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3 years ago

Simplify the expression. helppp me plz do it right (xy)3·(–3x4y2)

Mathematics

2 answers:

klemol [59]3 years ago

7 0

I suggest you get photo math

Maru [420]3 years ago

7 0

Answer:

-3x^7y^5

Step-by-step explanation:

I assume the number 3 after parentheses is an exponent. I also assume that 4 after x and 2 after y are exponents. If so, please use the symbol ^ to show clearly it's an exponent.

(xy)^3 · (-3x^4y^2) =

= x^3 * y^3 * (-3x^4y^2)

= -3x^7y^5

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Let p= x^2-7.

Advocard [28]

Since p = x² - 7, you can substitute/plug in p for x² - 7

So:

(x² - 7)² - 4x² + 28 = 5

(p)² - 4x² + 28 = 5 You can factor out -4 from (-4x² + 28)

p² - 4(x² - 7) = 5 Plug in p

p² - 4p = 5 Subtract 5

p² - 4p - 5 = 0 Your answer is C

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3 years ago

What is the solution set of the inequality?

jolli1 [7]

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Step-by-step explanation:

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4 years ago

Which equation has the solutions x=1+or-square root of 5?

stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

case a) $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=-4+1$

$x^{2}+2x+1=-3$

Rewrite as perfect squares

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

$(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i$

therefore

case a) is not the solution of the problem

case b) $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

$(x-1)^{2}=-3$

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

case c) $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

case d) $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

the answer is

$x^{2}-2x-4=0$

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Elizabeth has 140 citrus trees and 180 palm trees

lisabon 2012 [21]

You have to find the GCF of 140 and 180.

The GCF is 20 so thats your answer.

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Simplify −2(2h+10)+4h

Anuta_ua [19.1K]

First distribute: -4h - 20 + 4h
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