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adell [148]
4 years ago
10

Which table shows a linear function? Thank you!

Mathematics
2 answers:
tatiyna4 years ago
8 0
I think its not linear
Diano4ka-milaya [45]4 years ago
6 0
Between all points the slope(=(y2-y1)/(x2-x1)) has to be the same:
first graph: [-5,-3] has a slope of 0.5, [-1,1] has a slope of 2 -> not a linear function
second graph: [-4,-3] has a slope of -3, [-3,-2] has a slope of -1 -> not a linear function
third graph: slope matches for all subsection->is a linear function
fourth graph: [-4,-3] has a slope of +1, [-3,-2] has a slope of +2 -> not a linear function

-> the third graph is the only given linear function
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-5(2b + 7) + b < -b - 11
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5. The point (-2, -3) is rotated 90 degrees counterclockwise using center (0, 0).
SSSSS [86.1K]

B. (−3, 2)

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Read 2 more answers
(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2
Montano1993 [528]

Answer:

\dfrac{m^3}{16p^2q^2}

Step-by-step explanation:

Given:

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}

1.

m^{-1}=\dfrac{1}{m}

2.

q^0=1

3.

2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}

4.

(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}

5.

m^{-1}=\dfrac{1}{m}

6.

2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}

7.

\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}

8.

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}

8 0
3 years ago
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