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Semenov [28]
3 years ago
13

Please help, I need to find the area, circumference, and radius

Mathematics
1 answer:
Contact [7]3 years ago
4 0

... What do I do? Please help  Please help, I need to find the area, circumference, and radius

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Thirteen divided by 1,690
enyata [817]

Answer:

130

Step-by-step explanation:

13 goes into 16 how many times - 1

subtract- you get 390

13 goes into 390 how many times?- 30

Your answer - 130

7 0
2 years ago
Can anyone help with this math problem?
finlep [7]

ok hi im here to help not rlly just want my points up your useless

8 0
2 years ago
Please help, I will give Brainliest!!! :D​
anygoal [31]

Step-by-step explanation:

Firstly, notice that this shape is composed of ³/₄ of a circle and ¹/₂ of a square;

The lower left part is the ¹/₂ square, with only 2 sides (a left and lower side) of length 3 units;

The perimeter of this part of the shape is simply the sum of the two sides:

3 + 3 = 6

The remaining part is the ³/₄ circle, which has a radius of 3 units;

The circumference of a circle is found by the formula:

πd

d = diameter = 2 × radius

We only have ³/₄ of the circle, however, so we only have ³/₄ the circumference:

³/₄ × π(2(3)) = ³/₄ × 6π

= ⁹/₂π (or, equally, 4.5π)

So the total perimeter is the sum of the perimeter of the ¹/₂ square and the ³/₄ circle:

4.5π + 6

4 0
2 years ago
What is the answer to this 10+4(2+20)
Bogdan [553]

Answer:

98

Step-by-step explanation:

PEMDAS: Parenthesis, Exponents, Multiply, Divide, Add, Subtract, if tied, go left to right.

Elementary Algebra rhyme you should consider a requirement to remember because it's the base of a vast majority of problem's you'll encounter.

10+4(2+20)

10+4(22). . .distribute the 4 into the parenthesis aka multiply 22*4

10+88 = 98

7 0
3 years ago
Read 2 more answers
Help guys with the question
Naya [18.7K]

Answer:

\sin\Big(\dfrac{\theta }{2}\Big) = \dfrac{2\sqrt 5}{5}

Step-by-step explanation:

\cos(2x) = \cos^2 x-\sin^2 x = 1-2\sin^2 x \\ \\ \cos(x) = 1-2\sin^2 (\frac{x}{2}) \\ \\ \Rightarrow \sin^2 (\frac{x}{2}) = \dfrac{1-\cos(x)}{2}\\ \\ \sin(\frac{x}{2}) = \pm \sqrt{\dfrac{1-\cos(x)}{2}},\quad x\in [\frac{3\pi }{2},\pi] \Rightarrow \frac{x}{2}\in [\frac{3\pi}{4},\frac{\pi}{2}]\\ \\ \Rightarrow \sin(\frac{x}{2}) > 0 \Rightarrow \sin(\frac{x}{2}) = \sqrt{\dfrac{1-(-\frac{3}{5})}{2}} \Rightarrow \sin(\frac{x}{2}) = \sqrt{\dfrac{8}{10}}=\dfrac{2\sqrt 2}{\sqrt{10}} = \\ \\ =\dfrac{2\sqrt 5}{5}

7 0
3 years ago
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