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4 years ago

8

Expand & simplify 4(x+6)+4(x-5)

1 answer:

EleoNora [17]4 years ago

5 0

[Answer]

$\boxed{8x \ + \ 4}$

[Explanation]

4(x + 6) + 4(x - 5)

[Expand] 4(x + 6): 4x + 24

4x + 24 + 4(x - 5)

[Expand] 4(x - 5): 4x - 20

4x + 24 + 4x + 20

[Simplify] 4x + 24 + 4x - 20: 8x + 4

= 8x + 4

$\boxed{[ \ Eclipsed \ ]}$

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What is the equation of a line parallel to y=1/2×+6 that passes through (-4,1)

IgorC [24]

keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above

$y = \stackrel{\stackrel{m}{\downarrow }}{\cfrac{1}{2}}x+6\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

so we're really looking for the equation of a line whose slope is 1/2 and passes through (-4 , 1)

$(\stackrel{x_1}{-4}~,~\stackrel{y_1}{1})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{1}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{ \cfrac{1}{2}}(x-\stackrel{x_1}{(-4)}) \\\\\\ y-1=\cfrac{1}{2}(x+4)\implies y-1=\cfrac{1}{2}x+2\implies y=\cfrac{1}{2}x+3$

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2 years ago

Which term can be added to the list so that the greatest common factor of the three terms is 12h3? 36h3, 12h6, __________

topjm [15]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

If log.. (125) = 3, then x =

Naddik [55]

5x5x5=125 so it could be x=5

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3 years ago

Find the absolute maximum and minimum values of f on the set D. f(x, y) = x3 − 3x − y3 + 12y + 1, D is quadrilateral whose verti

mojhsa [17]

$D$ is the set of points,

$\left\{(x,y)\mid-2\le x\le2,x\le y\le3\right\}$

Check for critical points:

$f(x,y)=x^3-3x-y^3+12y+1\implies\begin{cases}f_x=3x^2-3=0\implies x=\pm1\\f_y=-3y^2+12=0\implies y=\pm2\end{cases}$

Of these 4 points, only 2 belong to $D$ , (-1, 2) and (1, 2), for which we have

$\begin{cases}f(-1,2)=19\\f(1,2)=15\end{cases}$

Now look for extrema along the boundary.

If $x=-2$ , then

$f(-2,y)=-y^3+12y-1\implies f'(-2,y)=-3y^2+12=0\implies y\pm2$

We have $f'(-2,y)>0$ for $-2$ and $f'(-2,y)$ for $2$ , which indicates a local maximum at $y=2$ and minima at the endpoints of this boundary. So

$\begin{cases}f(-2,2)=15\\f(-2,-2)=-17\\f(-2,3)=8\end{cases}$

If $x=2$ , then

$f(2,y)=y^3+12y+3\implies f'(2,y)=-3y^2+12=0\implies y=\pm2$

We have $f'(2,y)$ for $2$ , so we have extrema at the endpoints of this boundary.

$\begin{cases}f(2,2)=19\\f(2,3)=12\end{cases}$

If $y=x$ , then

$f(x,x)=9x+1\implies f'(x,x)=9>0$

which tells us $f$ is strictly increasing on this boundary, giving the extrema we already know about,

$\begin{cases}f(-2,-2)=-17\\f(2,2)=19\end{cases}$

If $y=3$ , then

$f(x,3)=x^3-3x+10\implies f'(x,3)=3x^2-3=0\implies x=\pm1$

We have $f'(x,3)>0$ for $-2$ and $1$ , and $f'(x,3)$ for $-1$ . This indicates a maximum at $x=-1$ and a minimum at $x=1$ , with

$\begin{cases}f(-2,3)=8\\f(-1,3)=12\\f(1,3)=8\\f(2,3)=12\end{cases}$

From this analysis, we find that $f$ attains an absolute maximum of 19 at (-1, 2) and (2, 2), and an absolute minimum of -17 at (-2, -2).

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topjm [15]

Answer: b i just took the test

Step-by-step explanation:

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3 years ago