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Leviafan [203]
3 years ago
15

an emu is 57 inches tall 75 inchess less than an 6 times height of an kakapoo what is the height of a kakapoo in inches

Mathematics
1 answer:
vredina [299]3 years ago
8 0

We can solve the following equation: 5 x - 70 = 60, where x is the height of a kakapo in inches : 5 x - 70 + 70 = 60 + 70 ( we will add 70 to the both sides of an equation ), 5 x = 130, x = 130 : 5, x = 26 inches. We can prove it: 26 * 5 - 70 = 130 - 70 = 60 inches ( the height of an emu in inches ). Answer: The height of a kakapo is 26 inches.

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Answer:
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Evaluate expression 20+3c+17d when c=8 and d=2.
Tems11 [23]
Just plug in the numbers for the variables.


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3 years ago
washington high school's head tennis coach, ms. racket, runs a tennis camp for middle school students every summer. the students
Kaylis [27]

Your question was incomplete. Please refer the content below:

Washington high school's head tennis coach, Ms. racket, runs a tennis camp for middle school students every summer. the students bring their own lunches, but Ms. racket provides them with snacks. there is a proportional relationship between the number of students who enroll in Ms. racket's tennis camp, x, and the total number of snacks she buys, y. what is the constant of proportionality? write your answer as a whole number or decimal. snacks per student.

X      Y

10     30

13     39

14     42

21     63

The constant of proportionality is 3 as for 1 student she buys 3 snacks.

The constant of proportionality means the ratio between the terms x and y.

It is calculated by dividing y by x.

Here we get that:

X      Y           constant of proportionality(Y/X)

10     30                               3

13     39                               3

14     42                               3

21     63                               3

Therefore, we get the constant of proportionality as 3.

Learn more about constant of proportionality here:

brainly.com/question/1835116

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Answer:

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8 0
3 years ago
Read 2 more answers
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oof

remember some rules

(a/b)(c/d)=(ac)/(bd)

\frac{x^a}{x^b}=x^{a-b}

x^0=1

(x^a)(x^b)=x^{a+b}

x^{-a}=\frac{1}{x^a}


\frac{(7^{-3})(7^{-2})(2^0)}{(7^{-1})(7^{-4})(2^3)}=

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\frac{1}{2^3}=

\frac{1}{8}

7 0
3 years ago
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