Question

Algebra 1 exponents - will award and fan asap (page 2) (questions 7, 8, and 9)

Answer 1

@rhettwalker14

Answer 2

Do you need help with all the above questions?

Answer 3

#6a) since x^3/x^8 can be written as x^3+x^-8 following basic exponent rules
you get 1/x^5 which can be put in ax^n form by writing it as \[x ^{-5}\]

#6b) \[\frac{ 6x }{ 2x^8 }\] here you can do it like this 6/2 =3 and x^1 +x^-8 = 1/x^7 which results in \[\frac{ 3 }{ x^7 }\] .

You can now write this in zx^n form by saying \[3x ^{-7}\]

Answer 4

no, just 7,8,and 9 @Lily2913

Answer 5

7a) \[n(d)=20(2)^{d}\] At n(0) = 20 because the value of any exponent at 0 is 1, therefore \[20*1 = 20\], and we also can tell from this that at d = 0 that is is Monday.

Answer 6

7b) since 2^-2 = 0.25 we know that n(-2) = 20*0.25 = 5 we can conclude that 5 people knew the rumor on Saturday, before the rumor went public on Monday.

Answer 7

thanks you're very helpful! @Lily2913 ..having trouble with 8 and 9

Answer 8

\[\frac{ (x^(2a+1))^3 }{ x^(a+3)^2 } = x^(4a-3) \]

Answer 9

\[x ^{4a-3}\]

Answer 10

great! @Lily2913 only 8b and 9 left :)

Answer 11

#9) f(x)=18(3)^-x if x = 1 then f(1) = 18*3^-1 = 18 * 1/3 = \[\frac{ 18 }{ 3 }\]
this is assuming that x=0 to start since it says that x is increasing, or else it would be 1/9 if x = 2 , but that is not an option.

Answer 12

\[(a+b)^{n} = \sum_{k=0}^{n} \left(\begin{matrix}n \\ b\end{matrix}\right)a ^{n-k}b ^{k}\]
\[(x ^{-3+4a}) = \sum_{n=0}^{\infty}\left(\begin{matrix}-3+4a \\ n\end{matrix}\right)(-1+x)^{n}\]

Answer 13

\[(0+x)^{4a-3}\] should also work