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il63 [147K]
2 years ago
6

Find an angle between 0 and 2π that is coterminal with 10pi/3

Mathematics
1 answer:
azamat2 years ago
8 0

In this question we have to find an angle between 0 and 2π that is coterminal with 10pi/3 .

To find the coterminal angle, we need to subtract 2 pi from 10 pi/3. That is

= \frac{10 \pi }{3} - 2 \pi

Multiplying numerator and denominator of 2 pi by 3

= \frac{10 \pi}{3} -\frac{6 \pi}{3} = \frac{4 \pi}{3}

And that's the required coterminal angle between 0 and 2pi.

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Let a be a rational number and b be an irrational number. Which of the following are true statements?(there is more than 1 answe
kramer
<span>A.) the sum of a and b is never rational.
This is a true statement. Since an irrational umber has a decimal part that is infinite and non-periodical, when you add a rational number to an irrational number, the result will have the same infinite non periodical decimal part, so the new number will be irrational as well.

</span><span>B.) The product of a and b is rational
This one is false. Zero is a rational number, and when you multiply an irrational number by zero, the result is always zero.

</span><span>C.) b^2 is sometimes rational
This one is true. When you square an irrational number that comes from a square root like </span>\sqrt{2}, you will end with a rational number: ( \sqrt{2} )^{2}=2, but, if you square rationals from different roots than square root like \sqrt[3]{2}, you will end with an irrational number: \sqrt[3]{2^{2} } = \sqrt[3]{2}. 

<span>D.) a^2 is always rational
This one is false. If you square a rational number, you will always end with another rational number.

</span><span>E.) square root of a is never rational
</span>This one is false. The square root of perfect squares are always rational numbers: \sqrt{64} =8, \sqrt{16} =4,...

F.) square root of b is never rational
This one is true. Since the square root of any non-perfect square number is irrational, and all the irrational numbers are non-perfect squares, the square root of an irrational number is always irrational.

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4 0
2 years ago
Can anyone plz help me
yaroslaw [1]

Answer:

They multiplied the exponents instead of adding.

Step-by-step explanation:

Essentially, 2^2 x 2^3 is (2)(2) x (2)(2)(2) which is 5, not 6

8 0
3 years ago
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Step-by-step explanation:

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Answer:

Midpoint of AB              = (0 + 2a / 2 , 0 + 0 / 2) = (2a / 2 , 0 / 2) = (a,0)

x coordinate of point c = a

                                  N = (0 + a / 2 , 0 + b / 2) = (a / 2 , b / 2)

                                  M = ( 2a + a / 2 , 0 + b / 2) = (3a / 2 , b / 2)

                                MA = √(3a / 2 - 0)² + b / 2 - 0)²

                                       = √(3a / 2 )² + (b / 2) = 9a² / 4 + b² / 4

                                NB = √(a / 2 - 2a)² + (b / 2 - 0 )²

                                      = √( a / 2 - 4a / 2)² + (b / 2 - 0)²

                                      = √(-3a / 2)² + (b / 2)² = √9a² / 4 + b² / 4

Step-by-step explanation:

I tried my best hope its correct :0

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