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2 years ago

Question is attached PLS HELP!

Mathematics

1 answer:

suter [353]2 years ago

6 0

Answer:

Step-by-step explanation:

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Solve the following simultaneous linear equations by the substitution method.

anyanavicka [17]

Answer:

p = 2 and q = 2

Step-by-step explanation:

hope this helps please like and mark as brainliest

5 0

3 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

2 years ago

What does 2.8- 4.4n- 2n + 7 equal?

boyakko [2]

Answer:

2.8- 4.4n- 2n + 7 = -6.4n + 9.8

Step-by-step explanation:

combined like term:

-4.4n - 2n = -6.4n

2.8 + 7 = 9.8

so all you do is combined together and got -6.4n + 9.8

5 0

2 years ago

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|f−6|=|f+8| please help

soldi70 [24.7K]

Answer:f=-1

Step-by-step explanation:

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3 years ago

Simplify 123 over 3

igor_vitrenko [27]

Answer:

The answer is 41

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Question is attached PLS HELP!​

Question is attached PLS HELP!