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Rashid [163]
3 years ago
8

The average cost per month of a 2-bedroom apartment in Grayson was $625 last year.This year,the average cost is $650.What is the

percent of increase from late year?
Mathematics
1 answer:
Ann [662]3 years ago
8 0

The amount increased 4% from last year.

Step-by-step explanation:

Given,

Average cost last year = $625

Average cost this year = $650

Increase amount = Average cost last year - Average cost this year

Increase amount = 650 - 625 = $25

Percent increase = \frac{Increase\ amount}{Average\ cost\ last\ year}*100

Percent\ increase=\frac{25}{625}*100\\Percent\ increase=\frac{2500}{625}\\Percent\ increase=4\%

The amount increased 4% from last year.

Keywords: percentages, subtraction

Learn more about subtraction at:

  • brainly.com/question/8805387
  • brainly.com/question/8806598

#LearnwithBrainly

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a_sh-v [17]

For the first question please find the attached diagram.

As per the diagram, P is the upstream point and Q is the downstream point. The distance between P and Q is 22.5 miles.

Let the speed of the boat in the still waters of the lake be represented by S.

Then, when the boat travels upstream, the net speed of the boat will be (S-6) miles per hour because the river flows downstream and thus the speed of the boat will have to be subtracted from the speed of the river.

Now, we know that the relationship between the net speed, distance and time of travel is give as:

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We can rearrange the above equation to be:

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By similar arguments we know that the downstream speed of the boat is S+6 and the distance travelled is the same and so the time taken to travel downstream (represented by T_{2}) will be:

T_{2}=\frac{22.5}{S+6}................(Equation 2)

Now, we know that the total time of travel should be 9 hours.

This means that: T_{1}+T_{2}=9............(Equation 3)

Plugging in the values of T_{1} and T_{2} from (Equation 1) and (Equation 2) into (Equation 3), we get:

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The roots of this quadratic equation are:

S=-1 and S=6

Since, speed cannot be negative, S=-1 is out of consideration.

The speed of the boat in the lake is thus S=6 miles per hour.

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Let us solve problem 2

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\frac{x-2}{x+3}+\frac{10x}{(x-3)(x+3)}

\frac{(x-3)(x-2)+10x}{(x-3)(x+3)} =\frac{x^2-5x+6+10x}{(x-3)(x+3)}

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\frac{x+2}{x-3}

The restriction on the variable x is that it cannot be equal to either +3 or -3 as that would make the denominator of the original question equal to zero.

Thus, the restriction is x\neq \pm 3

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