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3 years ago

When you mix two colors of paint in equivalent ratios, the resulting color is always the same.

Mathematics

1 answer:

Pavel [41]3 years ago

5 0

Answer:

3 blue cups and 9 red cups

Step-by-step explanation:

it needs to be in a ratio of 1:3 so for every 1 cup of blue there is 3 cups of red.

its the same ratio as the other two

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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

A massless beam supports two weights as shown. 685 n w l 4 l 4 l 2 l a b find w such that the supporting force at a is zero. ans

BabaBlast [244]

Idk idk idk Idk idk idk

5 0

3 years ago

A region R is enclosed by the coordinate axes and the graph of y=k(x-5)^2, k>0. When this region is revolved around the x-axi

r-ruslan [8.4K]

$y=k(x-5)^2$ touches the x-axis at $x=5$ and has a y-intercept when $x=0$ at

$y=k(0-5)^2=25k$

Using disks, the volume of the revolved region would be given by the integral

$\displaystyle\pi\int_0^5\bigg(k(x-5)^2\bigg)^2\,\mathrm dx=\pi k^2\int_0^5(x-5)^4\,\mathrm dx=\pi k^2\int_{-5}^0x^4\,\mathrm dx$
$\implies 2500\pi=625\pi k^2$
$\implies 4=k^2$
$\implies k=2$

6 0

3 years ago

Read 2 more answers

X 8 11 14 17, y 36 12 -12 -36, what is the slope

Mnenie [13.5K]

The slope is -24/3=-8

5 0

3 years ago

Tony's cat weighs c pounds. Three months from know, the cat's weight is expected to increase by 12.5% Write two expressions to r

horrorfan [7]

Answer:

Step-by-step explanation:

Tony's cat weighs c pounds. This means that the cat currently weighs c pounds.

Three months from know, the cat's weight is expected to increase by 12.5%. This means that in three months time, the weight of the cat is expected to be the sum of its current rate × 12.5 % of its current rate. It becomes

c + 12.5/100 × c = c + 0.125c

= 1.125c

So the two expressions that represent the cat's weight in three months would be

1) three months from now, the cat's weight will increase by 0.125c

2) three months from now, the new weight of the cat will be 1.125c

6 0

4 years ago