What is the domain and range of the function f(x)= √x

Maria has a bag with 6 red marbles , 6 blue marbles , and 4 green marbles . Without looking Maria pick one marble. She does not

podryga [215]

40%

The probability of her getting a red marble is the number of red marbles divided by the number of total marbles. Since she picked a green marble first, we must subtract one marble from the total number of marbles, so that the total # of marbles is 6 + 6 + 4 = 16 - 1 (green marble) = 15. Since there are 6 red marbles, our probability is 6/15 or 2/5. Converted to percentage form, this is 0.4 —> 40%

7 0

3 years ago

Find (if possible) the complement and the supplement of each angle. (if not possible, enter impossible.) (a) π 10 complement rad

blagie [28]

In this question, we have to find the complement and supplement of the given angles .

Complementary angles are those angles whose sum is 90 degree and supplementary angles are those angles whose sum is 180 degree.

So to find the complement and supplement angles, we need to subtract the given angles from pi/2 and pi respectively .

a.

$\frac{ \pi}{10}
\\
complement \ angle = \frac{ \pi}{2} - \frac{ \pi}{10} \\ = \frac{4 \pi}{10} = \frac{ 2 \pi}{5}
\\
Supplement \ angle = \pi - \frac{ \pi}{10} = \frac{9 \pi}{10}$

b.

$\frac{9 \pi}{10}
\\
Complement \ angle = \frac{ \pi}{2} - \frac{9 \pi}{10} = \frac{-4 \pi}{10} = -\frac{2 \pi}{5}
\\
Supplement \ angle = \pi - \frac{9 \pi}{10} = \frac{ \pi}{10}$

4 0

3 years ago

Which ordered pair is a solution to the system of equations?

tatyana61 [14]

2x - y = -2 . . . (1)
5x - 2y = -7 . . . (2)
From (1), y = 2x + 2 . . . (3)
Putting (3) into (2), gives:
5x - 2(2x + 2) = -7
5x - 4x - 4 = -7
x - 4 = -7
x = -7 + 4 = -3
y = 2(-3) + 2 = -6 + 2 = -4.

Therefore, the required answer is (-3, -4)

6 0

3 years ago

Read 2 more answers

What is greater 2 3/4 or 2 2/3

Ronch [10]

2 3/4 is greater then 2 2/3

5 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B5x%2F8y%7D" id="TexFormula1" title="\sqrt[4]{5x/8y}" alt="\sqrt[4]{5x/8y}" al

Furkat [3]

Answer: $\frac{\sqrt[4]{10xy^3}}{2y}$

where y is positive.

The 2y in the denominator is not inside the fourth root

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Work Shown:

$\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\$

The idea is to get something of the form $a^4$ in the denominator. In this case, $a = 2y$

To be able to reach the $16y^4$ , your teacher gave the hint to multiply top and bottom by $2y^3$

For more examples, search out "rationalizing the denominator".

Keep in mind that $\sqrt[4]{(2y)^4} = 2y$ only works if y isn't negative.

If y could be negative, then we'd have to say $\sqrt[4]{(2y)^4} = |2y|$ . The absolute value bars ensure the result is never negative.

Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.

3 0

2 years ago