If f(x)=3x^2+1 and g(x)=1-x, what is the value of (f-g)(2)?

2 answers:

madam [21]3 years ago

3 0

The answer is 14 you just have to substitute x with the given number and finish the equation

never [62]3 years ago

3 0

Answer:

(f – g)(2) = 12.

Step-by-step explanation:

Given : f(x) = 3x²+1 and g(x) =1-x .

To find : which expression is equivalent to (f – g)(2).

Solution : We have given f(x) = 3x²+1 and g(x) = 1-x .

(f – g)(2) = f (2 ) - g(2) .

g(x) = 1-x .

Plug x = 2 .

g(2) = 1 - 2 .

g(2) = - 1.

For f(x) =3x²+1

Plug x = 2

f( 2) =3 (2)²+1.

f(2) = 12 + 1.

f(2) = 13 .

Then , (f – g)(2) = f (2 ) - g(2) .

Plugging the values

(f – g)(2) = -1 + 13

(f – g)(2) = 12

(f – g)(2) = 12.

Therefore, (f – g)(2) = 12.

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OlgaM077 [116]

Answer:

The velocities after 739 s of firing of each engine would be 6642.81 m/s in the x direction and 5306.02 in the y direction

Step-by-step explanation:

For a constant acceleration: $v_{f}=v_{0}+at$ , where $v_{f}$ is the final velocity in a direction after the acceleration is applied, $v_{0}$ is the initial velocity in that direction before the acceleration is applied, a is the acceleration applied in such direction, and t is the amount of time during where that acceleration was applied.
<em>Then for the x direction</em> it is known that the initial velocity is $v_{0x} =$ 5320 m/s, the acceleration (the applied by the engine) in x direction is $a_{x}$ 1.79 m/s2 and, the time during the acceleration was applied (the time during the engines were fired) of the is 739 s. Then: $v_{fx}=v_{0x}+a_{x}t=5320\frac{m}{s} +1.79\frac{m}{s^{2} }*739s=6642.81\frac{m}{s}$
In the same fashion, <em>for the y direction</em>, the initial velocity is $v_{0y} =$ 0 m/s, the acceleration in y direction is $a_{y}$ 7.18 m/s2, and the time is the same that in the x direction, 739 s, then for the final velocity in the y direction: $v_{fy}=v_{0y}+a_{y}t=0\frac{m}{s} +7.18\frac{m}{s^{2} }*739s=5306.02\frac{m}{s}$