Question

On the number line, point B is the midpoint of line AC. If k is positive, what is the value of n?

Answer 1

8

Answer 2

@pgpilot326 How?

Answer 3

sorry. A is at \[2^{k}\]
C is at \[2^{k+4}=16\times2^{k}\]
\[B=\frac{ 2^{k}+16\times2^{k} }{ 2 }=\frac{ 17 }{ 2 }2^{k}\]
So B = 17/2 (not 8)

Answer 4

Note that if B is the midpoint of AC on the number line then B is the arithmetic mean of A and C. Hence:\[\bf B=\frac{ A \times C }{ 2 } \implies n(2)^k=\frac{ (2)^k(2)^{k+4} }{ 2 } \implies n(2)^{k-1}=2^{k+4}\]\[\bf \implies n = \frac{ 2^{k+4} }{ 2^{k-1} }=2^{(k+4)-(k-1)} \implies n = 2^5=32\]

Answer 5

@sakigirl
@pgpilot326

Answer 6

There is no 32 on the answer choices :(

Answer 7

oh i messed up just asecond

Answer 8

Arithmetic mean, meaning sum and divide. If you multiply and then take the root that would be the geometric mean. But that is not what is needed here.

Answer 9

brb

Answer 10

I think you did a mult when it should have been an addition.

Answer 11

@pgpilot326 I made an exponent law mistake. it's all good I corrected, ill post in a sec.

Answer 12

no worries

Answer 13

lmfao not only did i make an exponent law mistake, I also multiplied instead of adding in the arithmetic mean. That's hilarious. Well here is the corrected version:\[\bf n(2)^k=\frac{ 2^k+2^{k+4} }{ 2 }=\frac{ 2^k+(2^k)(2^4) }{ 2 }=\frac{ 2^k(1+2^4) }{ 2 }\]\[\bf \implies n(2^k)=2^{k-1}(17) \implies n = \frac{ 2^{k-1}(17) }{ 2^k }=2^{(k-1)-(k)}(17)\]\[\bf \implies n = \frac{ 17 }{ 2 }\]

Answer 14

@pgpilot326 there we go, exponent law/arithmetic mean both corrected. Don't know what I was thinking when I made those dumb mistakes lol...

Answer 15

@sakigirl