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3 years ago

12

What is the exact volume of a cylinder that has a height of 1.8 ft and a radius of 6.1 ft?

1 answer:

lozanna [386]3 years ago

3 0

Cylinder area=height times area of base (area of base=area of circle)

height =1.8
radius=6.1
area of circle=pi times r^2

area of circle=pi times 6.1^2=37.21π
times height

1.8 times 37.21π=66.978π

volume=66.978π ft^3

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By copying and completing the working below, work out 3 + 1 16 Give your answer as a fraction in its simplest form.

Ksenya-84 [330]

Answer:

Assuming the question is: 3 + 1 16 [I see no "working below."

Step-by-step explanation:

3 + 1 16

48/16 + (16/16 + 1/16) [Make all numbers into fractions using 16 as the denominator. E.g. 48/16 = 3.

Add: 48/16 + (16/16 + 1/16)

(48+16+1)/16

=63/16

[Also equal to 3 15/16]

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2 years ago

Given isosceles ABC, median BD to base AC. Prove ADB=CDB. AB=, AD=, BD=, ADB=

lesya692 [45]

Answer:

1. AB=CB def of isosceles triangle

2. AD=CD DEF of median

3. BD=DB reflexive property

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Step-by-step explanation:

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3 years ago

I NEED HELP ASAP!!!!

abruzzese [7]

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3 years ago

A box with a rectangular base and open top must have a volume of 128 f t 3 . The length of the base is twice the width of base.

noname [10]

Answer:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

Step-by-step explanation:

Given

$Volume = 128ft^3$

$L = 2W$

$Base\ Cost = \$9/ft^2$

$Sides\ Cost = \$6/ft^2$

Required

The dimension that minimizes the cost

The volume is:

$Volume = LWH$

This gives:

$128 = LWH$

Substitute $L = 2W$

$128 = 2W * WH$

$128 = 2W^2H$

Make H the subject

$H = \frac{128}{2W^2}$

$H = \frac{64}{W^2}$

The surface area is:

Area = Area of Bottom + Area of Sides

So, we have:

$A = LW + 2(WH + LH)$

The cost is:

$Cost = 9 * LW + 6 * 2(WH + LH)$

$Cost = 9 * LW + 12(WH + LH)$

$Cost = 9 * LW + 12H(W + L)$

Substitute: $H = \frac{64}{W^2}$ and $L = 2W$

$Cost =9*2W*W + 12 * \frac{64}{W^2}(W + 2W)$

$Cost =18W^2 + \frac{768}{W^2}*3W$

$Cost =18W^2 + \frac{2304}{W}$

To minimize the cost, we differentiate

$C' =2*18W + -1 * 2304W^{-2}$

Then set to 0

$2*18W + -1 * 2304W^{-2} =0$

$36W - 2304W^{-2} =0$

Rewrite as:

$36W = 2304W^{-2}$

Divide both sides by W

$36 = 2304W^{-3}$

Rewrite as:

$36 = \frac{2304}{W^3}$

Solve for $W^3$

$W^3 = \frac{2304}{36}$

$W^3 = 64$

Take cube roots

$W = 4$

Recall that:

$L = 2W$

$L = 2 * 4$

$L = 8$

$H = \frac{64}{W^2}$

$H = \frac{64}{4^2}$

$H = \frac{64}{16}$

$H = 4$

Hence, the dimension that minimizes the cost is:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

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Mrac [35]

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3 years ago