<span>w+6tw=11-w
</span>
<span>Let's solve for w.
</span><span><span>
w+<span><span>6t</span>w</span></span>=<span>11−w
</span></span>
Step 1: Add w to both sides.<span><span><span><span><span>
6t</span>w</span>+w</span>+w</span>=<span><span><span>−w</span>+11</span>+w</span></span><span><span><span><span>
6t</span>w</span>+<span>2w</span></span>=11</span><span>
Step 2: Factor out variable w.
</span><span><span>
w<span>(<span><span>6t</span>+2</span>)</span></span>=11</span><span>
Step 3: Divide both sides by 6t+2.
</span><span><span>
w<span>(<span><span>6t</span>+2</span>)</span></span><span><span>/6t</span>+2</span></span>=<span>11/<span><span>6t</span>+2</span></span>
Answer: 11/6t+2
Step-by-step explanation:
(a+bi)2
=a2 +2abi +b2i2
=a2 +2abi + b2. (-1)
=a2 -b2 +2abi
Answer = (a+b(a-b) +2abi
Answer:
2/5 or .4
Step-by-step explanation:
basically look at the points where the line meets the graph. two points would be 0,20 and 50,40. then count the amount that it changes vertically and then horizontally. +20vertical +50 horizonal giving the slope 20/50. this can then be simplified to 2/5 or the equivalent number .4
8/18 because just times them by 2 and it will be 8/28.
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