3x + 50 = 6x -10
add 10 to each side
3x +60 = 6x
subtract 3x from each side
60 = 3x
x = 60/3 = 20
x = 20
H = –16t2 + 156t + 105, when t = t', the time where the rocket hit the ground after it is launched, so 105= –16t'2 + 156t' + 105, so -16t'2 + 156t' = 0 implies t' =0 or -16t'+ 156= 0, equivalent to t'=9.75, so the answer is
b) <span>9.8 s</span>
The answer is -25. tried tested and approved. welcome
Answer:
Step-by-step explanation:
<u>Step 1: Multiply
</u>
Answer: The correct answer is -1
Answer:
Step-by-step explanation:
The point of this question is to find out the point where two lines intersect. First we need to get the equation of those lines
Slope of line 1:
(Yb -Ya)/(Xb - Xa) =
(-10 - (-14))/(-1 - (-3)) =
4/2 =
2
Use that slope to find the Y-intercept of line 1
y = 2x + b
-14 = 2(-3) +b
-14 = -6 + b
-8 = b
Therefore Line 1 is:
y = 2x - 8
Slope of line 2
(11 - 13)/(-1 - (-3)) =
-2/2 =
-1
Y-intercept of line 2
y = -x + b
13 = -(-3) +b
13 = 3 + b
10 = b
Therefore line 2 is
y = -x + 10
Now we have 2 equations to solve for the coordinates x and y
y = 2x - 8
y = -x + 10
Substitute y out in one of the equations
2x - 8 = -x + 10
3x = 18
x = 6
Plug x into one of the equations
y = 2(6) - 8
y = 12 - 8
y = 4
Therefore the solution is:
x=6, y=4
