Answer:
2 x 2 x 11
Step-by-step explanation:
Note that prime factors are numbers that cannot be reduced anymore, and, when multiplied together, would equal the original number.
44/2 = 22/2 = 11
2 x 2 x 11 = 44 ∴ 2 , 2, 11 are your answers.
~
Well the LCM (lowest common multiples) for 8 and 32 would be 4. So plug 4 into "b" -8×4-32. -8×4=32. -32-32. Since you can't subtract 32 and 32 you add! Which gives you -64.
Hope this helps (:
For
|a|<b
assume
-b<a<b
so
add 5 to both side
|-2-8x|<66
assume
-66<-2-8x<66
add 2 to both sides
-64<-8x<68
divide everyboy by -8, don't forget to flip sign
8>x>-8.5
-8.5<x<8
the solution is all numbers between -8.5 and 8, not including -8.5 and 8
in interval notation: (-8.5,8)
or
S={x|-8.5<x<8}
Answer:
Step-by-step explanation:
To solve this problem you need the function
h(t) = -16t2 + v0t + h0
where t = time
v0 is the initial velocity, which in our case is 0
h0 = initial height, which in our case is 256
h(t) = 0 since we want to know when the ball will hit the ground.
0 = -16 t2 + 256
And we can solve for t
If we rearrange the terms we see that this is a difference of 2 squares
0 = 256 - 16t2
0 = (16-4t)(16+4t)
Setting each factor = 0
16-4t=0 16+4t=4
t = 4 t = -4
The second solution is discarded as time cannot be negative.
So the ball will hit the ground in 4 seconds.
