(a) If the particle's position (measured with some unit) at time <em>t</em> is given by <em>s(t)</em>, where
then the velocity at time <em>t</em>, <em>v(t)</em>, is given by the derivative of <em>s(t)</em>,
(b) The velocity after 3 seconds is
(c) The particle is at rest when its velocity is zero:
(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:
In interval notation, this happens for <em>t</em> in the interval (0, √11) or approximately (0, 3.317) s.
(e) The total distance traveled is given by the definite integral,
By definition of absolute value, we have
In part (d), we've shown that <em>v(t)</em> > 0 when -√11 < <em>t</em> < √11, so we split up the integral at <em>t</em> = √11 as
and by the fundamental theorem of calculus, since we know <em>v(t)</em> is the derivative of <em>s(t)</em>, this reduces to
Answer:
i believe its 1,000
Step-by-step explanation:
HOPE THIS HELPED PLS GIVE ME BRAINILEST
Answer:
Subtract from both sides of the equation the term you don't want
Step-by-step explanation:
In solving equations, you generally want to "undo" operations that are done to the variable. Addition is "undone" by adding the opposite (that is, subtracting the amount that was added). Multiplication is "undone" by division.
If you have variables on both sides of the equation, pick one of the variable terms and subtract it from both sides of the equation.
<u>Example</u>
2x = x +1
If we choose to subtract x, then we will have a variable term on the left and a constant term on the right:
2x -x = x -x +1 . . . . . . . x is subtracted from both sides
x = 1 . . . . . . simplify
__
Note that we purposely set up this example so that removing the variable term from the right side caused the variable term and constant term to be on opposite sides of the equal sign. It may not always be that way. As long as you remember that an unwanted term can be removed by subtracting it (from both sides of the equation), you can deal with constant terms and variable terms no matter where they appear.
_____
<em>Additional Comment</em>
It usually works well to choose the variable term with the smallest (or most negative) coefficient. That way, when you subtract it, you will be left with a variable term that has a positive coefficient.
The Pythagorean Theorem formula is listed below. It can only be applied to right triangles.
(side1)^2 + (side2)^2 = (hypotenuse)^2
Since we are given that the base is 8cm, we know that the side of one triangle is 4cm. We are also given that the hypotenuse is 8cm. The height of the cone will be one of the short sides. If we plug these numbers into the formula, we get the following...
4^2 + height^2 = 8^2
If we solve for height, we get the following...
height = sqrt(8^2 - 4^2) = sqrt(64 - 16) = sqrt(48)
Since we are only asked for what is inside the square root, our answer is 48.
