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3241004551 [841]
4 years ago
7

I WILL GIVE BRAINLIEST The area of the surface of the trampoline (square) is equal to twice its perimeter. Find the dimensions.

length= 4x ft., width= (x+6) ft.
FACTORING POLYNOMIALS/SOLVE THE EQUATION BY FACTORING HOMEWORK
Mathematics
1 answer:
Brums [2.3K]4 years ago
4 0
Given that the area is equal to twice the perimeter and the dimensions are:
length=4x ft, width=(x+6) ft
the perimeter will be:
P=2(L+W)
P=2(4x+(x+6))
P=2(4x+x+6)
P=2(5x+6)
P=(10x+12)

Area of the rectangle is:
A=4x(x+6)
A=4x²+24x
but:
2P=A
thus
2(10x+12)=4x²+24x
20x+24=4x²+24x
thus
4x²+4x-24=0
x²+x-6=0
x²-2x+3x-6=0
x(x-2)+3(x-2)=0
(x-2)(x+3)=0
thus the answer is:
x=-3 or x=2
thus
length=4x=4*2=8 ft
width=(x+6)=(2+6)=8 ft

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Stella [2.4K]
First step you would have to do would be to times 35 by 12 to see how much you would have to pay for 12 months.

35 x 12 = 420

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8 0
3 years ago
If the domain of the function f(x)=2x2- 8is {-2, 3, 5}, then the range is
Simora [160]

Answer:

{0, 10, 42}

Step-by-step explanation:

<em>Domain is set of input values and range is set output values.</em>

For the function f(x) = 2x² - 8 and domain set of {-2, 3, 5}

<u>Range is:</u>

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<u>Range:</u> {0, 10, 42}

6 0
4 years ago
2c + 7.5 = 6.2 - 3c a. 0.26 b. -0.26 c. -2.74 d. 1.3
Maurinko [17]

b

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then divide it :)

6 0
3 years ago
CALC- limits<br> please show your method
gladu [14]
A. Factor the numerator as a difference of squares:

\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6

c. As x\to\infty, the contribution of the terms of degree less than 2 becomes negligible, which means we can write

\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4

e. Let's first rewrite the root terms with rational exponents:

\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}

Next we rationalize the numerator and denominator. We do so by recalling

(a-b)(a+b)=a^2-b^2
(a-b)(a^2+ab+b^2)=a^3-b^3

In particular,

(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3
(x^{1/2}-x)(x^{1/2}+x)=x-x^2

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For x\neq0 and x\neq1, we can simplify the first term:

\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x

So our limit becomes

\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43
3 0
3 years ago
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