It's 416,666,667/1,000,000,000. //// If you said that the 6s keep going and never end, then it would be the decimal form of 5/12 .
Answer: if you mean 2/1 the answer would be 4/2, 6/3 and 8/4 if you mean 1/2 the answer would be 2/4,3/6 and 4/8 good luck!
Step-by-step explanation:
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
Answer:
32
Step-by-step explanation:
The equation 
and the graph below describe the velocity V in feet per second of a ball <em>t</em> seconds after it is shot into the air.
To find the balls velocity after seven seconds, substitute t = 7 into the function expression:
<h3><u>The value of the larger number, x, is 57.</u></h3><h3><u>The value of the smaller number, y, is equal to 42.</u></h3>
x - y = 15
2x + 8 = 3y - 4
We can quickly get a temporary value for x by altering the original equation.
x - y = 15
<em><u>Add y to both sides.</u></em>
x = 15 + y
Now that we have a value of x, we can find the exact value of y.
2(15 + y) + 8 = 3y - 4
<em><u>Distributive property.</u></em>
30 + 2y + 8 = 3y - 4
<em><u>Combine like terms.</u></em>
38 + 2y = 3y - 4
<em><u>Subtract 2y from both sides.</u></em>
38 = y - 4
<em><u>Add 4 to both sides.</u></em>
y = 42
Now that we know the exact value of y, we can plug it back into the original equation.
x - 42 = 15
<em><u>Add 42 to both sides.</u></em>
x = 57
