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Reil [10]
3 years ago
13

2. The sum of two numbers is 30. Determine the two numbers if their product is a maximum.

Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
5 0

Answer:

X+y=30

X=30-y

Z=yx

Z=y(30-y)

D/dy

z=30–2y

At max 30–2y=0

2y=30 Y=15

Therefore x=15

Step-by-step explanation:

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Help me please, ps u don't have to answers all of them
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meaning you just have to calculate one side and multiply it by six.
541.5 is the surface area of the cube.
B) To find the surface area for just five sides, you'll be doing the same, except instead of 6 it'll be five.
so multiply 9.5 by itself
you get 90.25
Then you multiply 90.5 by 5
You get 451.25

Unfortunately I'm not sure how to do C. good luck

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Help needed! Plz A graph of a linear inequality in two variables is shown below:
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2 years ago
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Answer:

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Step-by-step explanation:

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2 years ago
If the cos of angle x is 8 over 17 and the triangle was dilated to be two times as big as the original, what would be the value
Viktor [21]

Answer:

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Step-by-step explanation:

i got it right on the quiz

6 0
3 years ago
A tank has capacity 6 L and initially contains 11 mg of salt dissolved in 3 L of water. A solution containing 1 mg/L of salt ent
BlackZzzverrR [31]

Answer:

The answer is below

Step-by-step explanation:

a) The maximum capacity of he tank is 6 L and initially it contains 11 mg of salt dissolved in 3 L of water. Solution enters the tank at a rate of 3 L/hr, therefore in x hours, the amount of water that have entered the tank = 3x.

Solution also leaves the tank at a rate of 2L/hr, therefore in x hours, the amount of water that have left the tank = 2x

Hence the amount of water present in the tank at x hours is given as:

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b)

\frac{dQ}{dx}=3-\frac{2Q}{3+x}\\  \\\frac{dQ}{dx}+\frac{2Q}{3+x}=3\\\\let\ u'=\frac{2u}{3+x}\\\\\frac{u'}{u}=\frac{2Q}{3+x}\\\\ln(u)=2ln(3+x)\\\\u=(3+x)^2\\\\(3+x)^2Q]'=3(3+x)^2\\\\(3+x)^2Q=(3+x)^3+c\\\\Q(0)=11\\\\(3+0)^2(11)=(3+0)^3+c\\\\x=72\\\\Q=x+3+\frac{72}{(x+3)^2}\\ \\Q(3)=3+3+\frac{72}{(3+3)^2}=8\ mg

8 mg/ 6 L = 4/3 mg/L

8 0
3 years ago
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