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Artist 52 [7]
2 years ago
7

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

Mathematics
1 answer:
Semenov [28]2 years ago
8 0

\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}

<u>Alternative</u>

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

<u>Answer</u>

  • (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!

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The radius of a cylindrical gift box is (2x+3) in. The height of the gift box is twice the radius. What is the surface area of t
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A right triangle has a leg length of 21 inches and a hypotenuse length of 29 inches. What is the length of the other leg?​
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20

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4 0
3 years ago
3.
loris [4]

(a) It looks like the ODE is

<em>y'</em> = 4<em>x</em> √(1 - <em>y </em>^2)

which is separable:

d<em>y</em>/d<em>x</em> = 4<em>x</em> √(1 - <em>y</em> ^2)   =>   d<em>y</em>/√(1 - <em>y</em> ^2) = 4<em>x</em> d<em>x</em>

Integrate both sides. On the left, substitute <em>y</em> = sin(<em>t </em>) and d<em>y</em> = cos(<em>t</em> ) d<em>t</em> :

∫ d<em>y</em>/√(1 - <em>y</em> ^2) = ∫ 4<em>x</em> d<em>x</em>

∫ cos(<em>t</em> ) / √(1 - sin^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>

∫ cos(<em>t</em> ) / √(cos^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>

∫ cos(<em>t</em> ) / |cos(<em>t</em> )| d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>

Since we want the substitutiong to be reversible, we implicitly assume that -<em>π</em>/2 ≤ <em>t</em> ≤ <em>π</em>/2, for which cos(<em>t</em> ) > 0, and in turn |cos(<em>t</em> )| = cos(<em>t</em> ). So the left side reduces completely and we get

∫ d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>

<em>t</em> = 2<em>x</em> ^2 + <em>C</em>

arcsin(<em>y</em>) = 2<em>x</em> ^2 + <em>C</em>

<em>y</em> = sin(2<em>x</em> ^2 + <em>C </em>)

(b) There is no solution for the initial value <em>y</em> (0) = 4 because sin is bounded between -1 and 1.

7 0
3 years ago
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