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3 years ago

Which is the value of this expression when p=-2 and q=-1?

Mathematics

1 answer:

saul85 [17]3 years ago

8 0

Answer:

D. 4

Step-by-step explanation:

$[(p^2) (q^{-3}) ]^{-2}.[(p)^{-3}(q)^5] ^{-2}\\\\=[(p^2) (q^{-3}) \times(p)^{-3}(q)^5 ]^{-2}\\\\=[(p^{2}) \times(p)^{-3} \times (q^{-3}) \times(q)^5 ]^{-2}\\\\=[(p^{2-3}) \times (q^{5-3}) ]^{-2}\\\\=[(p^{-1}) \times (q^{2}) ]^{-2}\\\\=(p^{-1\times (-2)}) \times (q^{2\times (-2) }) \\\\=p^{2}\times q^{-4} \\\\= \frac{p^2}{q^4}\\\\= \frac{(-2)^2}{(-1)^4}\\\\= \frac{4}{1}\\\\= 4$

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11111nata11111 [884]

7 1/2 * 2 = 15

The 3rd hour was 1/4 of the total distance in 3 hours.

1/4 = 0.25 as a decimal, and 25% as a percent.

15 (0.25) = 3.75
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0.75 = 3/4 as a fraction.

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(-2/3)^7*(3/5)^6*(5/6)^5*(1/3)^-7 simplify of law of indices

Wewaii [24]

Answer:

$\boxed {-\frac{12}{5}}$

Step-by-step explanation:

Solve the following expression:

$(-\frac{2}{3})^{7} \times (\frac{3}{5})^{6} \times (\frac{5}{6})^{5} \times (\frac{1}{3})^{-7}$

-Calculate $-\frac{2}{3}$ to the power of $7$ :

$(-\frac{2}{3})^{7} \times (\frac{3}{5})^{6} \times (\frac{5}{6})^{5} \times (\frac{1}{3})^{-7}$

$-\frac{128}{2187} \times (\frac{3}{5})^{6} \times (\frac{5}{6})^{5} \times (\frac{1}{3})^{-7}$

-Calculate $\frac{3}{5}$ to the power of $6$ :

$-\frac{128}{2187} \times (\frac{3}{5})^{6} \times (\frac{5}{6})^{5} \times (\frac{1}{3})^{-7}$

$-\frac{128}{2187} \times (\frac{729}{15625}) \times (\frac{5}{6})^{5} \times (\frac{1}{3})^{-7}$

-Multiply both $-\frac{128}{2187}$ and $\frac{729}{15625}$ :

$-\frac{128}{2187} \times (\frac{729}{15625}) \times (\frac{5}{6})^{5} \times (\frac{1}{3})^{-7}$

$-\frac{128}{46875} \times (\frac{5}{6})^{5} \times (\frac{1}{3})^{-7}$

-Calculate $\frac{5}{6}$ to the power of $5$ :

$-\frac{128}{46875} \times (\frac{5}{6})^{5} \times (\frac{1}{3})^{-7}$

$-\frac{128}{46875} \times (\frac{3125}{7776}) \times (\frac{1}{3})^{-7}$

-Multiply both $-\frac{128}{46875}$ and $\frac{3125}{7776}$ :

$-\frac{128}{46875} \times (\frac{3125}{7776}) \times (\frac{1}{3})^{-7}$

$-\frac{4}{3645} \times (\frac{1}{3})^{-7}$

-Calculate $\frac{1}{3}$ to the power of $-7$ :

$-\frac{4}{3645} \times (\frac{1}{3})^{-7}$

$-\frac{4}{3645} \times 2181$

-Multiply both the $-\frac{4}{3645}$ and $2187$ :

$-\frac{4}{3645} \times 2181$

$\boxed {-\frac{12}{5}}$

Therefore, the final answer is $-\frac{12}{5}$ .

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