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salantis [7]
3 years ago
13

HELP!!! Which of the following sets of ordered pairs is a function?

Mathematics
1 answer:
Fantom [35]3 years ago
7 0
C, because the x value doesn’t repeat
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Determine all intervals on which f(x) > 0.
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Answer: [-7, -4]u[0, 3]

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3 years ago
What is the simplified form of<br> (4ab)^2
Alex
Use\ (a\cdot b)^n=a^nb^n\\-----------------\\\\(4ab)^2=4^2a^2b^2=\huge\boxed{16a^2b^2}
7 0
3 years ago
2w^2-11w=-12 <br> a w=3/2 or 4<br> b w=1 or 12<br> c w=1 or 7<br> d w = 2.5 or 6
dlinn [17]
Hey mate !!
Here's your answer !!


2w^2 - 11w = -12
2w^2 -11w + 12 = 0
2w^2 - 8w - 3w + 12 = 0

2w ( w - 4) -3 ( w - 4) = 0
(2w - 3) ( w - 4) = 0

Hence

2w - 3 = 0
2w = 3
w = 3/2

w - 4 = 0
w = 4

Hence value of w are 4 and 3/2

Hope this helps!!
Cheers!!
7 0
3 years ago
Determine if the following functions are increasing or decreasing, and compare their rates of change. A. Both functions are incr
adoni [48]

Answer:

D is the answer

Step-by-step explanation:

The both are - 2 slopes hope this helps!:)

4 0
3 years ago
Read 2 more answers
If sin(theta) =4/5 and is in quadrant 2, the value of cot(theta
iren [92.7K]
\sin\theta= \frac{4}{5} &#10;\\&#10;\\\sin^2\theta+\cos^2\theta=1&#10;\\ \cos\theta=\pm \sqrt{1-\sin^2\theta} =\pm \sqrt{1-( \frac{4}{5})^2 } =\pm \sqrt{1- \frac{16}{25} } =\pm \sqrt{\frac{9}{25} }=\pm \frac{3}{5}}

\theta \in II \Rightarrow \cos\theta\ \textless \ 0 \Rightarrow \cos\theta=-\frac{3}{5}

\cot\theta= \frac{\cos\theta}{\sin\theta}= \frac{-\frac{3}{5}}{\frac{4}{5}}  =- \frac{3}{4}

5 0
3 years ago
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