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blagie [28]
3 years ago
12

What is the gcmf of 5x^2-10x^2100 points​

Mathematics
2 answers:
ruslelena [56]3 years ago
7 0

Answer:

Simplifying

5x2 + -10x = 2

Reorder the terms:

-10x + 5x2 = 2

Solving

-10x + 5x2 = 2

Solving for variable 'x'.

Reorder the terms:

-2 + -10x + 5x2 = 2 + -2

Combine like terms: 2 + -2 = 0

-2 + -10x + 5x2 = 0

Begin completing the square.  Divide all terms by

5 the coefficient of the squared term:  

Divide each side by '5'.

-0.4 + -2x + x2 = 0

Move the constant term to the right:

Add '0.4' to each side of the equation.

-0.4 + -2x + 0.4 + x2 = 0 + 0.4

Reorder the terms:

-0.4 + 0.4 + -2x + x2 = 0 + 0.4

Combine like terms: -0.4 + 0.4 = 0.0

0.0 + -2x + x2 = 0 + 0.4

-2x + x2 = 0 + 0.4

Combine like terms: 0 + 0.4 = 0.4

-2x + x2 = 0.4

The x term is -2x.  Take half its coefficient (-1).

Square it (1) and add it to both sides.

Add '1' to each side of the equation.

-2x + 1 + x2 = 0.4 + 1

Reorder the terms:

1 + -2x + x2 = 0.4 + 1

Combine like terms: 0.4 + 1 = 1.4

1 + -2x + x2 = 1.4

Factor a perfect square on the left side:

(x + -1)(x + -1) = 1.4

Calculate the square root of the right side: 1.183215957

Break this problem into two subproblems by setting  

(x + -1) equal to 1.183215957 and -1.183215957.

Step-by-step explanation:

elixir [45]3 years ago
3 0

5x^2 - 10x

= 5x(x - 2)

expand using the distributive property to see that it is correct.

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What is the shape of the cross-section formed when a plane containing line AC and line EH intersects this cube? What is the area
liq [111]

Answer:

Area of given figure = 96 unit²

Step-by-step explanation:

Given:

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May someone help me with this problem?
Katyanochek1 [597]

Answer:

a. \:  \:  \frac{ {4x}^{6} }{ {y}^{3} }  \\

b. \:  \:  \frac{ {8x}^{3} }{ {y}^{ 1\frac{1}{3} } }  \\

Step-by-step explanation:

<h3>a. </h3>

{( {4x}^{ - 2} y)}^{ - 3}  \\  {4x}^{ - 2 \times  - 3}  \:  \: {y}^{1 \times  - 3}  \\  {4x}^{6}  {y}^{ - 3}  \\  \frac{ {4x}^{6} }{ {y}^{3} }  \\

<h3>b.</h3>

{( {8x}^{6} {y}^{ - 3})  }^{ \frac{1}{2} }  \\  {8x}^{6 \times  \frac{1}{2} } \:  \:  {y}^{ - 3 \times  \frac{1}{2} }   \\  {8x}^{3}  {y}^{  - \frac {3}{2} }  \\  \frac{ {8x}^{3} }{ {y}^{ 1\frac{1}{3} } } \\

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