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just olya [345]
3 years ago
13

The area of a rectangle is x^2 - 6x - 7 with a width of x - 7. What is the lenght?

Mathematics
1 answer:
Sauron [17]3 years ago
4 0

{x}^{2} - 6x - 7 = (x - 7)(x + 1) \\

width = (x - 7)

So

length = (x  + 1)

Just it...

So we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

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Pls answer this question
Pavlova-9 [17]

Answer:

Question 37:

B) 21

Step-by-step explanation:

How I found the radius:

radius= 132/ 2 x π  

r=132 / 6.2831853071796

r=21.0084525

Divide 132 by 6.2831853071796

π = 3.1415926535898

Note: Hope this Helps!

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Step-by-step explanation:

6 0
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Read 2 more answers
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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