Question

HELP ASAP MEDAL! Please show steps I would like to understand.
Find the percent increase the following equation models? y=34(1.4)^x
a. 1.4%
b.34%
c.40%
d.140%

Answer 1

@jdoe0001 do you mind helping?

Answer 2

hmm do you know what they're asking for?

Answer 3

how much it is increasing by?

Answer 4

yes... I'm reading that too

Answer 5

Well i don't know how to solve it haha

Answer 6

hmmm not sure if we could get a fixed value for it though

Answer 7

So what should I do?

Answer 8

well ahemm so.... this is what I get from it, one sec

Answer 9

ok

Answer 10

\(\large y=34(1.4)^x\qquad
\begin{array}{ccllll}
x&y
\\\hline\\
1&34(1.4)^1\to 47.6\\
2&34(1.4)^2\to 66.94\\
3&34(1.4)^3\to 93.296\\
...&...
\end{array}
\\ \quad \\
\textit{by how much did "y" increase from one spot to the next in }\%?
\\ \quad \\
66.94-47.6={\color{brown}{ 19.34}}\implies
\begin{array}{llll}
amount&\%\\
\\\hline\\
47.6&100\\
{\color{brown}{ 19.34}}&x
\end{array}\implies \cfrac{47.6}{{\color{brown}{ 19.34}}}=\cfrac{100}{x}\implies x=?
\\ \quad \\
93.296-66.94={\color{brown}{ 26.356}}\implies
\begin{array}{llll}
amount&\%\\
\\\hline\\
66.94&100\\
{\color{brown}{ 26.356}}&x
\end{array}\implies \cfrac{66.94}{{\color{brown}{ 26.356}}}=\cfrac{100}{x}\implies x=?\)

Answer 11

Ugh...

Answer 12

So i found out the answer is 40% but i cant find how it is/how they got that answer

Answer 13

hehe, so there's a "difference" from one point to the next for "y" coordinate
so seems to me you're asked what's that difference in approximate percent

Answer 14

yeah 40% would be the approximate increase percentage

Answer 15

but how do you know that?

Answer 16

well..how did you get the 40% anyway?

Answer 17

I didn't get it. I asked my friend but I can't figure out step by step how they got that

Answer 18

well... check my posting, and solve those, maybe add more values to the table, say maybe x =4 and 5 and 6, and check \(y=34(1.4)^x\qquad
\begin{array}{ccllll}
x&y
\\\hline\\
1&34(1.4)^1\to 47.6\\
2&34(1.4)^2\to 66.94\\
3&34(1.4)^3\to 93.296\\
...&...
\end{array}
\\ \quad \\
\textit{by how much did "y" increase from one spot to the next in }\%?
\\ \quad \\
66.94-47.6={\color{brown}{ 19.34}}\implies
\begin{array}{llll}
amount&\%
\\\hline\\
47.6&100\\
{\color{brown}{ 19.34}}&x
\end{array}\implies \cfrac{47.6}{{\color{brown}{ 19.34}}}=\cfrac{100}{x}\implies x=?
\\ \quad \\
93.296-66.94={\color{brown}{ 26.356}}\implies
\begin{array}{llll}
amount&\%
\\\hline\\
66.94&100\\
{\color{brown}{ 26.356}}&x
\end{array}\implies \cfrac{66.94}{{\color{brown}{ 26.356}}}=\cfrac{100}{x}\implies x=?\)

Answer 19

seems in short, you just have to make a table of values, like above, for "x" and "y" and get the difference for the "y" values and then check by how much they went up, in percent

Answer 20

for example say

x y
-----
1 5
2 15

so... by how much "y" went up from 5 to 10? well it went up 15-5 = 10

so what's 10 in % of 5? well \(\begin{array}{llll}
amount&\%
\\\hline\\
5&100\\
10&x
\end{array}\implies \cfrac{5}{10}=\cfrac{100}{x}\implies x=\cfrac{10\cdot 100}{5}\)