(1/64)t^3=t
solve for t.
i have a 4 page packet..and i've only got a couple left, but my friends and i can't figure them out. help is greatly appreciated!!
ok
First, multiply both sides by 64 to remove the fraction. Remember, (1/X) x X = 1.
t^3 = 64t
then divide both sides by t
t^2 = 64
so, t = sq root of 64, = 8
There are three solutions for t:
t = -8, 0, and +8.
Given:
(1/64)t^3 = t
Subtract "t" from the right side:
(1/64)t^3 - t = 0
Now, factor out a "t" from both terms:
t * [(1/64)t^2 - 1] = 0
You now have:
t = 0 <--- First Factor
(1/64)t^2 - 1 = 0 <-- Second Factor
On the Second Factor, add 1 to both sides:
(1/64)t^2 = 1
Now, multiply both sides by 64, which gives you:
Taking the square root of both sides:
t = ±8
If you graph this equation, you'll see the x-intercepts are:
-8, 0 and +8.
Good luck in your studies,
~ Mitch ~
using a graphing calculator and finding the intersect of y=(1/64)t^3 and y=t the answers are -8,0 and 8
(1/64)t^3 -t =0
t{(1/64)t^2-1} = 0
so t = 0 or t = 64^0.5, t = 8
its 8 if u want an explantation tell me or else you are just doing them to get the packet done...
Comments
ok
First, multiply both sides by 64 to remove the fraction. Remember, (1/X) x X = 1.
t^3 = 64t
then divide both sides by t
t^2 = 64
so, t = sq root of 64, = 8
There are three solutions for t:
t = -8, 0, and +8.
Given:
(1/64)t^3 = t
Subtract "t" from the right side:
(1/64)t^3 - t = 0
Now, factor out a "t" from both terms:
t * [(1/64)t^2 - 1] = 0
You now have:
t = 0 <--- First Factor
(1/64)t^2 - 1 = 0 <-- Second Factor
On the Second Factor, add 1 to both sides:
(1/64)t^2 = 1
Now, multiply both sides by 64, which gives you:
t^2 = 64
Taking the square root of both sides:
t = ±8
If you graph this equation, you'll see the x-intercepts are:
-8, 0 and +8.
Good luck in your studies,
~ Mitch ~
using a graphing calculator and finding the intersect of y=(1/64)t^3 and y=t the answers are -8,0 and 8
(1/64)t^3 -t =0
t{(1/64)t^2-1} = 0
so t = 0 or t = 64^0.5, t = 8
its 8 if u want an explantation tell me or else you are just doing them to get the packet done...