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3 years ago

What are the exact values of sin(2π/3radians) and cos(2π/3radians)?

1 answer:

6 0

Sin(2π/3radians) = sin(π/3radians) = (√3)/2cos(2π/3radians) = - cos(π/3radians) = 1/2

becuase 2π/3 + π/3 = π

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Hi, thx for helping

Sliva [168]

Answer:

C. 8.52624...

Step-by-step explanation:

The reason that C is different is because all of the other decimals are terminating decimals (meaning that they end or stop), while C is a repeating decimal, because it never ends.

7 0

4 years ago

Expanded -8(-2x-5) please give answer

nika2105 [10]

Answer:

-80

Step-by-step explanation:

We have to do the parentheses first because it's part of PEDMAS.

-2x-5= 10

Next, we take 10 times -8

-8 x 10= -80

There you go! Hope this helps!

8 0

3 years ago

Read 2 more answers

After a 5% discount, a telephone bill is $79.50. How much was the bill originally?

ICE Princess25 [194]

The original bill is $ 83.475

7 0

4 years ago

6x - 2y = 10

pishuonlain [190]

Answer:

y = 3x -5

Step-by-step explanation:

6x - 2y = 10

-6x. -6x

Inverse Operations

-2y = -6x + 10

÷-2. ÷-2. ÷-2

y = 3x - 5

5 0

3 years ago

Read 2 more answers

Find the length of midsegment YZ in trapezoid ABCD when A(-5,-6) B(-6,-2) C(-4,0) and D(3,2)

Luda [366]

Answer:

Option C

Step-by-step explanation:

Since, Y and Z are the midpoints of sides AB and CD of the given trapezoid.

Segment YZ will the midsegment of trapezoid ABCD.

By the theorem of midsegment,

m(YZ) = $\frac{1}{2}(AD+BC)$

By using expression for the length of a segment between two points,

Length of a segment = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Distance between two points A(-5, -6) and D(3, 2),

AD = $\sqrt{(3+5)^2+(2+6)^2}$

AD = $\sqrt{64+64}$

AD = $\sqrt{128}$

AD = $8\sqrt{2}$

Distance between B(-6, -2) and C(-4, 0)

BC = $\sqrt{(-6+4)^2+(-2-0)^2}$

BC = $\sqrt{8}$

BC = $2\sqrt{2}$

Therefore, m(YZ) = $\frac{1}{2}(8\sqrt{2}+2\sqrt{2})$

= $\frac{1}{2} (10\sqrt{2})$

= $5\sqrt{2}$

Option C will be the answer.

7 0

3 years ago