A.) is the correct answer.
<span>B(n) = A(1 + i)^n - (P/i)[(1 + i)^n - 1]
where B is the balance after n payments are made, i is the monthly interest rate, P is the monthly payment and A is the initial amount of loan.
We require B(n) = 0...i.e. balance of 0 after n months.
so, 0 = A(1 + i)^n - (P/i)[(1 + i)^n - 1]
Then, with some algebraic juggling we get:
n = -[log(1 - (Ai/P)]/log(1 + i)
Now, payment is at the beginning of the month, so A = $754.43 - $150 => $604.43
Also, i = (13.6/100)/12 => 0.136/12 per month
i.e. n = -[log(1 - (604.43)(0.136/12)/150)]/log(1 + 0.136/12)
so, n = 4.15 months...i.e. 4 payments + remainder
b) Now we have A = $754.43 - $300 = $454.43 so,
n = -[log(1 - (454.43)(0.136/12)/300)]/log(1 + 0.136/12)
so, n = 1.54 months...i.e. 1 payment + remainder
</span>
1: 8 faces and 9 with the base 9 vertices and 16 edges
2: 3 faces and 5 with the bases 6 vertices and 9 edges
3: 3 faces and 4 with the base 4 vertices and 6 edges
Hope this can help you.
Answer:
Volume of original toolbox = 180 in³
Yes, doubling one dimension only would double the volume of the toolbox.
Step-by-step explanation:
Volume = L x W x H
10 x 6 x 3 = 180 in³
proof:
double length = 20 x 6 x 3 = 360 in³, which is double the original
double width = 10 x 12 x 3 = 360 in³, which is double the original
double height = 10 x 6 x 6 = 360 in³, which is double the original
