this is integrated math2, i’m so confused what does it mean and what is the answer help please (this is the last one lol sorry)

(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

2 years ago

A sample with mean of 85 and SD of 12 is transformed into z-scores. After the transformation, what are the values for the mean a

MrRa [10]

Answer:

The value remain unchanged.

Step-by-step explanation:

By Z score, we mean:

Z = $\frac{x-\mu}{\sigma}$ , if we substitute the given values of mean and standard deviation, we obtain the z score value. That is,

Z = $\frac{85 - 0}{12}$ = 85/12 = 7.083333.

NB: We assume that mean and standard deviation are in the same units.

5 0

2 years ago

Find the value of x.

erastovalidia [21]

Answer:

60 = x

Step-by-step explanation:

180-144 = 36

36+84 = 120

180-120 = 60

7 0

2 years ago

Read 2 more answers

From the statement select the related given statement. If B is between A and C, then AB + BC = AC

Bogdan [553]

<span>From the statement select the related given statement. If B is between A and C, then AB + BC = AC

answer
</span><span>C.) Point B is between points A and C.</span>

7 0

3 years ago

Read 2 more answers

Answer the following correctly do not give foolish answers or i will report your account

Anvisha [2.4K]

Answer:

I don't know the answer but I have a suggestion, have you tried using 900gle lens app putting it on homework setting then scanning it for the answer

8 0

2 years ago