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Mars2501 [29]
3 years ago
12

Hiii friends plzzzzz talk me i am alone​

Mathematics
2 answers:
Diano4ka-milaya [45]3 years ago
6 0

♥heyyyyyy no ur not alone!♥

nikdorinn [45]3 years ago
3 0

Answer:

Okay what do you want to talk about ?

Step-by-step explanation:

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Need help with this problem. ​
tigry1 [53]

Answer:

C) -\frac{47}{49}

Step-by-step explanation:

By the cos double angle formula, we know \cos(2\theta)=2\cos^2\theta-1. Substituting \cos \theta into the formula, we get

\cos(2\theta)=2(-\frac{1}{7})^2-1\\~~~~~~~~~~=\frac{2}{49}-1\\~~~~~~~~~~=\boxed{-\frac{47}{49}}.

The answer is C.

5 0
2 years ago
(2/5) to the 2 power
marta [7]

Answer:

0.16

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Circumference and Area questions look at the picture
Helga [31]

Note that the formula for the circumference of a circle is πd, while the formula for the area of a circle is πr².

π≈3.14

A. C=πd

Simply plug in the numbers into the formula.

Diameter=Radius*2

17*2=34

C=34(π)

B. (π)(5²)

Plug in the numbers into the formula. Remember that half of the diameter is the radius.

C.  (π)(4.5)

There are two possible formulas that could be used to calculate the circumference of a circle: πd and 2πr.

The expression above is simply multiplying the circle's radius times pi. Therefore, it is not a method that could be used to find the circumference of a circle.

D. (π)(6.5²)

Remember that the formula for calculating the area of a circle is πr².

Half of the diameter is 6.5 (13.5/2=6.5). 6.5 cm. is the radius. Now just plug the numbers into the formula.

(π)(6.5²)

Therefore, the last answer choice is the correct answer.

5 0
3 years ago
Brielle is making a mosaic picture that is ¼ foot by ¾ foot. What is the area of her mosaic?
fenix001 [56]
1/4 x 3/4 = 3/16 ft squared
6 0
3 years ago
What is the equation (simplified) the inverse of y=2x^2?
Yuki888 [10]
2x^2=y\ \ \ |divide\ both\ sides\ by\ 2\\\\x^2=\dfrac{y}{2}\\\\\boxed{x=\sqrt{\frac{y}{2}}}\\\\x=\dfrac{\sqrt{y}}{\sqrt2}=\dfrac{\sqrt{y}\cdot\sqrt2}{\sqrt2\cdot\sqrt2}\to\boxed{x=\dfrac{\sqrt{2y}}{2}}\\\\\\\boxed{f(x)=2x^2\to f^{-1}(x)=\dfrac{\sqrt{2y}}{2}}
3 0
3 years ago
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