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____ [38]
4 years ago
6

How do you find the surface area of a circle?

Mathematics
1 answer:
Blizzard [7]4 years ago
4 0

Answer:

πr²

Step-by-step explanation:

The surface area of a circle is πr²

Pie is 22/7 or 3.14

While the radius of a circle is any line from the centre of a circle to any part of the circumference of the circle. A diameter is any line that cuts the circle into two equal halves, cutting the circle from the centre.

To get the surface area of a circle, you draw a line from the centre of the circle, and then touch any part of the circumference of the circle. Then you measure the line, that's the radius.

If for instance, the radius gives 7cm, you find the Square, And then multiply it with pie.

22/7 X 7 ²

22/7 X 7 X 7

22 X 7

154cm²

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H Quarshie
Yuki888 [10]

Answer:

28

Step-by-step explanation:

f(6) = -3 * 6 + 10 = -18 + 10 = -8

g(f(6)) = g(-8)

g(-8) = (-8)^{2} + 3 * (-8) - 12 = 64 - 24 - 12 = 64 - 36 = 28

6 0
4 years ago
How I do this someone ples help I need help with 25 and 27.
marta [7]

Number 15 is no solution.

Number 27 is that x has to be between -2 or 2.

This could be explained as -2≤x≤2

8 0
4 years ago
Solve for x : 2x^2+4x-16=0
alekssr [168]
2x^2+4x-16=0\ \ \ /:2\\\\x^2+2x-8=0\\\\\underbrace{x^2+2x\cdot1+1^2}_{(*)}-1^2-8=0\\\\(x+1)^2-1-8=0\\\\(x+1)^2-9=0\\\\(x+1)^2=9\iff x+1=-3\ or\ x+1=3\\\\x=-3-1\ or\ x=3-1\\\\x=-4\ or\ x=2\\\\\\(*)\ (a+b)^2=a^2+2ab+b^2



2x^2+4x-16=0\ \ \ /:2\\\\x^2+2x-8=0\\\\a=1;\ b=2;\ c=-8\\\\\Delta=b^2-4ac\ if\ \Delta > 0\ then\ x_1=\frac{-b-\sqrt\Delta}{2a}\ and\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\Delta=2^2-4\cdot1\cdot(-8)=4+32=36;\ \sqrt\Delta=\sqrt{36}=6\\\\x_1=\frac{-2-6}{2\cdot1}=\frac{-8}{2}=-4;\ x_2=\frac{-2+6}{2\cdot1}=\frac{4}{2}=2
3 0
3 years ago
Read 2 more answers
Ayo could anyone please help with this<br><br><br><br><br><br><br><br><br><br>​
mestny [16]
The answer should be answer choice B have a good day! brainleist;) ?
7 0
3 years ago
Please help will mark brainliest
nataly862011 [7]
I think it’s A. If not, then B.
5 0
4 years ago
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