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Arisa [49]
3 years ago
13

HEEEELP ABC m a-58 b-42 c-55 d-41

Mathematics
1 answer:
Savatey [412]3 years ago
7 0
Your answer is going to be C
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Course Home
kolbaska11 [484]

Answer:

........... .........

3 0
3 years ago
Find all points on the portion of the plane x+y+z=5 in the first octant at which f(x, y, z) = xy2z2 has a maximum value.
irina1246 [14]
Lagrange multipliers:

L(x,y,z,\lambda)=xy^2z^2+\lambda(x+y+z-5)

L_x=y^2z^2+\lambda=0
L_y=2xyz^2+\lambda=0
L_z=2xy^2z+\lambda=0
L_\lambda=x+y+z-5=0

\lambda=-y^2z^2=-2xyz^2=-2xy^2z

-y^2z^2=-2xyz^2\implies y=2x (if y,z\neq0)

-y^2z^2=-2xy^2z\implies z=2x (if y,z\neq0)

-2xyz^2=-2xy^2z\implies z=y (if x,y,z\neq0)

In the first octant, we assume x,y,z>0, so we can ignore the caveats above. Now,

x+y+z=5\iff x+2x+2x=5x=5\implies x=1\implies y=z=2

so that the only critical point in the region of interest is (1, 2, 2), for which we get a maximum value of f(1,2,2)=16.

We also need to check the boundary of the region, i.e. the intersection of x+y+z=5 with the three coordinate axes. But in each case, we would end up setting at least one of the variables to 0, which would force f(x,y,z)=0, so the point we found is the only extremum.
4 0
3 years ago
I need,help plsssssss
Mila [183]

Answer:

(3 * 3^2)+(2 * 4^2)

combined area is 59 sq ft

8 0
3 years ago
If 3t=5t-8 evaluate -4t+5
Sergio [31]

3t = 5t - 8

-2t = -8

t = 4

Now put that into -4t + 5:

-4t + 5

-4(4) + 5

-16 + 5

-11

⭐ Please consider brainliest! ⭐

✉️ If any further questions, inbox me! ✉️

6 0
3 years ago
A line includes the points (-39,49) and (0,0). What is its equation in slope-intercept form?
malfutka [58]

Answer:

The equation in the slope-intercept form will be:

y=-\frac{49}{39}x+0

Step-by-step explanation:

Given the points

  • (-39,49)
  • (0,0)

Finding the slope between the points using the formula

\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}

\left(x_1,\:y_1\right)=\left(-39,\:49\right),\:\left(x_2,\:y_2\right)=\left(0,\:0\right)

m=\frac{0-49}{0-\left(-39\right)}

m=-\frac{49}{39}

We know that the point-slope of the line equation is

y-y_1=m\left(x-x_1\right)

substituting m=-\frac{49}{39} and (-39,49)  in the equation

y-y_1=m\left(x-x_1\right)

y-49=\frac{-49}{39}\left(x-\left(-39\right)\right)

Now writing the equation in slope-intercept form

y=mx+b

where m is the slope and b is the y-intercept

y-49=\frac{-49}{39}\left(x-\left(-39\right)\right)

y-49=\frac{-49}{39}\left(x+39\right)

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}

y-49=-\frac{49}{39}\left(x+39\right)

\mathrm{Add\:}49\mathrm{\:to\:both\:sides}

y-49+49=-\frac{49}{39}\left(x+39\right)+49

y=-\frac{49}{39}x+0              ∵ y=mx+b

Where

m=-\frac{49}{39} and the y-intercept i.e. b=0

Therefore, the equation in the slope-intercept form will be:

y=-\frac{49}{39}x+0

5 0
3 years ago
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