Here we want to have x on one side of equation so,
we put number 5 on right side of equation:
abx=5
then we divide both sides by ab to obtain x on left side:
Result:
x=<u />
Step-by-step explanation:
I'll do the first problem as an example.
∠P and ∠H both have one mark. That means they're congruent.
∠T and ∠G both have two marks. So they're congruent.
∠W and ∠D both have three marks. So they're congruent.
So we can write a congruence statement:
ΔPTW ≅ ΔHGD
We can write more congruence statements by rearranging the letter, provided that corresponding pairs have the same position (P is in the same place as H, etc.). For example:
ΔWPT ≅ ΔDHG
ΔTWP ≅ ΔGDH
Answer:
203/11x
Step-by-explanation: I am soo very sorry gave you the wrong answer before but going over it I see it is 203/11x
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:
If you want to know what b is then it is 16
Step-by-step explanation:
38 + b = 54
54- 38 = 16
38 + 16= 54
