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Mathematics

1 answer:

worty [1.4K]3 years ago

6 0

Answer: It should equal 10

Step-by-step explanation: X=10

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F(x)=2x^2-x-6 g(x)=4-x (f+g)(x)

Semenov [28]

$\large \boxed{(f + g)(x) = f(x) + g(x)}$

The property above is distribution property where we distribute x-term in the function.

Substitute both f(x) and g(x) in.

$\large{ \begin{cases} f(x) = 2 {x}^{2} - x - 6 \\ g(x) = 4 - x \end{cases}} \\ \large{f(x) + g(x) = (2 {x}^{2} - x - 6) + (4 - x)} \\ \large{f(x) + g(x) = 2 {x}^{2} - x - 6 + 4 - x}$

Évaluate/Combine like terms.

$\large{f(x) + g(x) = 2 {x}^{2} - 2x - 2}$

The function can be factored so there are two answers. (Both of them work as one of them is factored form while the other one is not.)

$\large{(f + g)(x) = 2 {x}^{2} -2x -2}$

Alternative

$\large{(f + g)(x) = 2({x}^{2} -x - 1)}$

Answer

(f+g)(x) = 2x²-2x-2
(f+g)(x) = 2(x²-x-1)

Both answers work. The second answer is in factored form.

Let me know if you have any doubts!

8 0

3 years ago

31 cars are in a race. In how many different ways can cars finish in order in the top 4 positions?

cupoosta [38]

Answer:

755,160 ways

Explanation:

We're told that 31 cars are in a race.

Considering the order in which the cars can finish in the top 4 positions, any of the 31 cars can finish in the first position, any of the remaining 30 cars can finish in the second position, any of the remaining 29 cars can finish in the third position and any of the remaining 28 cars can finish in the fourth position, so the different numbers of ways can be determined as seen below;