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GalinKa [24]
3 years ago
8

A water trough is

Mathematics
2 answers:
MissTica3 years ago
8 0

Answer is 0.0333m/min

olganol [36]3 years ago
6 0
Your wording is a little strange. can you show it so it is complete sentences? then I can help you!
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uranmaximum [27]
I think that it’s A, Perpendicular (correct me if I’m wrong)
4 0
3 years ago
Read 2 more answers
Please help me anyone ;(
Sergio039 [100]

Answer:

9 tables

Step-by-step explanation:

All we have to do is see how many time 6³/₄ ft. can fit in 60³/₄ ft. through division.

60 ³/₄ ÷ 6 ³/₄ =

²⁴³/₄ ÷ ²⁷/₄ =

²⁴³/₄ × ⁴/₂₇ =

²⁴³/₂₇ =

9 tables

5 0
3 years ago
Find the equations of the lines for the following. Your answer should be in the form y=mx+b.
spin [16.1K]

Answer:

1 ) y = 2x - 1

slope = 2

y intercept is (0,-1)

2) y = -1/2x + 5

slope = -0.5

y intercept  = (0,5)

Step-by-step explanation:

4 0
3 years ago
PLS HELP ME, I REALLY NEED THIS RIGHT NOW, 100 POINTS, IF YOU ANSWER WITH NO EXPLANATION, OR ANSWER FOR THE POINTS YOU WILL BE R
Reika [66]

9514 1404 393

Answer:

  • 0 ≤ m ≤ 7
  • 0.4541 cm/month; average rate of growth over last 4 months of study

Step-by-step explanation:

<u>Part A</u>:

The study was concluded after 7 months. The fish cannot be expected to maintain exponential growth for any significant period beyond the observation period. A reasonable domain is ...

  0 ≤ m ≤ 7

__

<u>Part B</u>:

The y-intercept is the value when m=0. It is the length of the fish at the start of the study.

__

<u>Part C</u>:

The average rate of change on the interval [3, 7] is given by ...

  (f(7) -f(3))/(7 -3) = (4(1.08^7) -4(1.08^3))/4 = 1.08^3·(1.08^4 -1)

   ≈ 0.4541 cm/month

This is the average growth rate of the fish in cm per month over the period from 3 months to 7 months.

3 0
2 years ago
A cylinder shaped can needs to be constructed to hold 400 cubic centimeters of soup. The material for the sides of the can costs
LenKa [72]

Answer:

The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.

Step-by-step explanation:

Volume of the Cylinder=400 cm³

Volume of a Cylinder=πr²h

Therefore: πr²h=400

h=\frac{400}{\pi r^2}

Total Surface Area of a Cylinder=2πr²+2πrh

Cost of the materials for the Top and Bottom=0.06 cents per square centimeter

Cost of the materials for the sides=0.03 cents per square centimeter

Cost of the Cylinder=0.06(2πr²)+0.03(2πrh)

C=0.12πr²+0.06πrh

Recall: h=\frac{400}{\pi r^2}

Therefore:

C(r)=0.12\pi r^2+0.06 \pi r(\frac{400}{\pi r^2})

C(r)=0.12\pi r^2+\frac{24}{r}

C(r)=\frac{0.12\pi r^3+24}{r}

The minimum cost occurs when the derivative of the Cost =0.

C^{'}(r)=\frac{6\pi r^3-600}{25r^2}

6\pi r^3-600=0

6\pi r^3=600

\pi r^3=100

r^3=\frac{100}{\pi}

r^3=31.83

r=3.17 cm

Recall that:

h=\frac{400}{\pi r^2}

h=\frac{400}{\pi *3.17^2}

h=12.67cm

The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.

3 0
3 years ago
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