The relevant rules of exponents are
.. (t^a)^b = t^(a·b)
.. t^a·t^b = t^(a+b)
You have
.. (t^-4)^-9·t^2
.. = t^((-4)*(-9) +2)
.. = t^38 . . . . . . . . . . . selection C
6y+1>=10
Subtract 1 from both sides
6y>=9
Divide 6 on both sides so that the only thing remaing on the left side is the variable y.
Final Answer: y>= 1.5 or 1 1/2
Answer:
19ft
Step-by-step explanation:
Given the height of a ball above the ground after x seconds given by the quadratic function y = -16x2 + 32x + 3, we can find the maximum height reached by the ball since we are not told what to look for.
The velocity of the ball is zero at maximum height and it is expressed as:
V(x) = dy/dx
V(x) = -32x+32
Since v(x) = 0
0 = -32x+32
32x = 32
x = 32/32
x = 1s
Get the height y
Recall that y = -16x² + 32x + 3.
Substitute x = 1
y = -16(1)²+32(1)+3
y = -16+32+3
y = -16+35
y = 19ft
Hence the maximum height reached by the ball is 19ft
40.32 mpg, because 36x.12=4.32 + 36= 40.32
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