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Likurg_2 [28]
2 years ago
12

1. What is the value of the expression ƒ2 + 8 if ƒ = 7?

Mathematics
1 answer:
r-ruslan [8.4K]2 years ago
6 0

Answer:

f=8/5

Step-by-step explanation:

f*2+=f7

2f+8=f*7

2f+8=7f

2f=7f-8

2f-7f=-8

-5f=-8

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The cross section perpendicular to the base is a rectangle. The cross section parallel to the base is a pentagon similar to the pentagonal base. The solid is a cylinder.

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3 years ago
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What is the solution to the equation startfraction y over y minus 4 endfraction minus startfraction 4 over y 4 endfraction = sta
zmey [24]

The solution to the given equation is 4 and -4.

<h3>What is a fraction?</h3>

Fractions are the numerical values that are a part of the whole.

A whole can be an object or a group of objects.

The given equation is;

\rm \dfrac{y}{y - 4}- \dfrac{4}{y + 4}= \dfrac{32}{y^2- 16}

A fraction is used to represent the portion/part of the whole thing.

The solution of the fraction is determined in the following steps given below.

\rm \dfrac{y}{y - 4}- \dfrac{4}{y + 4}= \dfrac{32}{y^2- 16}\\\\\rm \dfrac{y(y+4)-4(y-4)}{(y - 4)(y+4)}= \dfrac{32}{(y-4)(y+4)}\\\\y^2+4y-4y+16=32\\\\y^2=32-16\\\\y^2=16\\\\y=\pm 4

Hence the solution to the given equation is 4 and -4.

Learn more about fractions here;

brainly.com/question/17431959

6 0
1 year ago
The sum of two numbers is 43 and the difference is 13 what are the numbers
Zinaida [17]

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Hope this helps!

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6 0
3 years ago
Colin has a pad with x pieces of paper on it. For his first class, he wrote on 5 fewer than half of the pieces of paper in the p
PIT_PIT [208]

Answer:

Colin has <em>8 sheets </em>left for his third class.

Step-by-step explanation:

Given that:

Total Number of pieces of papers = x

Number of pieces of papers used for 1st class = 5 fewer than half of the pieces in the pad

Writing the equation:

\text{Number of pieces of papers used for 1st class =} \dfrac{x}{2} -5 ...... (1)

Also, Given that number of pieces of papers used for the 2nd class are 2 more than that of papers used in the 1st class.

\text{Number of pieces of papers used for 2nd class =} \dfrac{x}{2} -5+2 = \dfrac{x}2 -3 ...... (2)

Now, number of pieces of papers left for the third class = Total number of pieces of papers in the pad - Number of pieces of papers used in the first class - Number of pieces of papers used in the first class

\text{number of pieces of papers left for the third class = }x-(\dfrac{x}{2}-5)-(\dfrac{x}{2}-3)\\\Rightarrow x-\dfrac{x}2-\dfrac{x}2+5+3\\\Rightarrow x-x+5+3\\\Rightarrow 8

So, the answer is:

Colin has <em>8</em> <em>sheets </em>left for his third class.

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800 is 10 times as much as what
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800 is 10 times as much as 80
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