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kondor19780726 [428]
3 years ago
6

Which type of transformation is shown below?

Mathematics
2 answers:
Lina20 [59]3 years ago
7 0

Answer:

C.) A dilation. It got bigger so its a dilation

damaskus [11]3 years ago
6 0
Translation is your answer
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Chloe wants to buy monitor for her bedroom the height of the monitor is 18 inches in the width is 16 inches calculate the diagon
Akimi4 [234]
Equation: a^2 + b^2 = c^2

18^2 + 16^2 = c^2

324 + 256 = c^2

C^2 = square root 580

Square root 580 reduces to 2 square root 145.

As a decimal that is: 24.08

Rounded to the nearest tenth: 24 inches.

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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
Naily [24]

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

As

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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