Answer:
71.6056
Step-by-step explanation:
c² = a² + b² - 2abcosC°
29² = 30² + 15² - 2(30)(15)cosC°
841 = 900 + 225 - 900cosC°
-59 = 225 - 900cosC°
-284 = -900cosC°
71/225 = cosC°
cos⁻¹(71/225) = C°
C° = 71.6056
Answer:
I'm sorry I would help you I don't understand the second sentence.
Step-by-step explanation:
Answer:
20x + 18
Step-by-step explanation:
We need to use the distributive property, where we essentially take the sum of the product of the outside number with each of the inside terms.
In 7(4x - 2), 7 is the outside number and 4x and -2 are the inside numbers, so:
7(4x - 2) = 7 * 4x + 7 * (-2) = 28x - 14
In 4(2x - 8), 4 is the outside number and 2x and -8 are the inside numbers, so:
4(2x - 8) = 4 * 2x + 4 * (-8) = 8x - 32
Now, we have:
28x - 14 - (8x - 32) = 28x - 14 - 8x + 32 = 20x + 18
The answer is 20x + 18.
Answer:
3/500
Step-by-step explanation:
3/5 x 1%
=> 3/5 x 1/100
=> 3/500
Hope it helps you
In an installment loan, a lender loans a borrower a principal amount P, on which the borrower will pay a yearly interest rate of i (as a fraction, e.g. a rate of 6% would correspond to i=0.06) for n years. The borrower pays a fixed amount M to the lender q times per year. At the end of the n years, the last payment by the borrower pays off the loan.
After k payments, the amount A still owed is
<span>A = P(1+[i/q])k - Mq([1+(i/q)]k-1)/i,
= (P-Mq/i)(1+[i/q])k + Mq/i.
</span>The amount of the fixed payment is determined by<span>M = Pi/[q(1-[1+(i/q)]-nq)].
</span>The amount of principal that can be paid off in n years is<span>P = M(1-[1+(i/q)]-nq)q/i.
</span>The number of years needed to pay off the loan isn = -log(1-[Pi/(Mq)])/(q log[1+(i/q)]).
The total amount paid by the borrower is Mnq, and the total amount of interest paid is<span>I = Mnq - P.</span>
