2) 2x2 – 32 and 3x2 + 6x – 24

Mathematics

1 answer:

ankoles [38]3 years ago

6 0

4-32=-28 && i cant help with the other one

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1/4 divided by 5/6 = 1/a× b/c

motikmotik

Answer:

a= 24b/5c b= 5ac/24 c= 24b/5a

Step-by-step explanation:

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3 years ago

What is the answer? Please explain

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Answer:

Step-by-step explanation:

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3 years ago

Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

zloy xaker [14]

Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
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x = -4 / -2 = 2

The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
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4 years ago

3. Find the value of x. 4 3 6<br>

tester [92]

Answer:

Step-by-step explanation:

x in (-oo:+oo)

(x/4)/3 = 6 // - 6

(x/4)/3-6 = 0

1/12*x-6 = 0 // + 6

1/12*x = 6 // : 1/12

x = 6/1/12

x = 72

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3 years ago

5 + {[2 × (14 − 9)] − 1}

aivan3 [116]

It will be 14.
i’ll attach an image to show work

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3 years ago