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2 years ago

What is the domain of the function y= (square root) x+4

Mathematics

1 answer:

garri49 [273]2 years ago

6 0

Answer:

0 ≤ x ≤ ∞

Step-by-step explanation:

Square roots always need to be either equal to or greater than zero because it should be positive inside (unless we're working with imaginary numbers).

So the domain would be that x ≥ 0 (or the third choice ⇒ 0 ≤ x ≤ ∞).

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Select the correct answer.

Law Incorporation [45]

i think its b

Step-by-step explanation:

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2 years ago

Simplify (8) ÷ (2x3)

Aleks04 [339]

Answer:

1.33

Step-by-step explanation:

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1 year ago

What is the value of x a. 25 b. 12 c. 7 d. 5

tia_tia [17]

The answer is 5 because a^2 + b^2 = c^2

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2 years ago

Read 2 more answers

Carlos has six cards in a bag, numbered 2, 2, 3, 6, 8, and 9. What is the probability Carlos will randomly select a 2 or a 9

AfilCa [17]

Answer:

50 percent

Step-by-step explanation:

100 percent divided by 6 (the amount of data) is 16 2/3.

Each number represents 16 2/3 percent.

3 of the numbers are either 2 or 9.

16 2/3x3=50.001

The final answer is a 50% probability.

I think

8 0

2 years ago

Can someone help me find the equivalent expressions to the picture below? I’m having trouble

miss Akunina [59]

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is $\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$ .

Now we will solve this expression with the help of law of exponents.

$\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}$

$=2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}$

$=2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}$

$=2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}$

$=2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }$ [Option 2]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$ [Option 1]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$

$=(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }$

$=\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} }$ [Option 3]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }$

$=\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}}$ [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

6 0

2 years ago