Answer:
(160, 36)
Step-by-step explanation:
Distance is always positive because, well, you aren't -50 meters away from your house or something. You are just 50 meters away from your house. And 160 is the distance between -6 and 154, because there is one minus and one plus, you add them, because that is how far they are apart, and then-43 and -7 were easy just subtract them.
Answer:
$12.80
Step-by-step explanation:
Before markup: $8
After markup: $8 x 160% = $12.80
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Answer:
-2
Step-by-step explanation:
x^2 + x=2
Subtract 2 from each side
x^2 + x-2=2-2
x^2 + x-2=0
Factor
What 2 numbers multiply to -2 and add to 1
2*-1 = -2
2+-1 =1
(x-1)(x+2)=0
Using the zero product property
x-1 = 0 x+2 = 0
x=1 x = -2
Product of the roots
1*-2 = -2
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
