The amount of Tina money can be expressed in an exponent function like this:
an= $1100(1.0725)^n
The variable an represent the total money and variable n is the years needed to achieve that amount.
Then, the time needed for the money to reach $6,600 would be:
an= $1100(1.0725)^n
$6,600= $1100(1.0725)^n
$1,100(6)= $1100(1.0725)^n
6= (1.0725)^n
n= log1.075 6
n= 24.78
AP teachers for<span> course and exam preparation; permission </span>for<span> any other use ... 3. Question 2. Let f and g be the functions given by ( ). (. ) 2 1. </span>f x<span> x x. = − and. ( ) (. ) ..... </span>Let f be the function given by<span> ( ). ( ) </span>sin<span> 5. ,. </span>4<span>. </span>f x<span> x </span>π<span>. = + and let ( ). P x be the ...</span>
The closest one is 5 to the power of 1 over 3.
sq rt of 3 = 1.732
sq rt of 5 = 2.236
5^1/3 = 1.667
5^1/6 = 0.833
5^2/3 = 8.333
5^3/2 = 62.5
Y-5=-2(x+4)
Y-5=-2x-8
Add 5 to both sides
Y=-2x-3
D
We have that
<span>p(t)=-3t^2+18t-4
using a graphing tool, we can see the maximum of the graph
(see the attached figure)
A) </span><span>In what year of operation does Mr. Cash’s business show maximum profit?
</span>
Mr. Cash’s business show maximum profit at year 3 (maximum in the parabole)
<span>B) What is the maximum profit?
23 (hundred of thousand of dollars) = 2.300.000 dollars
</span>c) What time will it be two late?
(This is the time when the graph crosses zero and the profits turn into losses )
5.77 years, or an estimate of about 69 months.