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saul85 [17]
3 years ago
11

On average, a commercial airplane burns about 3.9 × 10³ milliliters of gasoline per second. There can be as many as 5.1 × 10³ ai

rplanes in the air in the United States at any given time. About how many milliliters per minutes of gasoline are burned by commercial airplanes in the United States at any given time?
Mathematics
2 answers:
Pavel [41]3 years ago
5 0
1 plane burns 60x3.9x10^3
= 234x10^3 mls per minute
Required amount =
234x10^3 x 5.1x10^3
= 11934x10^6
= 1.1934x10^10 MLS Answer
Tresset [83]3 years ago
3 0

60 x 3.9x10-3 = 234x10e3ml/minute/plane

You have 5.1x10e3 planes in the air

(234 x 5.1) (10e6) ---you add the two exponents together for this one

1193.4 x 10e6...which turns into 1.2 x 10e9

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The diameter of a circular garden is 18 feet. What is the approximate area of the garden?
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Travis had a rectangular garden that measured 10feet by 12 feet. He planted pumpkins in his garden. Each pumpkin olant took up 2
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30

dividing the length and width by 2

10 ÷ 2 = 5 and 12 ÷ 2 = 6

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You throw a ball up and its height h can be tracked using the equation h=2x^2-12x+20.
postnew [5]

<em><u>This problem seems to be wrong because no minimum point was found and no point of landing exists</u></em>

Answer:

1) There is no maximum height

2) The ball will never land

Step-by-step explanation:

<u>Derivatives</u>

Sometimes we need to find the maximum or minimum value of a function in a given interval. The derivative is a very handy tool for this task. We only have to compute the first derivative f' and have it equal to 0. That will give us the critical points.

Then, compute the second derivative f'' and evaluate the critical points in there. The criteria establish that

If f''(a) is positive, then x=a is a minimum

If f''(a) is negative, then x=a is a maximum

1)

The function provided in the question is

h(x)=2x^2-12x+20

Let's find the first derivative

h'(x)=4x-12

solving h'=0:

4x-12=0

x=3

Computing h''

h''(x)=4

It means that no matter the value of x, the second derivative is always positive, so x=3 is a minimum. The function doesn't have a local maximum or the ball will never reach a maximum height

2)

To find when will the ball land, we set h=0

2x^2-12x+20=0

Simplifying by 2

x^2-6x+10=0

Completing squares

x^2-6x+9+10-9=0

Factoring and rearranging

(x-3)^2=-1

There is no real value of x to solve the above equation, so the ball will never land.

This problem seems to be wrong because no minimum point was found and no point of landing exists

3 0
3 years ago
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