Answer:
.. q T 0 = (q/p)a (q/p)a − (q/p)T−10 if p ≠ q and qT0 = 1 − T0/a if p = q = 1/2.
Step-by-step explanation:
Suppose that there are two different solutions, p and q, in [a, b]. Thus p =g(p) q =g(q) p ≠ q The function g(x) satisfies the hypotheses of the mean-value ... that g(p) –g(q) = (p – q) g׳(t) Because g(p) =p and g(q) = q, the left side of Eq. (1-3) may...
Answer:
6x^2+17x+9
Step-by-step explanation:
you have to subtract the area of the white rectangle from the area of the entire rectangle
to find the area of the inner rectangle do (2x-3)*(x+1)=2x^2-3x+2x-3=2x^2-x-3
to find the area of the entire rectangle do (4x+6)*(2x+1)=8x^2+12x+4x+6=8x^2+16x+6
(8x^2+16x+6)-(2x^2-x-3)=8x^2+16x+6-2x^2+x+3=6x^2+17x+9
ince the problem is only asking for 4 years, we can just calculated it out year by year. Recall the formula for compounding interest: A = P(1+r)n, where A is the total amount, P is the principle (amount you start with), r is the interest rate per period of time, and n is the number of periods (in this case, r is annual interest rate, so n is number of years). At the beginning (Year 0), Lou starts off with 10000: A = 10000 At the end of Year 1, Lou earned interest on that amount, plus he has deposited another 5000: A = 10000(1.08) + 5000 End of Year 2, Lou's interest from the year 0 amount has compounded, he has started earning interest on the amount deposited last year, and he deposits another 5000: A = 10000(1.08)2 + 5000(1.08) + 5000 End of Year 3, same idea. Lou has earned compounding interest on all existing deposits, and deposits another 5000: A = 10000(1.08)3 + 5000(1.08)2 + 5000(1.08) + 5000 End of Year 4, same idea: A = 10000(1.08)4 + 5000(1.08)3 + 5000(1.08)2 + 5000(1.08) + 5000 = 36135.45
Answer:
A) 1
Step-by-step explanation:
2(1+x)= x+3
2(1+1)= 1+3
2×2= 4
4=4
Hence proven
