Answer:
6194.84
Step-by-step explanation:
Using the formula for calculating accumulated annuity amount
F = P × ([1 + I]^N - 1 )/I
Where P is the payment amount. I is equal to the interest (discount) rate and N number of duration
For 40 years,
X = 100[(1 + i)^40 + (1 + i)^36 + · · ·+ (1 + i)^4]
=[100 × (1+i)^4 × (1 - (1 + i)^40]/1 − (1 + i)^4
For 20 years,
Y = A(20) = 100[(1+i)^20+(1+i)^16+· · ·+(1+i)^4]
Using X = 5Y (5 times the accumulated amount in the account at the ned of 20 years) and using a difference of squares on the left side gives
1 + (1 + i)^20 = 5
so (1 + i)^20 = 4
so (1 + i)^4 = 4^0.2 = 1.319508
Hence X = [100 × (1 + i)^4 × (1 − (1 + i)^40)] / 1 − (1 + i)^4
= [100×1.3195×(1−4^2)] / 1−1.3195
X = 6194.84
Answer:
????
Step-by-step explanation:
Answer:
h(-8)=-15
Step-by-step explanation:
Substitute t=-8
h(-8) =-2(-8+5)^2+3
=-2(-3)^2+3
=-2(9)+3
=-18+3
=-15
Pretty sure its the 3rd one
Answer:
x = 10
Explanation:
Using Pythagoras Theorem:
Solve:
8² + 6² = x²
x² = 64 + 36
x² = 100
x = √100
x = 10
