Answer:
(a) A = 700×1.015^t
(b) 36.2 years
Step-by-step explanation:
(a) Each year, the account value is multiplied by (1 + 1.5%) = 1.015. Repeated multiplication is signified using an exponent. In t years, when the account has been multiplied by 1.015 t times, the account value will be ...
A = $700×1.015^t
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(b) You want to find t when A=$1200. Logarithms are involved.
1200 = 700×1.015^t . . . . use 1200 for A
1200/700 = 1.015^t . . . . . divide by 700
log(12/7) = t×log(1.015) . . . . . take logarithms
log(12/7)/log(1.015) = t ≈ 36.2 . . . . divide by the coefficient of t
It will take about 36.2 years for the account balance to reach $1200.
X = -4
As g(x) is replaced by g(-4)
plug in -4 for x inside the equation.
g(-4) = (-4) - 1g(-4) = -5
Plug in -4 for x in f(x)f(-4) = 4(-4) - 4f(-4) = -16 - 4f(-4) = -20
Now solve for f(x) over g(x)
remember, f(x) = -20, and g(x) = -5
-20/-5 = 4
4 should be your answer
hope this helpsClick to let others know, how helpful is it
Answer:
THE ANSWER IS A
Step-by-step explanation:
Composition of a function is done by substituting one function into another function. For example, f [g (x)] is the composite function of f (x) and g (x). The composite function f [g (x)] is read as “f of g of x”. The function g (x) is called an inner function and the function f (x) is called an outer function.
The notation used for the composition of functions looks like this, (f g)(x). ... (f g)(x) and(g f)(x) are often different because in the composite(f g)(x), f(x) is the outside function and g(x) is the inside function. Whereas in the composite(g f)(x), g(x) is the outside function and f(x) is the inside function.
GIVE BRAINLIST IF HELPED
Calculating
the value of f(x) for the given interval.
For x = -
4, f(x) = f(- 4) = (- 4)^2 + 2 (- 4) + 3 = 11
For x =
6, f(x) = f(6) = (6)^2 + 2 (6) + 3 = 51
Now using
formula for the calculation of average rate of change of f(x) over the given
interval of [- 4, 6];
(f(b) –
f(a)) / b – a = (f(6) – f(- 4)) / 6 – (- 4) = (51 -11) / 10 = 4
<span>So option “E” is
correct.</span>
