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marysya [2.9K]
3 years ago
12

Eric plotted the graph below to show the relationship between the temperature of his city and the number of cups of lemonade he

sold daily. (see the graph)
Answer the following questions:

Part A:
Describe the relationship between the temperature of the city and the number of cups of lemonade sold. (I already have this question done, I only need help on part B)

Part B:
Describe how you can make the line of best fit. Write the approximate slope and y-intercept of the line of best fit. Show your work, including the points that you use to calculate the slope and y-intercept. (3 points)

Mathematics
2 answers:
alexandr1967 [171]3 years ago
8 0

Answer:

The answer for part A: If the city is hotter in temperature more people will want the cold refreshing lemonade.

Part B: The line of best fit is a line that will have the same amount of point on each side so you should draw a straight line and then make sure it has the same amount of point on each side. To calculate the slope use Y2-Y1/X2-X1, Y-intercept is the point that intercepts the Y-intercept.

Step-by-step explanation:

Part A: Because if it is hot outside you will sweat more being less hydrated which is why more people would want a drink especially lemonade.

AysviL [449]3 years ago
3 0

Answer:

give that man brainlist

Step-by-step explanation:

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If the number of nickels is 7 less than three times the number of dimes, the answers should be 29 nickels and 12 dimes. Hope it helps!

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22 8(11 + 2r) = 126r + 3
natulia [17]

8(11 + 2r) = 126r + 3​

first open the parenthesis

88 + 16r = 126r + 3

88 - 3 = 126r - 16 r

85 = 110r

divide both-side of the equation by 110

85/110 = r

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[(7 + 3) • 5 – 4] ÷ 2 + 2 <br><br> is the answer 17?<br> show how you got your answer.
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Okay! Math isn't my strongest subject. However, though, I think the answer is 25.
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Kazeer [188]

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Step-by-step explanation:

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Can someone explain step by step how to do this problem? Thanks! Calculus 2
Ganezh [65]

Answer:

  1.314 MJ

Step-by-step explanation:

As water is removed from the tank, decreasing amounts are raised increasing distances. The total work done is the integral of the work done to raise an incremental volume to the required height.

There are a couple of ways this can be figured. The "easy way" involves prior knowledge of the location of the center of mass of a cone. Effectively, the work required is that necessary to raise the mass from the height of its center to the height of the discharge pipe.

The "hard way" is to write an expression for the work done to raise an incremental volume, then integrate that over the entire volume. Perhaps this is the method expected in a Calculus class.

<h3>Mass of water</h3>

The mass of the water being raised is the product of the volume of the cone and the density of water.

The cone volume is ...

  V = 1/3πr²h . . . . . . for radius 2 m and height 8 m

  V = 1/3π(2 m)²(8 m) = 32π/3 m³

The mass of water in the cone is then ...

  M = density × volume

  M = (1000 kg/m³)(32π/3 m³) ≈ 3.3510×10^4 kg

<h3>Center of mass</h3>

The center of mass of a cone is 1/4 of the distance from the base to the point. In this cone, it is (1/4)(8 m) = 2 m from the base.

<h3>Easy Way</h3>

The discharge pipe is 2 m above the base of the cone, so is 4 m above the center of mass. The work required to lift the mass from its center to a height of 4 m above its center is ...

  W = Fd = (9.8 m/s²)(3.3510×10^4 kg)(4 m) = 1.3136×10^6 J

<h3>Hard Way</h3>

As the water level in the conical tank decreases, the remaining volume occupies a space that is similar to the entire cone. The scale factor is the ratio of water depth to the height of the tank: (y/8). The remaining volume is the total volume multiplied by the cube of the scale factor.

  V(y) = (32π/3)(y/8)³

The differential volume at height y is the derivative of this:

  dV = π/16y²

The work done to raise this volume of water to a height of 10 m is ...

  (9.8 m/s²)(1000 kg/m³)(dV)((10 -y) m) = 612.5π(y²)(10 -y) J

The total work done is the integral over all heights:

  \displaystyle W=612.5\pi\int_0^8{(10y^2-y^3)}\,dy=\left.612.5\pi y^3\left(\dfrac{10}{3}-\dfrac{y}{4}\right)\right|_0^8\\\\W=612.5\pi\dfrac{2048}{3}\approx\boxed{1.3136\times10^6\quad\text{joules}}

It takes about 1.31 MJ of work to empty the tank.

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