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Ronch [10]
3 years ago
5

Simplify the expression below. 3x^2 - 5x^2 + 4x^2 (Please explain how you get the answer)

Mathematics
1 answer:
Talja [164]3 years ago
4 0

Answer:

The answer is -4x^2

Step-by-step explanation:

-3x^2 - 5x^2 + 4x^2

= -3x^2 + -5x^2 + 4x^2

Combine like terms:

-3x^2 + -5x^2 + 4x^2

(-3x^2 + -5x^2 + 4x^2)

= -4x^2

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Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Calculate the following limit:
aleksklad [387]
\lim_{x\to\infty}\dfrac{\sqrt x}{\sqrt{x+\sqrt{x+\sqrt x}}}=\\
\lim_{x\to\infty}\dfrac{\dfrac{\sqrt x}{\sqrt x}}{\dfrac{\sqrt{x+\sqrt{x+\sqrt x}}}{\sqrt x}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{\dfrac{x+\sqrt{x+\sqrt x}}{x}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{x}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{\sqrt{x^2}}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{x+\sqrt x}{x^2}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\dfrac{\sqrt x}{\sqrt{x^4}}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{x}{x^4}}}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{1}{x^3}}}}}=\\
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1


8 0
3 years ago
A farmer wants to enclose a rectangular field that borders a river with 400 feet of fencing material. if the farmer does not fen
ICE Princess25 [194]
Let each side perpendicular to the river be "x".Then the side parallel to the river is "400-2x".
The formula for Area = width*length
A(x) = x(400-2x)
A(x) = 400x - 2x^2
You have a quadratic with a = -2 and b = 400
Maximum Area occurs where x = -b/2a = -400/(2*-2) = 100 ft. (width)
length = 400-2x = 400-2*100 =  200 ft (length)
3 0
3 years ago
Help me PLEASEEEEEEEEEEEEE
qwelly [4]

Answer:

D. 144

Step-by-step explanation:

find the volume of the shipping container and divide it by the volume of a box.

2\frac{2}{3} × 1 × 2 = \frac{16}{3}

\frac{1}{3} × \frac{1}{3} × \frac{1}{3} = \frac{1}{27}

\frac{16}{3} divided by \frac{1}{27} is 144.

4 0
2 years ago
Plz I need help this is my last question plz help me
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3 years ago
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