Y=3x-5
y-3x=-5
Times 2 from both side
y(2)-3x(2)=-5(2)
2y-6x=-10
Times negative from both side
(-)2y-6x(-)=(-)-10
-2y+6x=10
or
6x-2y=10
6x=2y+10
6x-2y=10
Furthermore, we see that both equation have the same 6x-2y=10 which means that both of them have infinity solutions, not a solution. Hope it help!
(1/36) = 0.0277777777778
(1/108)^3 = 7.9383224102<span> x 10^-7 </span>
(1/9)^4 = 0.000152415790276
(1/6)^2 = 0.0277777777778
(1/2)^5 = <span>0.03125
The only one that matches with the value of 1/36 is (1/6)^2. Therefore, your answer is C. (1/6)^2
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The answer will be t=2.50g I think
We are to find how long will it take to return to the ground?
Answer:
t = 0.48 sec
Step-by-step explanation:
We are given;
initial velocity; vi = 40 ft/s
hi = 70 ft
acceleration due to gravity; a = 32 ft/s²
Now, we are given that the rocket's height as a function of time is;
h = -12at² + vit + hi
At ground, h = 0
Plugging in the relevant values to obtain ;
0 = -12(32)t² + 40t + 70
-384t² + 40t + 70 = 0
The roots of the equation gives; t = 0.48 sec
Answer:
help me first on this question
Step-by-step explanation: