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Solve. An investment is worth $3486 in 1995. By 2000 it has grown to $6441. Let y be the value of the investment in the year x, where x = 0 represents 1995. Write a linear equation that models the value of the investment in the year x.
`x = 0 represents 1995` in the year 2000, what is x?

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Other answers:

x is 5
yes
`An investment is worth $3486 in 1995` x = 0 and y = 3486 the point (0,3486) is on the graph
`By 2000 it has grown to $6441` the point (5,6441) is on the graph
use the slope formula to find the slope through (0,3486) and (5,6441).
Okay.
591
yes
m = 591 is the slope b = 3486 since we know (0,3486) is on the graph
plug those into y = mx+b
The equation is y=591x+3486
yes
Okay. One more.
38. Find the domain of the indicated combined function. Find the domain of (fg)(x) when f(x) = square root 5x+8 and g(x) = .square root 7x-9
Ive never done a problem like this with square root so im not sure where to start
Normally id plug them in next to each other and multiple.
first solve 5x+8 >= 0 for x
x>/ -8/5
type `>=` to mean "greater than or equal to"
I agree, it's `x >= -8/5`
now solve `7x-9 > 0` for x
9/7
there is NO underline because we don't want 7x-9 to be 0
`x > 9/7` I think you mean
plot `x >= -8/5` and `x > 9/7` on the same number line then shade the region where they both overlap
Yes.
The domain is [9/7 , infinity)
if x = 9/7, then the denominator would be 0. But we cannot divide by 0
Um okay..
so we have to kick out 9/7 from the domain
[0,infinity)
you were on the right track, just change the `[` to `(`
Domain: `(9/7, infinity)`
I only have options of- [9/7, infinity) [0,infinity)
hmm so strange. That has to be a typo because 9/7 isn't part of the domain
oh wait, it's (fg)(x) and not (f/g)(x) so yeah 9/7 is allowed
Okay. Thank you.
no problem

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