A projectile is fired vertically upward with an initial velocity of 210 m/sec from a point 2 m above the ground. Ifs meters is the height of the projectile above the ground t seconds after being fired, under the assumption that the only force acting on it is acceleration due to gravity, how high above the ground (in meters) will the projectile go? Note: Use a = 9.8m/sec² for acceleration due to gravity.

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A: C=x,y: 1≤x≤2, y=lnx+x2-1

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A:

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A: The points on the curve y=x33-x where the tangent line is parallel to the line y=3x

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A:

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A: SOLUTION-

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A:

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A:

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A: Solution:

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A: Since you have posted multiple questions i can do first question as per our company guidelines…

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A: We have to find the solution

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A:

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A: det-46127Use deta  b c  d  = ad-bc=-4· 7-6·12=-31

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A: using the simpson's method ∫01xdx for n=4.

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A:

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A: Let's solve given integration.

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A:

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A:

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A: We find the line integral in the following way

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A: ∫-11x7-4x9+sinx1+x22dx

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A:

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A:

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A:

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A: Evaluate the ff. by integration by parts.

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A: We can answer the question as below.

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A: Please see the below picture for detailed solution.

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A:

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A:

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A: I have appended the formula used in the solution part of the question

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A:

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A: Solution

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A:

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A: All Option are  correct So we have to choose Option 'A'

Q: Addition:Subtraction as___:antidifferentiation

A: We can answer the question as below.