It would be 24 units^2 because the middle is 16 and the two triangles are 4 each
Answer:
Order from Least to Greatest
-4/7 < -1/19 < 7/4
Showing Work
Rewriting as fractions or any negatives if necessary:
-1/19, 7/4, -4/7
The least common denominator (LCD) is: 532.
Rewriting as equivalent fractions with the LCD:
-28/532, 931/532, -304/532
Ordering these fractions by the numerator:
-304/532 < -28/532 < 931/532
Therefore, the order of your input is:
-4/7 < -1/19 < 7/4
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: X = 3t, Y =2 - t, Z =2
Step-by-step explanation: the plane
x + y + z =4has normal vector
M =<1,1,1> and the line
x = 1 + t, y = 2 − t, z = 2t has direction
v =<1, −1, 2>. So the vector
A= n × v
=<1, 1, 1> × <1, −1, 2>
=<2−(−1),1−2,−1−1>
=<3,−1,−2>
