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3 years ago

1) Simplify the product using the distributive property.

Mathematics

1 answer:

uysha [10]3 years ago

5 0

X^2 + 2x -8
Hope this helps :))

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Solve for x 4(x+9)= 3(2x - 7)

ElenaW [278]

Answer: X= 57/2

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3 years ago

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Solve equation 13=x/7+5

yuradex [85]

Answer:

Step-by-step explanation:

Isolate the variable (x). Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.

First, subtract 5 from both sides:

13 (-5) = (x/7) + 5 (-5)

13 - 5 = x/7

8 = x/7

Isolate the variable x. Multiply 7 to both sides:

8(7) = (x/7)(7)

8(7) = x

x = 8 * 7

x = 56

56 is your answer for x.

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2 years ago

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volunteers collected non perishable food items and packed 14 items per box. If the volunteers collected a total fo 200 items, ho

klio [65]

Do 200/14 and get 14.2857142857 so about 14 boxes

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11 months ago

The ratio of girls to boys in a a gardening club was 6:2 there were 16 boy how many are girls

marishachu [46]

Answer:

I don't know

Step-by-step explanation:

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1 year ago

What is the solution of

kobusy [5.1K]

Answer:

Third option: x=0 and x=16

Step-by-step explanation:

$\sqrt{2x+4}-\sqrt{x}=2$

Isolating √(2x+4): Addind √x both sides of the equation:

$\sqrt{2x+4}-\sqrt{x}+\sqrt{x}=2+\sqrt{x}\\ \sqrt{2x+4}=2+\sqrt{x}$

Squaring both sides of the equation:

$(\sqrt{2x+4})^{2}=(2+\sqrt{x})^{2}$

Simplifying on the left side, and applying on the right side the formula:

$(a+b)^{2}=a^{2}+2ab+b^{2}; a=2, b=\sqrt{x}$

$2x+4=(2)^{2}+2(2)(\sqrt{x})+(\sqrt{x})^{2}\\ 2x+4=4+4\sqrt{x}+x$

Isolating the term with √x on the right side of the equation: Subtracting 4 and x from both sides of the equation:

$2x+4-4-x=4+4\sqrt{x}+x-4-x\\ x=4\sqrt{x}$

Squaring both sides of the equation:

$(x)^{2}=(4\sqrt{x})^{2}\\ x^{2}=(4)^{2}(\sqrt{x})^{2}\\ x^{2}=16 x$

This is a quadratic equation. Equaling to zero: Subtract 16x from both sides of the equation:

$x^{2}-16x=16x-16x\\ x^{2}-16x=0$

Factoring: Common factor x:

x (x-16)=0

Two solutions:

1) x=0

2) x-16=0

Solving for x: Adding 16 both sides of the equation:

x-16+16=0+16

x=16

Let's prove the solutions in the orignal equation:

1) x=0:

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(0)+4}-\sqrt{0}=2\\ \sqrt{0+4}-0=2\\ \sqrt{4}=2\\ 2=2$

x=0 is a solution

2) x=16

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(16)+4}-\sqrt{16}=2\\ \sqrt{32+4}-4=2\\ \sqrt{36}-4=2\\ 6-4=2\\ 2=2$

x=16 is a solution

Then the solutions are x=0 and x=16

5 0

3 years ago