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Rasheda

Questions (4)

A particle starts at the point (5,0) at t=0 and moves along the x-axis in such a way that at time t>0 its velocity v(t) is given
1. 3 answers
2. 5,548 views
In the interval (0 is less than or equal to x which is less than or equal to 5), the graphs of y=cos(2x) andy=sin(3x) intersect
1. 0 answers
2. 916 views
What is the integral of (x*f(x))dx?
1. 2 answers
2. 317 views
I'm a little confused with this integration problem: If the definite integral from 0 to 2 of (e^(x^2)) is first approximated by
1. 5 answers
2. 5,004 views

a). 4^(-1) is the same as 1/4. Since this is the diameter, you divided it by the 2 to get the radius. So the radius is 1/8. Now the expression is: A= pi*(1/8)^2 or A=(1/64)*pi b). In part a, the expression is for the area of one punched hole. Since you
1. The first thing you need to do is convert 60km to meters, or 10,000m to kilometers; whichever you prefer. Next,it helps if you draw a right triangle. The plane is at a height of 10,000m; this is the vertical leg of the triangle. It is 60 km away from
Whenever you have a number raised to a negative power, you have to divide it. So 6y^-3 becomes (6)/(y^3)
So you have (8^6)/((8^4)*(8^2)). I would start with the denominator. When you are multiplying exponents with the same base, you add the exponents. So the above equation becomes (8^6)/(8^6). This simplifies to 1. For the second problem, it's slightly
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