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alexandr402 [8]
3 years ago
11

The water level in ricky lake changes at an average of -7/48 inch for 1 year.Based on the rate above how much will the water lev

el change after 5 years?Show your calcutations.
Mathematics
1 answer:
bearhunter [10]3 years ago
3 0

we are given

The water level in Ricky lake changes at an average of -7/48 inch for 1 year

It means that

In 1 year , the water level changes at \frac{-7}{48} inches

and we have to find the water level change after 5 years

so, we can multiply both sides by 5

In 5*1 year , the water level changes at 5*\frac{-7}{48} inches

In 5 years , the water level changes at \frac{-35}{48} inches

Hence,

the water level change after 5 years is \frac{-35}{48} inches........Answer

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If you would like to know what is the area of the paper in square centimeters, you can calculate this using the following steps:

1 inch equals to 2.54 centimeters.

11 inches = 11 * 2.54 = <span>27.94 centimeters long
8.5 inches = 8.5 * 2.54 = </span><span>21.59 centimeters wide

11 inches long * 8.5 inches wide = 27.94 centimeters long * 21.59 centimeters wide = </span>27.94 * 21.59 = 603.22 square centimeters
<span>
The correct result would be </span>603.22 square centimeters.<span>
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Mary is an years old. What expression represents her age 7 years from now? Two years ago?
Rudiy27

Answer

What expression represents her age 7 years from now?

y =x+7x

Two years ago?

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Step-by-step explanation:

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Janeel has a 10-inch by 12-inch photograph. She wants to scan the photograph, then reduce the result
Vaselesa [24]

<u>Complete Question:</u>

Janeel has a 10 inch by 12 inch photograph. She wants to scan the photograph, then reduce the results by the same amount in each dimension to post on her Web site. Janeel wants the area of the image to be one eight of the original photograph. Write an equation to represent the area of the reduced image. Find the dimensions of the reduced image.

<u>Correct Answer:</u>

A) (10-x)(12-x)=15

B) Dimensions are : Length = 10-x = 3 inch , Breadth = 12-x = 5 inch

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

a. Write an equation to represent the area of the reduced image.

Let the reduced dimensions is by x , So the new dimensions are

length=10-x\\breadth=12-x

According to question , Area of new image is :

⇒ Area = \frac{1}{8}Length(breadth)

⇒ Area = \frac{1}{8}(10)(12)

⇒ Area = 15

So the equation will be :

⇒ (10-x)(12-x)=15

b. Find the dimensions of the reduced image

Let's solve :  (10-x)(12-x)=15

⇒ 120-10x-12x+x^2=15

⇒ 120-22x+x^2=15

⇒ x^2-22x+105=0

By Quadratic formula :

⇒ x = \frac{-b \pm \sqrt{b^2-4ac} }{2a}

⇒ x = \frac{22 \pm8 }{2}

⇒ x = \frac{22 +8 }{2} , x = \frac{22 -8 }{2}

⇒ x = 15 , x =7

x = 15 is rejected ! as 15 > 10 ! Side can't be negative

⇒ x =7

Therefore, Dimensions are : Length = 10-x = 3 inch , Breadth = 12-x = 5 inch

5 0
3 years ago
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