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2 years ago

Factor Completely: 2x2 − 18

2 answers:

3 0

To factor the given function, find a term that is divisible by both terms. For this problem, the factor would be 2, since 2x^2 and -18 are both divisible by 2.

2(x^2 - 9)

You could further factor out x^2 - 9 because it is factorable by (x - 3) and (x + 3). Thus, it would be

2(x -3)(x+3)

Darya [45]2 years ago

3 0

Answer:

The correct answer you are looking for is answer C

Step-by-step explanation:

2(x + 3)(x − 3)

if we factor both polynomials and if we factor the exponents, take away both sides to get 2x^2 then multiply the terms and simplify

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Please help me. It’s due in 20 minutes.

egoroff_w [7]

Answer:whatever the first number was times 2

Step-by-step explanation:

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2 years ago

During a sale at the mal, all items were marked down 40% off of the original price. Lily purchased a purse that was originall $2

NNADVOKAT [17]

Answer:

$132

Step-by-step explanation:

40% off means it is actually 60% of the regular price.

Thus, you will multiply the original (220) by .6 (this means 60 percent)

and you will get 220*.6=132

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3 years ago

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David’s watch broke. He decides to get it fixed instead of replacing it. The total cost to repair the watch can be represented b

ivann1987 [24]

0.07

45h represents how much it costs per hour to get it fixed and how long it takes to get it fixed
0.07(45h) would represent the tax because it is 7% of the cost pretax

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2 years ago

Write two expressions where the solution is 4

Luden [163]

7x=28
What is x.

n+2=6

8 0

2 years ago

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From the side view, a gymnastics mat forms a right triangle with other angles measuring 60° and 30°. The gymnastics mat extends

valkas [14]

Answer: The mat is 4.33 ft high off the ground.

Step-by-step explanation:

Since we have given that

Angle of elevation with the first triangle = 30°

Angle of elevation with the second triangle = 60°

Length at which gymnastics mat extends across the floor = 5 feet

so, As shown in the figure:

We need to find the height of the mat off the ground.

If CD = 5 ft,

Let, AB = y, DC = x.

In Δ ABC,

$\tan 60^\circ=\frac{AB}{BC}\\\\\frac{\sqrt{3}}{2}=\frac{y}{x}\\\\x=\frac{y}{\sqrt{3}}$

Similarly, in Δ ACD,

$\tan 30^\circ=\frac{AB}{BD}\\\\\frac{1}{\sqrt{3}}=\frac{y}{x+5}\\\\\frac{1}{\sqrt{3}}=\frac{y}{\frac{y}{\sqrt{3}}+5}\\\\\frac{1}{\sqrt{3}}=\frac{y\sqrt{3}}{y+5\sqrt{3}}\\\\3y=y+5\sqrt{3}\\\\2y=5\sqrt{3}\\\\y=\frac{5\sqrt{3}}{2}\\\\y=4.33\ ft$

Hence, the mat is 4.33 ft high off the ground.

5 0

2 years ago

Read 2 more answers