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4 years ago

Pls help me!!! I'm timed

Mathematics

1 answer:

Aleonysh [2.5K]4 years ago

3 0

The domain is the x-axis and the range is the y-axis.

So the range of the function would be the Y values given.

So the range would be (3, 6, 9, 12, 15).

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Type the correct answer in the box. Round off your answer to the nearest integer.

Mkey [24]

Answer:

∠ p ≈ 59°

Step-by-step explanation:

Using Pythagoras' identity in right triangle ABD to find DB

DB² = 5² + 6² = 25 + 36 = 61 ( take square root of both sides )

DB = $\sqrt{61}$

---------------------------------------

Using the cosine ratio in right triangle DBC

cos p = $\frac{adjacent}{hypotenuse}$ = $\frac{DC}{DB}$ = $\frac{4}{\sqrt{61} }$ , thus

p = $cos^{-1}$ ( $\frac{4}{\sqrt{61} }$ ) ≈ 59°

6 0

3 years ago

Given: 3x < -6.

In-s [12.5K]

3x < -6
x < -6/3
x < -2
Therefore, the solution set is {x | x < -2}.

3 0

3 years ago

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What is the measure of a Pentagon

Mashutka [201]

Answer:

5 sides.

Step-by-step explanation:

Unlike a triangle, which is 180°, also equal to a straight line, the length is equal to 540°.

6 0

3 years ago

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arrange the expressions in the correct sequence to rationalize the denominator of the expression -(2)/(\sqrt(x+y-2)-\sqrt(x+y+2)

cupoosta [38]

We have to rationalize the denominator:
$\frac{-2}{ \sqrt{x+y-2} - \sqrt{x+y+2} } = \\ \frac{-2}{ \sqrt{x+y-2} - \sqrt{x+y+2} }* \frac{ \sqrt{x+y-2}+ \sqrt{x+y+2} }{ \sqrt{x+y-2}+ \sqrt{x+y+2} }= \\ \frac{-2*( \sqrt{x+y-2}+ \sqrt{x+y+2}) }{x+y-2-(x+y+2)}= \\ \frac{-2*( \sqrt{x+y-2}+ \sqrt{x+y+2}) }{x+y-2-x-y-2}= \\ \frac{-2*( \sqrt{x+y-2}+ \sqrt{x+y+2} }{-4}= \\ \frac{ \sqrt{x+y-2}+ \sqrt{x+y+2} }{2}$

6 0

3 years ago

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Something divided by4 =9

coldgirl [10]

9 * 4 = 36 sooo 36 / 4 = 9

so the answer is 36 .

i hope this helps.

4 0

3 years ago

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