Answer:
It's position at time t = 5 is 593.
Step-by-step explanation:
The velocity v(t) is the integral of the acceleration a(t)
The position s(t) is the integral of the velocity v(t)
We have that:
The acceleration is:
Velocity:
K is the initial velocity, that is v(0). Since V(0) = 13, K = 13
Then
Position:
Since s(0) = 3
What is its position at time t=5?
This is s(5).
It's position at time t = 5 is 593.
Hello:
<span>the symmetries : axis- x </span>
Uh it can really be anything. You should try and be more specific: see if you did the whole problem.
Note that x² + 2x + 3 = x² + x + 3 + x. So your integrand can be written as
<span>(x² + x + 3 + x)/(x² + x + 3) = 1 + x/(x² + x + 3). </span>
<span>Next, complete the square. </span>
<span>x² + x + 3 = x² + x + 1/4 + 11/4 = (x + 1/2)² + (√(11)/2)² </span>
<span>Also, for the x in the numerator </span>
<span>x = x + 1/2 - 1/2. </span>
<span>So </span>
<span>(x² + 2x + 3)/(x² + x + 3) = 1 + (x + 1/2)/[(x + 1/2)² + (√(11)/2)²] - 1/2/[(x + 1/2)² + (√(11)/2)²]. </span>
<span>Integrate term by term to get </span>
<span>∫ (x² + 2x + 3)/(x² + x + 3) dx = x + (1/2) ln(x² + x + 3) - (1/√(11)) arctan(2(x + 1/2)/√(11)) + C </span>
<span>b) Use the fact that ln(x) = 2 ln√(x). Then put u = √(x), du = 1/[2√(x)] dx. </span>
<span>∫ ln(x)/√(x) dx = 4 ∫ ln u du = 4 u ln(u) - u + C = 4√(x) ln√(x) - √(x) + C </span>
<span>= 2 √(x) ln(x) - √(x) + C. </span>
<span>c) There are different approaches to this. One is to multiply and divide by e^x, then use u = e^x. </span>
<span>∫ 1/(e^(-x) + e^x) dx = ∫ e^x/(1 + e^(2x)) dx = ∫ du/(1 + u²) = arctan(u) + C </span>
<span>= arctan(e^x) + C.</span>
Similar polygons only differ by a scaling factor. In other words, two polygons are similar if one is the scaled version of the other.
In particular, this implies that the angles are preserved, and the correspondent sides are in proportion.
These two polygons are both rectangles, so the angles are preserved. We must check the sides, and we have to check if the smaller sides are in the same proportion as the bigger sides.
So, the two rectangles are similar if the following is true.
In any proportion, the product of the inner terms must be the same as the product of the outer terms:
This is clearly false, and thus the two rectangles are not similar.
