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3 years ago

A circle has a circumference of 50.2450.24 units. What is the radius of the circle?

Mathematics

1 answer:

Roman55 [17]3 years ago

8 0

The formula for circumference is 2 x PI x r

Using the circumference we can find the radius:

50.24 = 2 x PI x r

Multiply 2 by PI:

2 x 3.14 = 6.28

50.24 = 6.28 x r

Divide both sides by 6.28:

R = 50.24 / 6.28

R = 8

The radius is 8 units.

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Solve the following and explain your steps. Leave your answer in base-exponent form. (3^-2*4^-5*5^0)^-3*(4^-4/3^3)*3^3 please st

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Answer:

$\boxed{2^{\frac{802}{27}} \cdot 3^9}$

Step-by-step explanation:

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$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

$\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }$

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

$\left(\dfrac{1}{3^2} \cdot \dfrac{1}{4^5} \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3$

Also,

$\boxed{a^{0} = 1, a\neq 0 }$

$\left(\dfrac{1}{9} \cdot \dfrac{1}{1024} \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

Note

$\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq 0}$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

Note

$\boxed{\dfrac{1}{\dfrac{1}{a} } = a}$

$9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)$

Once

$9216=2^{10}\cdot 3^2 \implies 9216^3=2^{30}\cdot 3^6$

$\boxed{(a \cdot b)^n=a^n \cdot b^n}$

And

$4^{\frac{4}{27}} = 2^{\frac{8}{27}$

We have

$\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)$

Also, once

$\boxed{\dfrac{c^a}{c^b}=c^{a-b}}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27$

$30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27} =\dfrac{802}{27}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3$

$2^{\frac{802}{27}} \cdot 3^9$

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Wich relation is a function of x

faust18 [17]

Answer:

the first one

Step-by-step explanation:

In the other 2 i can see the x repeats itself meaning it isn't a function.

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3 years ago

Use the function g(x)-x+2 with initial values of 4, 2, and 1. What happens

kolezko [41]

Answer:

Step-by-step explanation:

The given function is , g(x)= - x + 2

For values of , is the sequence, -2,4,-2,4,-2,4........

For values of , is the sequence, 0,2,0,2........

For values of , is the sequence, 1,1,1,1,........

For initial value, of 4,the sequence converges to two values that is -2, 4.

For initial value of 2, the sequence again converges to two values which are 0,2.

For initial value of 1, the sequence converges to one value which is 1.

→→First Iterative value that is (-2)=Second iterative value (0) + 2 ×Third iterative value(1)

→→ First Iterative value that is (4)=Second iterative value (2) + 2 ×Third iterative value(1)

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4 years ago