Step-by-step explanation:
5. 6x-8=9x+4
6x - 9x = 4+8
-3x = 12
x = -4
check, 6(-4) - 8=9(-4) + 4
-24-8 = -36+4
-32 = -32
6. 5a - 10 = 9a - 6
5a-9a = -6+10
-4a = 4
a = -1
check, 5(-1) -10 = 9(-1) -6
-5-10 = -9-6
-15 = -15
7. 8m-24 = 9-3m
8m + 3m = 9+24
11m = 33
m = 3
check, 8(3)-24 = 9- 3(3)
24-24 = 9-9
0 = 0
8. 10- 3z = z + 30
10-30 = z +3z
-20=4z
-5 = z
check, 10 - 3(-5) = (-5)+30
10+15 = 25
25=25
9. 3k - 14 = 10 +5k
3k - 5k = 10+14
-2k = 24
k = -12
check, 3(-12)-14 = 10+5(-12)
-36-14 = 10-60
-50 = -50
10. 6c + 8 = 2c -16
6c - 2c = -16-8
4c = -24
c = -6
check, 6(-6)+8 = 2(-6) - 16
-36 + 8 = -12 - 16
-28 = -28
Answer:when x is isolated in the first equation or when a equation is solved for the variable.
Step-by-step explanation:
Answer:
fg(×)=22fx
Step-by-step explanation:
I hope it help
Answer:
Step-by-step explanation:
With a factor of (t - 1) we know that zero (ground level) is reached at 1 second from an initial height of (0 - 1)(0 - 1)(0 - 11)(0 - 13)/3 = -1•-1•-11•-13 / 3 = 47⅔ meters at t = 0
As we have <em>two </em>factors of (t - 1) we know the track does not go underground at t = 1, but rises again.
At t = 11 seconds, the car has again returned to ground level, but as we only have a single factor of (t - 11) the car plunges below ground level and returns to above ground level at t = 13 seconds due to the single factor of (t - 13)
we can estimate that the car is the deepest below ground level halfway between 11 and 13 s, so at t = 12. At that time, the depth will be about (12 - 1)(12 - 1)(12 - 11)(12 - 13) / 3 = -(11²/3) = - 40⅓ m.
we can estimate that the car is the highest above ground level halfway between 1 and 11 s, so at t = 6s. At that time, the height will be about (6 - 1)(6 - 1)(6 - 11)(6 - 13) / 3 = 5²•-5•-7 / 3 = 291⅔ m.
It's obvious that the roller coaster car had significant initial velocity at t = 0 to achieve that altitude from an initial height of 47⅔ m
