Ask your own question, for FREE!
Mathematics 16 Online
OpenStudy (anonymous):

Which of the following functions has the largest value when x = 4? e(x) = x2 + 6x + 21 m(x) = 8x v(x) = 31x A)e(x) D)v(x) C)m(x) D)All the functions are equal at x = 4.

OpenStudy (anonymous):

@misty1212

OpenStudy (anonymous):

@ganeshie8

OpenStudy (anonymous):

@jagr2713

OpenStudy (butterflydreamer):

you just want to plug in " x = 4" for each of the functions :) Let's start with \[e(x) = x^2 + 6x + 21\] So when we replace all the x's with 4, we get this equation. \[e(4) = 4^2 + 6(4) + 21 = ?\]

OpenStudy (anonymous):

WAIT

OpenStudy (butterflydreamer):

okey dokey

OpenStudy (anonymous):

61?

OpenStudy (butterflydreamer):

yep :) good job! so, e(4) = 61 Now we do the same thing with the other functions :) \[m(x) = 8x \] plug in x = 4 and we get: \[m(4) = 8(4) = ?\]

OpenStudy (anonymous):

32?

OpenStudy (butterflydreamer):

yes good! Now next function: \[v(x) = 31(x)\] plug in x = 4 \[v(4) = 31(4) = ?\]

OpenStudy (anonymous):

124?

OpenStudy (butterflydreamer):

yeeees :D So now which of the functions had the largest value when x = 4?

OpenStudy (anonymous):

its A?

OpenStudy (butterflydreamer):

not quite... we already worked out what EACH function equals when x = 4, e(4) = 61 m(4) = 32 v(4) = 124 So which one is has the largest value?

OpenStudy (anonymous):

V(4)

OpenStudy (butterflydreamer):

yesss so that means v(x) would be the largest function when x = 4

OpenStudy (anonymous):

SO the answer its B? RIGHT

OpenStudy (butterflydreamer):

yes...

OpenStudy (anonymous):

Can you help with two more please?

OpenStudy (butterflydreamer):

i can try lol

OpenStudy (anonymous):

Indicate whether the following statement is true or false: An exponential growth function never exceeds a quadratic function with a positive leading coefficient. A)TRUE B)FALSE

OpenStudy (butterflydreamer):

ohh xD i can't answer that question, sorry!

OpenStudy (anonymous):

ONE MORE?

OpenStudy (butterflydreamer):

LOL close this question and post it as a new one so other people can help you answer them xD I'm not very good with this stuff hahaha.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!