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Lynna [10]
3 years ago
14

Show expression of -2(4 - 3x) + (5x - 2)

Mathematics
1 answer:
jeka57 [31]3 years ago
6 0

Answer:

11x-10

Step-by-step explanation:

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Evaluate u + xy, for u = 18, x = 10, and y = 8.
Brrunno [24]
Your answer is 98

This is because. u (18) + xy (10 × 8 = 80) so
18 + 80 = 98

Hope this helped
7 0
3 years ago
Which equation is represented by the phrase “one-fourth of a number, increased by eight equals sixteen”?
Alexandra [31]
(1/4)x + 8 = 16

or...
(x/4) + 8 = 16
7 0
4 years ago
Read 2 more answers
F ( x ) = -x^3 + 3x^2 + 4x -5 x = -3<br> Answear ?
Ivan

Answer: u spelt answer wrong f ( x f ( x ) = -x^3 + 3x^2 + 4x -5 x = -3

Step-by-step explanation:

7 0
3 years ago
Helpppp anyoneee please!!
pshichka [43]

Answer:x=6

Step-by-step explanation:

So if y is 0, that’s 8(y) which is 0. That means the equation is 2x=12, which means x is 6

My comments don’t work! But it’s (6,0)

4 0
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).

4 0
3 years ago
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