Answer:
Step-by-step explanation:
The expression shows the sum of two terms. The first term has coefficient 1. The second term is the product of 12 and x.
Answer:
There are 122 one dollar bills, 11 five dollar bills and 5 ten dollar bills.
Step-by-step explanation:
There are bills of one dollar, five dollars and ten dollars on the cash drawer, therefore the sum of all of them multiplied by their respective values must be equal to the total amount of money on the drawer. We will call the number of one dollar bills, five dollar bills and ten dollar bills, respectively "x","y" and "z", therefore we can create the following expression:
We know that there are six more 5 dollar bills than 10 dollar bills and that the number of 1 dollar bills is two more than 24 times the number of 10 dollar bills, therefore:
Applying these values on the first equation, we have:
Applying z to the formulas of y and x, we have:
There are 122 one dollar bills, 11 five dollar bills and 5 ten dollar bills.
Answer:
I don't know how to explain the math, but I know the answer is 115 women for every 100 men.
Hello,
I suppose you want the equation
There are 2 infinite solutions:
y=k(x-5)²(x+5)
or
y=k(x-5)(x+5)²
Answer:
R3 <= 0.083
Step-by-step explanation:
f(x)=xlnx,
The derivatives are as follows:
f'(x)=1+lnx,
f"(x)=1/x,
f"'(x)=-1/x²
f^(4)(x)=2/x³
Simialrly;
f(1) = 0,
f'(1) = 1,
f"(1) = 1,
f"'(1) = -1,
f^(4)(1) = 2
As such;
T1 = f(1) + f'(1)(x-1)
T1 = 0+1(x-1)
T1 = x - 1
T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2
T2 = 0+1(x-1)+1(x-1)^2
T2 = x-1+(x²-2x+1)/2
T2 = x²/2 - 1/2
T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3
T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3
T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)
Thus, T1(2) = 2 - 1
T1(2) = 1
T2 (2) = 2²/2 - 1/2
T2 (2) = 3/2
T2 (2) = 1.5
T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)
T3(2) = 4/3
T3(2) = 1.333
Since;
f(2) = 2 × ln(2)
f(2) = 2×0.693147 =
f(2) = 1.386294
Since;
f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).
Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,
Since;
f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2
Finally;
R3 <= |2/(4!)(2-1)^4|
R3 <= | 2 / 24× 1 |
R3 <= 1/12
R3 <= 0.083
