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Talja [164]
3 years ago
10

Can someone plsss help with these 2 questions thank u!!!

Mathematics
1 answer:
bixtya [17]3 years ago
4 0

#2: formula: V = (3.14) x radius (1/2 diameter) and height over 3

Answer is A

#3: Answer is 314.159 which rounds up to 314.160

Step by Step: easy

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OMG PLEASE Help me !!!! ILL GIVE YOU BRAINLIEST !!! Explain in order to get it
Doss [256]

Answer:

I'm not sure but I believe the correct answer is line L2.

Step-by-step explanation:

It seems to be closer to more points than the other lines.

If it is not L2, it must be L3.

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3 years ago
Please answer asap. thank you
DochEvi [55]

Answer:

7 Seconds

Step-by-step explanation:

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3 years ago
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How much did it cost summer to to buy 2.6 pounds of granola at the old price
coldgirl [10]
It was $6.00 per pound.  So 2 pounds would be $12.  But there is still that .6 of a pound left.  If we know 1 pound costs $6, then we know 1/10th of a pound will cost $0.60.  So 6 x .60 = 3.60.  Add 12 and 3.60, and you get 15.60 for 2.6 pounds of granola at the old price.
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3 years ago
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A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat
jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.  

We are given a function (corrected)

N(t) = 20(t^2-lnt^2)+ 30

N(t) = 20(t^2-2lnt)+ 30

(a)

First, we take its derivative

N'(t) = 20(2t-\frac{2}{t})

Solve N'(t)=0

20(2t-\frac{2}{t})=0

Simplifying

2t^2-2=0

Solving for t

t=1\ ,t=-1

Only t=1 belongs to the valid interval 1\leqslant t\leqslant 15

Taking the second derivative

N''(t) = 20(2+\frac{2}{t^2})

Which is always positive, so t=1 is a minimum

(b)

N(1)=20(1^2-2ln1)+ 30

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

N(15)=20(15^2-2ln15)+ 30

N(15)=4391.7 is a maximum

7 0
3 years ago
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andrey2020 [161]

Answer:

3/4

Step-by-step explanation:

Rise over run.

Go up 3 units and 4 units to the right to find the next point

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