!!! :(
lana66690 [7]

Answer:

the greatest common factor is 5^2 * 7^3

(I think!!!)

3 0
3 years ago
What addition statement does this show?
Anuta_ua [19.1K]

Answer:

D. –5.2 + (–6.4) = –11.6

Step-by-step explanation:

First, you can eliminate A and C because 5.2 is positive. Since the red arrow is pointing left, it indicates that the number (-5.2) is negative.

Then, you look at the second arrow, the green one. It is pointing left, so it would be -6.4.

I hope this helps :)

3 0
2 years ago
Help asap will give brainliest to whoever answer's correctly.
nydimaria [60]

Answer:

cant see the image

Step-by-step explanation:

8 0
2 years ago
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