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Orlov [11]
3 years ago
5

What is the circumference of a circle

Mathematics
1 answer:
borishaifa [10]3 years ago
5 0
C=2πr

Hoped i helped BYEEEEEEEEEEEEEEEEEEEEEEEEE
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Solve the system of linea<br> 1. y=x<br> y=2x - 11
aksik [14]

\left \{ {{y=x} \atop {y=2x-11}} \right. \\x=2x-11\\x=11\\\\y=x\\y=11\\\\\\Answer: (11;11)

P.S. Hello from Russia :^)

7 0
3 years ago
Need help! Not sure what exactly is going on here!
marishachu [46]

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idk

Step-by-step explanation:

4 0
3 years ago
Please help IM OFFERING ALOT OF POINTS !!!!
VMariaS [17]

Answer:

cos 2Ф = - 161/289 , tan 2Ф = - 240/161

Step-by-step explanation:

* Lets explain how to solve the problem

∵ cos Ф = - 8/17

∵ Ф lies in the 3rd quadrant

- In the 3rd quadrant sin and cos are negative values, but tan is

 a positive value

∵ sin²Ф + cos²Ф = 1

∴ sin²Ф + (-8/17)² = 1

∴ sin²Ф + 64/289 = 1

- Subtract 64/289 from both sides

∴ sin²Ф = 225/289 ⇒ take √ for both sides

∴ sin Ф = ± 15/17

∵ Ф lies in the 3rd quadrant

∴ sin Ф = -15/17

∵ cos 2Ф = 2cos²Ф - 1 ⇒ the rule of the double angle

∵ cos Ф = - 8/17

∴ cos 2Ф = 2(-8/17)² - 1 = (128/289) - 1 = - 161/289

* cos 2Ф = - 161/289

∵ tan 2Ф = sin 2Ф/cos 2Ф

∵ sin 2Ф = 2 sin Ф × cos Ф

∵ sin Ф = - 15/17 and cos Ф = - 8/17

∴ sin 2Ф = 2 × (-15/17) × (-8/17) = 240/289

∵ cos 2Ф = - 161/289

∴ tan 2Ф = (240/289)/(-161/289) = - 240/161

* tan 2Ф = - 240/161

3 0
3 years ago
Y varies directly with the square of X and when Y= 80, X=4
schepotkina [342]

Answer:

Whats the question?

Step-by-step explanation:

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5 0
3 years ago
For every 6 students who purchase a bag of M&amp;M’s, there are 5 students who purchase a pack of gum. Use M for the number of s
Assoli18 [71]

Answer:

  M/6 = G/5

Step-by-step explanation:

We are given a relation between the numbers of students in two different groups. That relation can be used to write an equation.

<h3>Groups who bought M&Ms</h3>

If we consider M&M buyers to be 6 in a group, then the number of those groups is ...

  M/6

<h3>Groups who bought gum</h3>

Similarly, the number of groups who bought gum will be ...

  G/5

where there are 5 gum-buyers in each group.

<h3>Equation</h3>

The problem statement tells us that for each group of one kind, there is a matching group of the other kind. That is, the numbers of groups are equal:

  M/6 = G/5

4 0
2 years ago
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