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➷ Form an equation with 'x' as the unknown number
7x + 3 = -74
Subtract 3 from both sides:
7x = -77
Divide both sides by 7:
x = -11
The number is -11
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Answer:
t = 0.293 s and 14.52 s
Step-by-step explanation:
The heights of the ball in feet is given by the equation as follows :
Where t is the number of seconds after the ball was thrown.
We need to find is the ball 18 ft above the moon's surface.
Put S = 18 ft
It is a quadratic equation. Its solution is given by :
So, the ball is at first 0.293 s and then 14.52 s above the Moon's surface.
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:
point B
Step-by-step explanation:
First you find (-2, -1) on the graph then you go up two units. there is not a point directly above (-2, -1) but if you look to the side B is exactly 2 units above which is what the question asked for.
Answer:
The coordinate r is the length of the line segment from the point (x,y) to the origin and the coordinate θ is the angle between the line segment and the positive x-axis.
Step-by-step explanation:
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