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4 years ago

15

There is no answer choices. I really need help with this! prime factorize (expand) the expression 12a^3bd^5

1 answer:

Molodets [167]4 years ago

3 0

$\bf 12a^3bd^5\qquad 
\begin{cases}
12=2\cdot 2\cdot 3\\
a^3=a\cdot a\cdot a\\
d^5=d\cdot d\cdot d\cdot d\cdot d
\end{cases}\implies (2)(2)(3)aaabddddd$

is all I can see, I mean, you can't kill a dead horse, you can hit it, it won't make it deader, just flatter.

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Determine three numbers a , b , c

natta225 [31]

Since $a,b,c$ are in geometric progression, if $r$ is the common ratio between consecutive terms, then

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$b = ar$

$c=ar^2$

Since $a,b,c$ are also in arithmetic progression, if $d$ is the common difference between consecutive terms, then

$a = a$

$b = a + d \implies d = b-a$

$c = b + d = a + 2d \implies c = a + 2(b-a) = 2b-a$

Given that $abc=27$ , we have

$abc = a\cdot ar\cdot ar^2 = (ar)^3 = 27 \implies ar = 3 \implies a = \dfrac3r$

$b = \dfrac3r \cdot r = 3$

$c = \dfrac3r \cdot r^2 = 3r$

It follows that

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Solve for $r$ .

$3r - 6 + \dfrac3r = 0$

$3r^2 - 6r + 3 = 0$

$r^2 - 2r + 1 = 0$

$(r-1)^2 = 0$

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so the only possible sequence is {3, 3, 3, …}.

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Answer:

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Step-by-step explanation:

Solution for What is 10 percent of 80:

10 percent * 80 =

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