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2 years ago

If f(x)=x/3 -2 and g(x)=3x^2+2x-6 find (f+g)(x)

Mathematics

1 answer:

umka21 [38]2 years ago

6 0

Answer:

(f + g)(x) = 3x² + (7/3)x - 8

Step-by-step explanation:

To find (f + g)(x), you need to add both the f(x) and g(x) equations together.

f(x) = x/3 - 2 ..... which is equal to ... f(x) = (1/3)x - 2

g(x) = 3x² + 2x - 6

(f + g)(x) = ((1/3)x - 2) + (3x² + 2x - 6) <----- Add both equations

(f + g)(x) = 3x² + (1/3)x + 2x - 2 - 6 <----- Rearrange (2 = 6/3)

(f + g)(x) = 3x² + (7/3)x - 8 <----- Simplify similar terms

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3 years ago

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Nimfa-mama [501]

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The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

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$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

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$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

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