Nine plus a unknown number equals unknown.
9514 1404 393
Answer:
- $137.90 more each month
- $246.00 less total interest
Step-by-step explanation:
The amortization formula is ...
A = P(r/12)/(1 -(1 +r/12)^(-12t))
for the monthly payment on principal P at annual rate r for t years. Here, we have P=3300, r = 0.14, and t=1, so the monthly payment is ...
A = $3300(0.14/12)/(1 -(1 +0.14/12)^-12) ≈ $296.30
The payment of $296.30 is ...
$295.30 -158.40 = $137.90 . . . more each month
The total amount paid is 12×$296.30 = $3555.60, so 255.60 in interest. This amount is ...
$501.60 -255.60 = $246.00 . . . less total interest
Markup =35%
Model train cost =100
So 135%=100
1%=100/135
100%=100/35)*100=74.07
Original cost =74.07
(tan²(<em>θ</em>) cos²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>))
Recall that
tan(<em>θ</em>) = sin(<em>θ</em>) / cos(<em>θ</em>)
so cos²(<em>θ</em>) cancels with the cos²(<em>θ</em>) in the tan²(<em>θ</em>) term:
(sin²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>))
Recall the double angle identity for cosine,
cos(2<em>θ</em>) = 2 cos²(<em>θ</em>) - 1
so the 1 in the denominator also vanishes:
(sin²(<em>θ</em>) - 1) / (2 cos²(<em>θ</em>))
Recall the Pythagorean identity,
cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1
which means
sin²(<em>θ</em>) - 1 = -cos²(<em>θ</em>):
-cos²(<em>θ</em>) / (2 cos²(<em>θ</em>))
Cancel the cos²(<em>θ</em>) terms to end up with
(tan²(<em>θ</em>) cos²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>)) = -1/2
