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Irina-Kira [14]
3 years ago
10

Thank u :)!! for the help

Mathematics
1 answer:
Alla [95]3 years ago
5 0
11 pls mark me brainliest
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A rectangular box has three of its faces on the coordinate planes and one vertex in the first octant on the paraboloid z=100−x2−
LuckyWell [14K]

The volume as a function of the location of that vertex is

... v(x, y, z) = x·y·z = x·y·(100-x²-y²)


This function is symmetrical in x and y, so will be a maximum when x=y. That is, you wish to maximize the function

... v(x) = x²(100 -2x²) = 2x²(50-x²)


This is a quadratic in x² that has zeros at x²=0 and x²=50. It will have a maximum halfway between those zeros, at x²=25. That maximum volume is

... v(5) = 2·25·(50-25) = 1250


The maximum volume of the box is 1250 cubic units.

7 0
3 years ago
Pls helppp me with this!!
zepelin [54]

Answer:4

Step-by-step explanation: Because 4 goes in 3/4 better and more eqvilent to that answer.

7 0
2 years ago
Read 2 more answers
A newborn baby has about 26,000,000,000 cells. An adult has about 4.94 x 13 cells. How many times as many cells does an adult ha
finlep [7]

an adult has 4.94 x 10^13 cells = 49,400,000,000,000

49,400,000,000,000 - 26,000,000,000 = 49,374,000,000,000

scientific notation form :  4.9374 x 10^13



3 0
3 years ago
7. In A.ABC. JB and KA are medians, JK = 10x - 12, AB = 9x + 18, JM = 21, KM = 23. AJ = 60. and
Juliette [100K]

Answer:

A) JK and AB =

[tex] \boxed{( \theta) = \f{12}{( \theta)} } So, =》 \dfrac{10}{5} = \dfrac{1}{ \( \theta) } =》 \( \theta) = \dfrac{5}{2}[/tex]

B) AABC

[tex] \boxed{( \theta) = {1}{( \theta)} } So, 23}{-18=\\Ans } \boxed

C) AABM:

[tex] e \fbox{: } The value of k for which \sin(jm) = \cos(x) is \dfrac{\p}{2} ans[/tex]

----------

3 0
3 years ago
What is the value of x? x=int(13.25 + 4/2) *​
julia-pushkina [17]

15.25 is the answer of the question

8 0
3 years ago
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