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3 years ago

The domain of a function g(x) is x < 2, and the range is y> 0. What are the

Mathematics

1 answer:

zavuch27 [327]3 years ago

8 0

Answer:

Step-by-step explanation:

Domain and ranges of inverse functions are swapped.

Domain of the original is the range of the inverse and vice versa.

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Which of these triangles are translations of Triangle A?

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A B C D E F are all the same shape

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Find perimeter pt4 brainliest

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Perimeter= X + Y + W. Perimeter is the sum of all sides

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3 years ago

Dana is buying Grandparents' Day gifts at the mall. She spends $13.75 on a teacup for her grandmother. She spends 4/5 of that on

marysya [2.9K]

The total money Dana spent in all is $24.75

<h3>Ratio and proportions</h3>

Ratio are written as fractions of two integers. For instance a/b.

Amount spent for grandmother = $13.75

Amount spent on grand father = 4/5 * 13.75

Amount spent on grand father = $11

Take the sum

Total money spent = $13.75 + $11

Total money spent = $24.75

Hence the total money Dana spent in all is $24.75

Learn more on proportion here: brainly.com/question/19994681

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2 years 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

1,304,000,000 move the decimal left 9 places

Viefleur [7K]

thats true... why did you ¨ask a question¨"thats not a question?

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3 years ago

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