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zalisa [80]
2 years ago
15

Need help with this problem on my hw

Mathematics
1 answer:
Andrew [12]2 years ago
6 0

Answer:

157

Step-by-step explanation:

C = \pi x diameter.

C = 3.14 x 50

C = 157 units

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3 years ago
Anne is currently h years old. bill is currently 2h years old and Charles is currently eight years old. find an expression for e
Margarita [4]

Ann's age after h years is 2h

Bill's age after h years is 3h

Charles' age after h years is 8 + h

The sum of their ages after h years is 6h + 8

Step-by-step explanation:

The given is:

Anne is currently h years old

Bill is currently 2h years old

Charles is currently eight years old

We need to find an expression for each person’s age after h years and then find an expression for the sum of their ages after h years

∵ Ann is h years old now

∵ Her age after h years = her age now + h

∴ Her age after h years = h + h

∴ Her age after h years = 2h

Ann's age after h years is 2h

∵ Bill is 2h years old now

∵ His age after h years = his age now + h

∴ His age after h years = 2h + h

∴ His age after h years = 3h

Bill's age after h years is 3h

∵ Charles is 8 years old now

∵ His age after h years = his age now + h

∴ His age after h years = 8 + h

∴ His age after h years = 8 + h

Charles' age after h years is 8 + h

Add their ages after h years

∵ The sum of their ages after 8 years = 2h + 3h + 8 + h

- Add like terms

∴ The sum of their ages after 8 years = (2h + 3h + h ) + 8

∴ The sum of their ages after 8 years = 6h + b

The sum of their ages after h years is 6h + 8

Learn more:

You can learn more about word problems in brainly.com/question/10557938

#LearnwithBrainly

7 0
3 years ago
Anyone know the answer to this algebra problem?
ikadub [295]

Answer:  \bold{a=1\qquad b=\dfrac{1}{16}\qquad c=\dfrac{1}{64}\qquad d=1\qquad e=\dfrac{4}{9}\qquad f=\dfrac{16}{81}}

<u>Step-by-step explanation:</u>

\begin{array}{c|l}\underline{\quad x\quad}&\underline{\quad 4^{-x}\qquad \qquad}\\-1&4^{-(-1)}=4^1=4\\\\0&4^{-(0)}=4^0=1\\\\2&4^{-(2)}=\dfrac{1}{4^2}=\dfrac{1}{16}\\\\4&4^{-(4)}=\dfrac{1}{4^4}=\dfrac{1}{64}\end{array}

\begin{array}{c|l}\underline{\quad \bigg x \quad}&\underline{\quad \bigg(\dfrac{2}{3}\bigg)^x\qquad \qquad}\\\\-1&\bigg(\dfrac{2}{3}\bigg)^{-1}=\dfrac{3}{2}\\\\0&\bigg(\dfrac{2}{3}\bigg)^{0}=1\\\\2&\bigg(\dfrac{2}{3}\bigg)^{2}=\dfrac{2^2}{3^2}=\dfrac{4}{9}\\\\4&\bigg(\dfrac{2}{3}\bigg)^{4}=\dfrac{2^4}{3^4}=\dfrac{16}{81}\end{array}

3 0
2 years ago
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