Item 25
25) Solve the problem.
An investment is worth $3733 in 1994. By 1997 it has grown to $5425. Let y be the value of
the investment in the vear x, where x 0 represents 1991. Write a linear equation that
models the value of the investment in the vear x.
y = -564x + 7117
1 x + 3733
564

Now we can use the value for 1994 and we can find the intercept like this:

And solving for b we got:

So then oir model would be given by:

Option C

Step-by-step explanation:

For this case we want to create a linear function for the the value of  the investment in the vear x, where x =0 represents 1994.

We know that for 1994 (x= 0) the value is y = $3733 and for 1997 (x=3) the value of y = $5425

We want to find a model given by:

Where m is the slope and b the intercept. We can find the slope with this formula:

Now we can use the value for 1994 and we can find the intercept like this:

And solving for b we got:

So then oir model would be given by:

Option C

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3