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Firdavs [7]
3 years ago
11

Can i please get help with this question?

Mathematics
1 answer:
AysviL [449]3 years ago
8 0
In addition, the order of the variables does not matter in the expression.
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Please help
denis-greek [22]

Answer:

answer is B

Step-by-step explanation:

7 0
3 years ago
3. A bag contains 4 white, 3 blue, and 6 red marbles. A marble is drawn from the bag, replaced,
gladu [14]

Answer:

Hope this helps! If I could have brainliest I'm trying to rank up.

Step-by-step explanation:

First find the probability of each color being drawn.

We need to find the total number of marbles first.

4 + 3 + 6 = 13

Everything will be over 13.

Probability of a white marble being drawn --> 4/13

Probability of a blue marble being drawn --> 3/13

Probability of a red marble being drawn --> 6/13

Now to find the probabilities of more than one being drawn, we can multiply.

Part A:

6/13*6/13=36/169

Part B:

3/13*3/13=9/169

Part C:

6/13*6/13=18/169

Part D:

7/13*7/13=49/169

8 0
3 years ago
Help plz! Please explain how u solved it i need as much help as possible
Stells [14]
What do you need help with?

7 0
3 years ago
Linda is making a mosaic stepping stone to place in
Lorico [155]

Answer:

D 565.2 in

Step-by-step explanation:

First you need to find the area of the stepping stone. The rule for area of circle is pi times radius squared. To find the radius you divide 12 by 2 = 6. After you do 6 times 6= 36 you multiply 36 times pi (3.14) = 113.04. After we find the are of the stepping stone we multiply it by five to get how many inches we have to cover. 113.04 times 5= 565.2 in.

3 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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