Pretty sure it’s B but I don’t rlly know
Here is the compound interest formula solved for years:
<span>Years = {log(total) -log(Principal)} ÷ log(1 + rate)
</span>Years = {log(800) - log(600)} <span>÷ log(1.025)
</span><span>Years = {2.903089987 -2.7781512504} / 0.010723865392
</span>Years = {
<span>
<span>
<span>
0.1249387366
} / </span></span></span><span><span><span>0.010723865392
</span>
</span>
</span>
Years =
<span>
<span>
<span>
11.6505319708
</span>
</span>
</span>
That's how many years it takes for the $600 to become exactly $800.00
The question specifically asks how long for the money to be MORE than $800.00?
So, if we enter 800.01 into the equation, then the answer is
Years = {log(800.01) - log(600)} <span>÷ log(1.025)
</span><span>Years = {2.9030954156 -2.7781512504} / 0.010723865392
</span>Years =
<span>
<span>
<span>
0.1249441652
</span>
</span>
</span>
/ 0.010723865392
<span>
<span>
<span>
Years = 11.6510381875
</span>
</span>
</span>
<span><span> </span></span>
Answer:
x^4 + 8x
-----------------
(4-x^3)^2
Step-by-step explanation:
d /dx (x^2/(4-x^3))
When we differentiate a fraction u/v
df/dx = u/v
= v du/dx-u dv/dx
---------------------------
v^2
we know u = x^2 so du/dx = 2x
v = (4-x^3) so dv/dx = -3x^2
d dx = (4-x^3) (2x)- x^2 ( -3x^2)
-------------------------------------
(4-x^3)^2
Combining terms
(8x-2x^4) --3x^4
-------------------------------------
(4-x^3)^2
8x-2x^4 +3x^4
-------------------------------------
(4-x^3)^2
x^4 + 8x
-------------------------------------
(4-x^3)^2
Slope is known as rise over run. Because the line is pointing "\", it is negative.
The rise, the distance from one point to another, specifically from (0,4) to (1,1) is 3, as 4-1=3. your run is 1-0=1.
So your rise over run is -3/1 or, -3.
Your y-int is where when x=0, y=?
In this case y=4 when x=0.
Your equation is
y=-3x+4
Tenths is after the decimal so the tenths is 5.
