Answer:
The number of fiction books is <u>500</u> and non fiction books is <u>1500</u>.
Step-by-step explanation:
Given:
Total book library contains = 2000.
There are 3 times as many non fiction books as fiction books.
Now, to determine the number of non fiction books and fiction books.
Let the fiction books be 
And the non fiction books be 
Total books = 2000.
Now, to get the number of non fiction books and fiction books we put an equation:
Dividing both sides by 4 we get:
Fiction books = 500.
Non fiction books = 
Therefore, the number of fiction books is 500 and non fiction books is 1500.
Answer:
Total 390 hardcover books were sold
Step-by-step explanation:
Complete question
The author received a total of $195,000 for the book sales. How many hardcover books were sold if the cost of one hardcover book is $500. Type in numbers only.
Solution
Given-
Total value of sale of hard cover = $195,000
Cost of one hardcover book = $500
Number of books sold = Total value of sale of hard cover/Cost of one hardcover book
Number of books sold = $195,000/$500
Number of books sold = 390
Total 390 hardcover books were sold.
Answer:
-13
Step-by-step explanation:
Subsitude -2 to where x is and you get -1o then subtract by -3 and the total is -13
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
