Sin51=y/12
y=12sin51 units
y≈9.33 units (to nearest hundredth of a unit)
...
tanα=12/5
α=arctan2.4°
α≈67.38° (to nearest hundredth of a degree)
...
tan13=x/24
x=24tan13 units
x≈5.54 units (to nearest hundredth of a unit)
...
sin20=10/x
x=10/sin20 units
x≈29.24 units (to nearest hundredth of a unit)
I think it is already reduced.
How you a doin?
Answer:
Step-by-step explanation:
Mrs. Siebenaller bought a bus for 25,000 with a 7% interest rate and she gets a loan payoff of 60 months,
We know that,
![\text{PV of annuity}=P\left[\dfrac{1-(1+r)^{-n}}{r}\right]](https://tex.z-dn.net/?f=%5Ctext%7BPV%20of%20annuity%7D%3DP%5Cleft%5B%5Cdfrac%7B1-%281%2Br%29%5E%7B-n%7D%7D%7Br%7D%5Cright%5D)
Where,
PV = Present value of annuity = 25000,
r = rate of interest of each period = 
% monthly
n = number of periods = 60 months,
Putting the values,
![\Rightarrow 25000=P\left[\dfrac{1-(1+\frac{0.07}{12})^{-60}}{\frac{0.07}{12}}\right]](https://tex.z-dn.net/?f=%5CRightarrow%2025000%3DP%5Cleft%5B%5Cdfrac%7B1-%281%2B%5Cfrac%7B0.07%7D%7B12%7D%29%5E%7B-60%7D%7D%7B%5Cfrac%7B0.07%7D%7B12%7D%7D%5Cright%5D)
![\Rightarrow P=\dfrac{25000}{\left[\dfrac{1-(1+\frac{0.07}{12})^{-60}}{\frac{0.07}{12}}\right]}](https://tex.z-dn.net/?f=%5CRightarrow%20P%3D%5Cdfrac%7B25000%7D%7B%5Cleft%5B%5Cdfrac%7B1-%281%2B%5Cfrac%7B0.07%7D%7B12%7D%29%5E%7B-60%7D%7D%7B%5Cfrac%7B0.07%7D%7B12%7D%7D%5Cright%5D%7D)
Hence total amount paid is,
Therefore interest amount is,
The discounts on a pair of shoes with a regular price of $48 is 24
Answer:
see below the first three problems
Step-by-step explanation:
f(g(-2))
First, find g(-2) using function g(x). Then use that value as input for function f(x).
g(x) = -2x + 1
g(-2) = -2(-2) + 1
g(-2) = 5
f(x) = 5x
f(5) = 5(5)
f(5) = 25
f(g(-2)) = 25
g(h(3))
First, find h(3) using function h(x). Then use that value as input for function g(x).
h(x) = x^2 + 6x + 8
h(3) = 3^2 + 6(3) + 8 = 9 + 18 + 8
h(3) = 35
g(x) = -2x + 1
g(35) = -2(35) + 1 = -70 + 1
g(35) = -69
g(h(3)) = -69
f(g(3a))
First, find g(3a) using function g(x). Then use that value as input for function f(x).
g(x) = -2x + 1
g(3a) = -2(3a) + 1
g(3a) = -6a + 1
f(x) = 5x
f(-6a + 1) = 5(-6a + 1)
f(-6a + 1) = -30a + 5
f(g(3a)) = -30a + 5
