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Oliga [24]
3 years ago
9

/5×+4/ =/×+9/ solve for x​

Mathematics
1 answer:
Bumek [7]3 years ago
4 0

Answer:

\large\boxed{x=\dfrac{5}{4}\ \vee\ x=-\dfrac{13}{6}}

Step-by-step explanation:

|5x+4|=|x+9|\iff5x+4=x+9\ or\ 5x+4=-(x+9)\\\\5x+4=x+9\qquad\text{subtract 4 from both sides}\\5x=x+5\qquad\text{subtract x from both sides}\\4x=5\qquad\text{divide both sides by 4}\\x=\dfrac{5}{4}\\\\5x+4=-(x+9)\\5x+4=-x-9\qquad\text{subtract 4 from both sides}\\5x=-x-13\qquad\text{add x to both sides}\\6x=-13\qquad\text{divide both sides by 6}\\x=-\dfrac{13}{6}

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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
Can I get help and no links pls
kodGreya [7K]
The answer is 30, can I get the brainliest answer
5 0
3 years ago
Read 2 more answers
Please help with 6 and 7
Maru [420]

Answer:

6a) i- 2hrs 36mins  ii- 3hrs 12mins

b) car A≈ 76.9km/h  car B≈ 62.5km/h

c)------

7a) 35km

b) car A=75km  car B=60km

c) 30km

d) car A≈36mins  car B≈48mins

Step-by-step explanation:

6a) Using the graph follow the lines until they finish then go downwards until you get to the x-axis. The x-axis is going up by 12mins for each square.

b) Using the answer from a, you divide 200km by the time.

For car A 2hrs 36mins becomes 2.6 because 36mins/60mins=0.6

∴ car A: 200/2.6≈ 76.92km/h

For car B 3hrs 12mins becomes 3.2 because 12mins/60mins=0.2

∴ car B: 200/3.2≈ 62.5km/h

7a) Using the graph go down from where the line of car A finished to meet car B. The y-axis is going up by 5km for each square.

b) Starting from the x-axis at 1 hour go upwards to see where you meet the car B line (60km) and car A line(75km). (sorry if that does not really make sense).

c) Difference from car A line to car B:

155km-125km=30km

d) Going across from 50km meet car A line and go down to see it has been travelling for approx. 36mins. Then continue across to car B line, go down to see it reached 50km at approx. 48mins.

Hope this helps.

5 0
3 years ago
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gayaneshka [121]
Simply plug in the value for x.

If x is equal to -18, then plug in -18 for x.

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f(x) = 15 + x/5. 
f(5) = 15 + 5/5.

Then solve the equation.

A function is just something that receives some value and then barfs a value out. This is an IO relationship. (Input/Output).
6 0
3 years ago
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laiz [17]

Answer:

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Step-by-step explanation:

 (3m⁴n)²     Remove parentheses

= 3²m⁸n²     Square the 3

= 9m⁸n²

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3 years ago
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