To answer this question we will use the following formula for compounded monthly interest:

where A₀ is the initial amount, r is the annual interest as a decimal number and 12 is the number of years.
Notice that:

Therefore, substituting t=42, r=0.02, and A₀=120000 we get:

Simplifying the above result we get:

Answer: $277,769.75.
Answer:
(gºf)(0) is 6
Step-by-step explanation:
It is a composition of functions problem, where
(gºf) = g(f(x)) is the composition of the functions g(x) and f(x). (gºf)(0) is the composition of the functions at the point x = 0. First, we find that function that is the g composite of x, then we replace x by 0.
g(4), for example, is g when x = 4, right? So g(4) = 4.
So,
g(f(x)) is going to represent g when x = f(x). x = f(x) when x = 2x + 6, so:
g(f(x)) = g(2x+6) = 2x + 6
We have that
g(f(x)) = gºf(x) = 2x + 6
Now, we just have to find the value of the function at x = 0. So
gºf(0) = 2(0) + 6 = 0 + 6 = 6.
(gºf)(0) is 6
take the √(-1-3)^2+(3-(-4))^2
Which should give you √65 or about 8.062
9514 1404 393
Answer:
4) 6x
5) 2x +3
Step-by-step explanation:
We can work both these problems at once by finding an applicable rule.

where O(h²) is the series of terms involving h² and higher powers. When divided by h, each term has h as a multiplier, so the series sums to zero when h approaches zero. Of course, if n < 2, there are no O(h²) terms in the expansion, so that can be ignored.
This can be referred to as the <em>power rule</em>.
Note that for the quadratic f(x) = ax^2 +bx +c, the limit of the sum is the sum of the limits, so this applies to the terms individually:
lim[h→0](f(x+h)-f(x))/h = 2ax +b
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4. The gradient of 3x^2 is 3(2)x^(2-1) = 6x.
5. The gradient of x^2 +3x +1 is 2x +3.
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If you need to "show work" for these problems individually, use the appropriate values for 'a' and 'n' in the above derivation of the power rule.