Answer:
1 5/9
Step-by-step explanation:
9 1/3 can be rewritten as 28/3. The question is asking for you to divide 28/3 by 6. 6 is 6/1, so the expression is 28/3 divided by 6/1. You then do the reciprocal of 6/1 and the expression is
. This gets you 28/18, which is simplified to 14/9 or 1 5/9. Hope this helps!
Answer:
1.Conduction
2.the second one
sorry if these are wrong
Step-by-step explanation:
Answer:
a) 72.25sec
b) 6.25secs
c) after 10.5secs and 2 secs
Step-by-step explanation:
Given the height reached by the rocket expressed as;
s(t)= -4t^2 + 50t - 84
At maximum height, the velocity of the rocket is zero i.e ds/dt = 0
ds/dt = -8t + 50
0 = -8t + 50
8t = 50
t = 50/8
t = 6.25secs
Hence it will reach the maximum height after 6.25secs
To get the maximum height, you will substitute t - 6.25s into the given expression
s(t)= -4t^2 + 50t - 84
s(6.25) = -4(6.25)^2 + 50(6.25) - 84
s(6.25) = -156.25 + 312.5 - 84
s(6.25) = 72.25feet
Hence the maximum height reached by the rocket is 72.25feet
The rocket will reach the ground when s(t) = 0
Substitute into the expression
s(t)= -4t^2 + 50t - 84
0 = -4t^2 + 50t - 84
4t^2 - 50t + 84 = 0
2t^2 - 25t + 42 = 0
2t^2 - 4t - 21t + 42 = 0
2t(t-2)-21(t-2) = 0
(2t - 21) (t - 2) = 0
2t - 21 = 0 and t - 2 = 0
2t = 21 and t = 2
t = 10.5 and 2
Hence the time the rocket will reach the ground are after 10.5secs and 2 secs
Answer:
(x+12)(x+5)
Step-by-step explanation:
Formula use: a²+bx+c
- Make one side equal to zero:
Original:
-7x-60 =x² +10x
New:
x² + 17x + 60
New:
(1)x x 60 = 60
- Find factors of 60 that when added, equal to 17.
New:
10 × 6, 60 × 1, 20 × 3, <u>5 × 12</u>, 4 × 15
5 times 12 equal 60, but when added equal to 17.
- Replace the 17 with 5 and 12
New:
x² + 5x + 12x + 60
- Break them off into two equations
New:
x² + 5x l 12x + 60
- Divide each equation into it's simpilest form. Make sure the numbers in the ( ) are the same.
New:
x(x + 5) l +12(x+5)
Students in 2014 = (100% + 33%) × 2,300
students in 2014 = 133% × 2,300
students in 2014 = 133/100 × 2,300
students in 2014 = 3,059
There are 3,059 students in 2014