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pentagon [3]
2 years ago
11

A four-sided figure is resized to create a scaled copy. The proportional relationship between any given side length in the origi

nal figure, f, and the corresponding side length in the scaled copy, s, can be represented by the equation s=1/7f. What is the constant of proportionality from side lengths in the original figure to side lengths in the scaled copy?
Mathematics
1 answer:
boyakko [2]2 years ago
5 0

Answer:

and I have

Step-by-step explanation:

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A restaurant earns $1073 on Friday and $1108 on Saturday. Write and solve an equation to find the amount xx (in dollars) the res
alekssr [168]
You start by adding 1073 and 1108 together to get 2181 then subtract 2181 from the 3 thousand to get 819. the restaurant needs to earn 819 dollars (as a minimum) on Sunday.
3 0
3 years ago
Evaluate <br> 3^-3*3^4*3*3^-5
miss Akunina [59]
To multiply exponents with like bases, simply add the exponents...

3^-3 × 3^4 × 3 × 3^-5 =

add exponents across, remember just 3 means 3^1.

-3 + 4 + 1 + -5 = -3

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8 0
3 years ago
Math homework help exponents
Len [333]

Answer + Step-by-step explanation:

\frac{5^{10}}{5^{5}} =5^{10-5} = 5^5

\text{used property :} \  \  \  \frac{a^{m}}{a^{n}} =a^{m-n}

_______________________

(4^8)^3 = a^{8 \times 3}=a^{24}

\text{used property :} \  \  \  (a^n)^m = a^{n \times m}

_______________________

10^{-4}=\frac{1}{10^{4}} \  \text{not} \  \frac{1}{4^{10}}

_______________________

15⁶ × 15³ = 15⁶⁺³ = 15⁹

_______________________

<u>Recall</u> : If a ≠0  ⇒ a⁰ = 1

Then

Since 6⁸ ≠ 0  ⇒ (6⁸)⁰ = 1  

6 0
1 year ago
Aurora has been babysitting for 30 minutes, which is 20% of the total time she must babysit. What is the total amount of time in
atroni [7]
150 minutes i believe
3 0
2 years ago
I don't want the answer, I want to know how to do it. What is the ratio of 4/5 to 7/15?
Ostrovityanka [42]

Answer:

\displaystyle 1\frac{5}{7}

Step-by-step explanation:

\displaystyle \frac{4}{5} \div \frac{7}{15} = \frac{4}{5} \times \frac{15}{7} = \frac{60}{35} = 1\frac{5}{7}

I can assist whenever I can.

6 0
3 years ago
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