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3 years ago

The altitude of an airplane

Mathematics

2 answers:

NeX [460]3 years ago

8 0

Answer:

33,000 feet to 42,000 feet.

Step-by-step explanation:

But the typical “cruising altitude” – that is, the highest altitude reached during a flight and sustained between the ascent of takeoff and the descent of landing – is around 35,000 feet. That's nearly 7 miles up in the air. However, the number generally varies from about 33,000 feet to 42,000 feet.

kirill [66]3 years ago

5 0

Answer:

it depends

Step-by-step explanation:

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The top and bottom margins of a poster are 6 cm and the side margins are each 8 cm. If the area of printed material on the poste

Degger [83]

Answer:

The total height and width of the poster is approximately 28.93 and 38.57.

Step-by-step explanation:

Consider the provided information.

Let us consider the width and length of the printed part of the poster is x and y respectively.

For better understanding refer the figure 1:

The area of the printed part is 382 cm²

Area = xy = 382

y = 382/x

Now the total height of the poster is y + 6 + 6 = y + 12

Total width of the poster is x + 8 + 8 = x + 16

Thus the total area of the poster is:

A = (x + 16)(y + 12)

Now substitute the value of y in above equation.

$A = (x+16)(\frac{382}{x}+12)$

$A = 382+12x+\frac{6112}{x}+192$

$A = 12x+\frac{6112}{x}+574$

Now differentiate the above equation with respect to x.

$A' = 12-\frac{6112}{x^2}$

Now, substitute A'= 0 and solve for x.

$0 = 12-\frac{6112}{x^2}$

$12=\frac{6112}{x^2}$

$x^2=\frac{6112}{12}$

$x=22.57$ Ignore the negative value of x as width can't be a negative number.

Now find A''

$A'' = \frac{2(6112)}{x^3}$

Here A" is positive for x>0 and x = 22.57 is minimum.

Use the value of x to find all the respective dimensions of the poster.

Substitute the value of x in y = 382/x

y = 382/22.57 = 16.93

The total height of the poster is y + 6 + 6 = y + 12 = 16.93 + 12 = 28.93

Total width of the poster is x + 8 + 8 = x + 16 = 22.57 + 16 = 38.57

Hence, the total height and width of the poster is approximately 28.93 and 38.57.

6 0

2 years ago

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An oblique cylinder has a radius of 9 centimeters and a volume of 486 cubic centimeters. Use Cavalieri’s Principle to calculate

igor_vitrenko [27]

Cavalieri's Principle states that the volume of an oblique cylinder is equal to the product of the area of the base and the height. Mathematically, this can be expressed as,
volume = Bh
V = (πD²/4)h
where D is diameter and h is the height. Substituting the known values from the given above,
486 cm³ =((π)(9 cm)²/4) x h
The value of h from the generated equation is equal to 7.64 cm. Thus, the height of the oblique cylinder is approximately 7.64 cm.

8 0

3 years ago

You are installing new carpeting in a family room. The room is rectangular with dimensions 2012feet × 1318feet . You intend to i

Pavlova-9 [17]

Answer: 1. When you estimate, it is not an exact measurement. 3ft 8 in gets rounded to 4ft and 12 ft 3 in rounds to 12ft. now find the perimeter. P=2l+2w P= 2*12 +2*4 P=32feet

2. 3ft 8in = 3 8/12 or reduced to 3 2/3 12ft 3in = 12 3/12 or reduced to 12 1/4 The fractional part is referring to a fraction of a foot.

3. The perimeter of the room is P=2l+2w or P=2(12 1/4) + 2(3 2/3) p=24 1/2 + 7 1/3 P= 31 5/6 feet

4. The estimate and the actual are very close. They are 1/6 of a foot apart.

5a. Total baseboard 31 5/6ft - 2 1/4 ft = 29 7/12 feet needed.

5b. Take the total and divide it by 8ft = 29 7/12 divided by 8= 3.7 You are not buying a fraction of a board so you would need 4 boards.

7 0

3 years ago

A car is on a driveway that is inclined 10 degrees to the horizontal. A force of 490 lb is required to keep the car from rolling

lara31 [8.8K]

Answer:

a) weight of the car = 2816,1 lbs

b) 2773 lbs

Step-by-step explanation:

The equilibrium force is 490 lbs. That force keep the car at rest, then

∑ Fy = 0 and ∑Fx = 0

Forces acting on the car:

The external force 490 lbs

weight of the car uknown

Normal force

sin∠10° = 0,174

cos∠10° = 0.985

∑Fx = 0 mg*sin10°- 490 = 0 ∑Fy = 0 mg*cos10° - N = 0

mg*0,174= 490

mg = 490 / 0,174

mg = 2816,1 lbs

weight of the car = 2816,1 lbs

The Normal force

mg*cos10° - N = 0 2816,1 * 0,985 = N

N = 2773 lbs

Then equal force in magnitude and in opposite direction will car exets on the driveway

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3 years ago

Jackson has twice as many cards as his friends Alex and Chase combined. Alex has 55 Cards in his collection. Jackson Has 200 car

Zina [86]

Answer:

Step-by-step explanation:

200 divided by 2 = 100-55=45

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3 years ago

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