MATHEMATICS COLLEGE

Kendra is participating in a fundraiser walk-a-thon she walks 4 miles in 60 minutes how many minutes will it take to walk 7 miles

Answers

Answer 1

Answer:

Answer:

105 min

Step-by-step explanation:

60 minutes ÷ 4 miles= 15 minutes per mile

15min per mile x 7 miles = 105 minutes

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I need help with this problem, please help!!!

Answers

Answer:

Yes. (See below for explanation.)

Step-by-step explanation:

The number of servings is found by dividing the quantity available by the size of a serving. The quantity of punch is the sum of the quantities of the juices that go into the punch. The serving size of 3/4 cup is the same as 6 ounces, since a cup is 8 ounces. (3/4 × 8 oz = 6 oz)

The quantity available is (64 oz + 28 oz + 76 oz). The serving size is 6 oz. Since the units of numerator and denominator are the same, they cancel, leaving ...

... number of servings = (quantity available)/(serving size)

... = (64 +28 +76)/6 . . . . as shown in the problem statement

_____

It might not be obvious that the above ratio gives the number of servings. However, if you look at the real units, you see how it happens.

$(oz)/(((oz)/(serving)))= oz(serving)/(oz)=(oz)/(oz)serving=serving$

Find the radius and height of a cylindrical soda can with a volume of 256cm^3 that minimize the surface area.B: Compare your answer in part A to a real soda can, which has a volume of 256cm^3, a radius of 2.8 cm, and a height of 10.7 cm, to conclude that real soda cans do not seem to have an optimal design. Then use the fact that real soda cans have a double thickness in their top and bottom surfaces to find the radius and height that minimizes the surface area of a real can (the surface area of the top and bottom are now twice their values in part A.

B: New radius=?

New height=?

Answers

Answer:

A) Radius: 3.44 cm.

Height: 6.88 cm.

B) Radius: 2.73 cm.

Height: 10.92 cm.

Step-by-step explanation:

We have to solve a optimization problem with constraints. The surface area has to be minimized, restrained to a fixed volumen.

a) We can express the volume of the soda can as:

$V=\pi r^2h=256$

This is the constraint.

The function we want to minimize is the surface, and it can be expressed as:

$S=2\pi rh+2\pi r^2$

To solve this, we can express h in function of r:

$V=\pi r^2h=256\n\nh=(256)/(\pi r^2)$

And replace it in the surface equation

$S=2\pi rh+2\pi r^2=2\pi r((256)/(\pi r^2))+2\pi r^2=(512)/(r) +2\pi r^2$

To optimize the function, we derive and equal to zero

$(dS)/(dr)=512*(-1)*r^(-2)+4\pi r=0\n\n(-512)/(r^2)+4\pi r=0\n\nr^3=(512)/(4\pi) \n\nr=\sqrt[3]{(512)/(4\pi) } =\sqrt[3]{40.74 }=3.44$

The radius that minimizes the surface is r=3.44 cm.

The height is then

$h=(256)/(\pi r^2)=(256)/(\pi (3.44)^2)=6.88$

The height that minimizes the surface is h=6.88 cm.

b) The new equation for the real surface is:

$S=2\pi rh+2*(2\pi r^2)=2\pi rh+4\pi r^2$

We derive and equal to zero

$(dS)/(dr)=512*(-1)*r^(-2)+8\pi r=0\n\n(-512)/(r^2)+8\pi r=0\n\nr^3=(512)/(8\pi) \n\nr=\sqrt[3]{(512)/(8\pi)}=\sqrt[3]{20.37}=2.73$

The radius that minimizes the real surface is r=2.73 cm.

The height is then

$h=(256)/(\pi r^2)=(256)/(\pi (2.73)^2)=10.92$

The height that minimizes the real surface is h=10.92 cm.

Final answer:

The minimal surface area for a cylindrical can of 256cm^3 is achieved with radius 3.03 cm and height 8.9 cm under uniform thickness, and radius 3.383 cm and height 7.14 cm with double thickness at top and bottom. Real cans deviate slightly from these dimensions possibly due to practicality.

Explanation:

For a cylinder with given volume, the surface area A, radius r, and height h are related by the formula A = 2πrh + 2πr^2 (if the thickness is uniform) or A = 3πrh + 2πr^2 (if the top and bottom are double thickness). By taking the derivative of A w.r.t r and setting it to zero, we can find the optimal values that minimize A.

For a volume of 256 cm^3, this gives us r = 3.03 cm and h = 8.9 cm with uniform thickness, and r = 3.383 cm and h = 7.14 cm with double thickness at the top and bottom. Comparing these optimal dimensions to a real soda can (r = 2.8 cm, h = 10.7 cm), we see that the real can has similar but not exactly optimal dimensions. This may be due to practical considerations like stability and ease of holding the can.