Answer:
B
Step-by-step explanation:
Of you put the values of c and y from B in the equation, then the result is,
here, 3 is less than 4, which is true.
But the result will come out as greater than 4 if you put other points in the equation, which is not true for the given condition.
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
All i know is that if the radius of a circle is 4 feet, the the diameter is 8 feet. Sorry, I don't know the rest.
Answer:
x = -8
y = 7
Step-by-step explanation:
5x + 7y = 9
2x - 3y = -37
1.) First, multiply each side to match either y-values or x-values. (In this example, we'll use match x-values)
10x + 14y = 18 (multiplied by 2)
10x - 15y = -185 (multiplied by 5)
2.) Then, subtract the entirety of one equation to isolate the y-value.
10x + 14y = 18
-10x + 15y = 185
3.) Add and subtract values and divide to find y.
29y = 203
y = 7
4.) Plug-in y to solve for x into one equation, or repeat steps 1-3.
15x + 21y = 27
14x - 21y = -259
29x = -232
x = -8
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