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: Equilateral triangles are also right triangles sometimes. All angles of an equilateral are congruet sometimes.
Answer:
a Euclidean space, the sum of angles of a triangle equals the straight angle (180 degrees, π radians, two right angles, or a half-turn). A triangle has three angles, one at each vertex, bounded by a pair of adjacent sides.
Original coordinates of the points:
A (8,15) ; B (12,13) ; C (8,10)
Dilated scale factor of 3.
A ⇒ 3x = 3(8) = 24 ; 3y = 3(15) = 45 ⇒ A' (24,45)
B ⇒ 3x = 3(12) = 36 ; 3y = 3(13) = 39 ⇒ B' (36, 39)
C ⇒ 3x = 3(8) = 24 ; 3y = 3(10) = 30 ⇒ C' (24, 30)
The given image forms a right triangle. So, I'll get the short leg and long leg of the right triangle to solve for the hypotenuse, length of CB.
Short leg: y value of B and C
39 - 30 = 9
Long leg: x value of B and C
36 - 24 = 12
a² + b² = c²
9² + 12² = c²
81 + 144 = c²
225 = c²
√225 = √c²
15 = c
The length of CB is 15 units.
