If there is 1000$ in the lockbox what year will it be worth 64,000$ at 24% interest?
1 answer:
Hey there!
First, set up an exponential equation that represents the rate at which your original amount, 1000, gains interest:
y = 1000(1.24)^x
Y represents the value after X years. 1.24 represents the rate at which the money gains interest, 1 + 0.24 (your 24% interest rate in decimal form). 1000 is your original amount.
Now, set this equation equal to 64000, graph y = 64000 and y = 1000(1.24)^x on a graphing calculator, and see where the two equations intersect in order to solve for x.
They intersect when x is about 19.334, as seen in the graph below (it is very zoomed in so that you can see where the two functions intersect). Therefore, it will be about 19 years after the year in which you deposited the 1000 dollars before the money is worth 64000 dollars.
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Answer:
y = x + 8
Step-by-step explanation:
the equation of a line in slope intercept form is
y = mx + c ( m is the slope and c the y-intercept )
rearrange 2x + 3y = 9 into this form
subtract 2x from both sides
3y = - 2x + 9 ( divide all terms by 3 )
y = - x + 3 ← in slope-intercept form
with slope m = -
given a line with slope m then the slope of a line perpendicular to it is
= -
= -
=
, hence
y = x + c ← is the partial equation
to find c substitute (- 2, 5) into the partial equation
5 = - 3 + c ⇒ c = 5 + 3 = 8
y = x + 8 ← equation of perpendicular line