Answer:
5/4k^2
Step-by-step explanation:
P=5\dfrac{k}{6}\times \dfrac{3}{2k^3}.
We will be using the following property of exponents:
\dfrac{a^x}{a^y}=a^{x-y}.
We have
P\\\\\\=5\dfrac{k}{6}\times\dfrac{3}{2k^3}\\\\\\=\dfrac{5}{6}\times\dfrac{3}{2}k^{1-3}\\\\\\=\dfrac{5}{4}k^{-2}=\dfrac{5}{4k^2}.
Thus, the required product is \dfrac{5}{4k^2}.
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Answer:
14
Step-by-step explanation:
Next time, please share the possible answer choices.
The absolute value of any quantity, whether it be a + or - quantity, is zero or positive.
Thus, |-14| = the absolute value of -14 = +14
PQ is bisected by RS at R. It means that PRQ is a line segment with PR = RQ and each of half of PQ.
RQ = PQ/2 = 1/2 [ 4 3/4 ] = 1/2 * 19/4 = <em>19/8 </em> = 2 3/4
Answer:
4 units count the boxes from a to b and that's 4
Answer:
huh
Step-by-step explanation:
