Y = t*e^(-t/2)
y' = t' [e^(-t/2)] + t [e^(-t/2)]' = e^(-t/2) + t[e^(-t/2)][-1/2]=
y' = [e^(-t/2)] [1 - t/2] = (1/2)[e^(-t/2)] [2 - t] = - (1/2) [e^-t/2)] [t -2]
Answer:
A. a red marble
there's more red marbles than blue and white marbles
Answer:
(200)+(50)+(0)
Step-by-step explanation:
you just take all the numbers and break them up
Answer:
(17)
Sum of interior angles of a quadrilateral is 360°
- 110° + 130° + x + x - 3° = 360°
- 2x = 360° - 237°
- 2x = 123°
- x = 61.5°
(18)
Sum of interior angles of a hexagon is 180°*(6 - 2) = 720°
- 2*90° + 2x + 2(x + 22°) = 720°
- 90° + x + x + 22° = 360°
- 2x = 360° - 112°
- 2x = 248°
- x = 124°
(19)
Interior angles of a given pentagon are all marked as congruent, so the exterior angles are congruent too.
Sum of exterior angles is 360°.
First step is to factor. With a polynomial function in the form ax² + bx + c = f(x), we have to find what factors of term C have a sum of term B.
So with this, we need factors of -90 add up to become -1. Your factors are - 10 and 9.
f(x) = x² + 9x - 10x - 90
Now we group together and pull out GCFs.
f(x) = (x² + 9x) + (10x - 90)
f(x) = x(x² + 9) - 10(x + 9)
f(x) = (x - 10)(x + 9)
Now, set each factor equal to zero.
x - 10 = 0, x + 9 = 0
For the first equation you are going to add 10 to both sides to get x by itself. Subtract 9 from both sides in the second equation for the same reason.
x = 10, x = -9
Your zeros are at x = -9, 10 or at the ordered pairs (-9, 0) and (10, 0).