Which expression is not equivalent to 6^3/6^6? 1)1/6^2 2)6^-3 3)1/216 4)1/6^3

Mathematics

2 answers:

Nadusha1986 [10]3 years ago

8 0

Remember
(x^m)/(x^n)=x^(m-n)
and
x^(-m)=1/(x^m)
so

(6^3)/(6^3)=6^(3-6)=6^-3=1/(6^3)
first one is wrong
2nd is right
3rdone is right
4th is right

first one is wrong 1/(6^2)

andriy [413]3 years ago

6 0

I hope this helps you

A) 1/6^2=6^-2

B) 6^-3

C) 1/216=1/6^3=6^-3

D) 1/6^3=6^-3

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Raul is paid $75 per week plus $5 for each new gym membership he sells. He may switch to a gym that pays $50 per week and $7.50

rosijanka [135]

Answer:
11 memberships

Step-by-step:
Gym 1 = 75+5m
Gym 2 = 50+7.5m

75+5m=50+7.5m
-50 from both sides
25+5m=7.5m
-5m from both sides
25=2.5m
divide both sides by 2.5
10=m
for 10therefor 75+5m<50+7.5m he must sell 11 memberships.

5 0

3 years ago

Evaluate using integration by parts

PolarNik [594]

Rather than carrying out IBP several times, let's establish a more general result. Let

$I(n)=\displaystyle\int x^ne^x\,\mathrm dx$

One round of IBP, setting

$u=x^n\implies\mathrm du=nx^{n-1}\,\mathrm dx$

$\mathrm dv=e^x\,\mathrm dx\implies v=e^x$

gives

$\displaystyle I(n)=x^ne^x-n\int x^{n-1}e^x\,\mathrm dx$

$I(n)=x^ne^x-nI(n-1)$

This is called a power-reduction formula. We could try solving for $I(n)$ explicitly, but no need. $n=5$ is small enough to just expand $I(5)$ as much as we need to.