A=p (1+I/k^kn
A future value ?
P principle 3700
i interest rate 0.0184
K compounded quarterly 4
N time 6 years
A=3,700×(1+0.0184÷4)^(4×6)
A=4,130.84
Interest earned=A-p
4,130.84−3,700=430.84
Hope it helps:-)
Answer:
21
Step-by-step explanation:
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Answer:
Option D: is the function
Explanation:
Let the general form of quadratic equation be The function passes through the intercepts and and also passes though the point Substituting the points , and in the equation , we get, -----------(1) ----------(2) -----------(3)
Subtracting (1) and (2), we get, -----------(4)
Subtracting (2) and (3), we get, ------------(5)
Multiplying equation (4) by 5 and equation (5) by 4, to cancel the term b when adding, we get,
Thus, the value of a is Substituting in equation (4), we get,
Thus, the value of b is Now, substituting the value of a and b in equation (1), we have,
Thus, the value of c is Now, substituting the value of a,b and c in the general formula , we get,
Taking out the common term as -2 we get,
Factoring , we get,
Thus, the function is
I think it would be A because out of all of the graphs that I can see only A has (3,3) plotted on it
