Answer:
4. $648
5. $86.40
Step-by-step explanation:
You take the percent (we'll take 4 for example, so 8%) and change it to a decimal: .08. You then multiply the decimal (.08) by your other number (in this case $600) and the resulting number is your answer, but not final. Take whatever the answer was (in this case 48) and add it to the orignal number (600) to get $648.
2×2=4
4÷2=2
4+2=6
4+2+6=12
we know that,
put a number you like anything I will write x°
x°+12°=90°
90°-12°=78°,,
again,
78°-12°=66°,,
now,
remove 6° from 60°
60
-----
12
=5,,
The value of n given the form of the function is (b) positive odd number
<h3>How to interpret the graph?</h3>
The form of the graph is given as:
f(x) = a(x + k)^1/n + c
For the given graph, we have the following features:
a > 0 --- a is positive
k > 0 --- k is positive
c < 0 --- c is negative
If n is an even number, the function would be undefined because the even root of a number is undefined
However, the function is defined if n is an odd number,
Hence, the value of n given the form of the function is a positive odd number
Read more about functions at:
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Step-by-step explanation:
$800 × 5% = $40
for 1 year, I earn $40
100 ÷ 40 = 2 remainder 2
Therefore I would have to leave the money in for at least 3 years to earn $100
Answer:
Using either method, we obtain: 
Step-by-step explanation:
a) By evaluating the integral:
![\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cint%5Climits%5Et_0%20%7B%5Csqrt%5B8%5D%7Bu%5E3%7D%20%7D%20%5C%2C%20du)
The integral itself can be evaluated by writing the root and exponent of the variable u as: ![\sqrt[8]{u^3} =u^{\frac{3}{8}](https://tex.z-dn.net/?f=%5Csqrt%5B8%5D%7Bu%5E3%7D%20%3Du%5E%7B%5Cfrac%7B3%7D%7B8%7D)
Then, an antiderivative of this is: 
which evaluated between the limits of integration gives:

and now the derivative of this expression with respect to "t" is:

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:
"If f is continuous on [a,b] then

is continuous on [a,b], differentiable on (a,b) and 
Since this this function
is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:
