2 years ago

In the figure, Line a is parallel to line b. Name the angles congruent to angle 3.

Mathematics

1 answer:

disa [49]2 years ago

3 0

Go with what u think will help u

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ankoles [38]

3(x-8)=12
I got this equation since there are three of the "x-8" sections that add up to the twelve length bar above

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2 years ago

(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2

Montano1993 [528]

Answer:

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

Step-by-step explanation:

Given:

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

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

$q^0=1$

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

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

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

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

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

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

8 0

3 years ago

(4x+3) = (7x-6)<br><br> more math :)

Ilia_Sergeevich [38]

Answer:

X=3

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Apparently I'm supposed to do one at a time so here we go again

Nat2105 [25]

A) 60mph
B) 209.7 miles
C) 2 hours and 45 minutes

3 0

2 years ago

Please please answer this correctly I have to finish this today as soon as possible

Simora [160]

Answer:

Step-by-step explanation:

Look at where -2 is on the x-axis (the left side of the horizontal line), and go another half a box in, that is -2.5. But then go down one box from there, and plot the point. that is (-2.5,-1).

Next go half a box to the right and two boxes up, and plot your point there. that is (.5,2).

8 0

3 years ago