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1 year ago

Please help ASAP!!!!!!!!

Mathematics

1 answer:

leva [86]1 year ago

4 0

The greatest common factor is 1 :)

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I need help with this and I have a lot more to go

enyata [817]

Answer:

under root 2, 1.5, 2π

Step-by-step explanation:

under under root 2 is equals to 1.414

1.5

2×22/7

44/7

6.2

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1 year ago

If you know the diameter of a circle, how can you find the area?

svet-max [94.6K]

Answer:

Divide the diameter by 2, and use the area formula A=pi(r^2)

Step-by-step explanation:

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1 year ago

Read 2 more answers

For what value of x is the parallelogram a rhombus.

Stells [14]

Answer:

Step-by-step explanation:

2 × ( 3x + 6 )° + ( 16x + 14 )° = 180°

22x + 26 = 180

22x = 154

x = 7

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1 year ago

(-2ab-a+2)-(-ab-2) help

BARSIC [14]

This is the answer, ab-a+4

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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

1 year ago