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Slav-nsk [51]
3 years ago
7

A plane flies 1640 miles in 4 hours . What is the average speed of the plane ??

Mathematics
2 answers:
tankabanditka [31]3 years ago
8 0

Answer:

410 mph

Step-by-step explanation:

average \: speed  \\ =  \frac{1640}{4}  \\  = 410 \: mph

Mariana [72]3 years ago
6 0

Answer:

410 miles

Step-by-step explanation:

total distance covered (d) = 1640 miles

total time taken (t) = 4 hrs

average distance per hour= d/t

= 1640/4

= 410 miles

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Here are two steps from the derivation of the quadratic formula. What room place between the first step and the second step?
Vaselesa [24]

Answer:

Step-by-step explanation:

The problem solver "completed the square."  He or she took half of the coefficient of x and squared the result, and then added this square and subtracted this square to/from both sides of the equation.

8 0
3 years ago
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Manita carries a box of mass 40 kg. What is the weight of the box?( g=9.8m/s)<br>​
Lilit [14]

Explanation:

Mass of boy = 40 kg

Mass of box = 20 kg

Total mass = 60 kg

Initial potential energy = 0 (since height would be 0 initially)

Final potential energy = mgh

Taking g as 10 m/s²,

=> 60 × 10 × 15

= 9,000 J

So, work done = final potential energy - initial potential energy

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=> Power = 9000/25

=> Power = 360 J/s

or 360 watt.

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3 years ago
Fill in the table what is the constant rate?
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4 years ago
What is the area of the shaded region? Explain please for both #9 and #12
Alja [10]
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5 0
3 years ago
Gravel is being dumped from a conveyor belt at a rate of 20 ft3/min, and its coarseness is such that it forms a pile in the shap
11Alexandr11 [23.1K]

Answer:

0.25 feet per minute

Step-by-step explanation:

Gravel is being dumped from a conveyor belt at a rate of 20 ft3/min. Since we are told that the shape formed is a cone, the rate of change of the volume of the cone.

\dfrac{dV}{dt}=20$ ft^3/min

\text{Volume of a cone}=\dfrac{1}{3}\pi r^2 h

Since base diameter = Height of the Cone

Radius of the Cone = h/2

Therefore,

\text{Volume of the cone}=\dfrac{\pi h}{3} (\dfrac{h}{2}) ^2 \\V=\dfrac{\pi h^3}{12}

\text{Rate of Change of the Volume}, \dfrac{dV}{dt}=\dfrac{3\pi h^2}{12}\dfrac{dh}{dt}

Therefore: \dfrac{3\pi h^2}{12}\dfrac{dh}{dt}=20

We want to determine how fast is the height of the pile is increasing when the pile is 10 feet high.

h=10$ feet$\\\\\dfrac{3\pi *10^2}{12}\dfrac{dh}{dt}=20\\25\pi \dfrac{dh}{dt}=20\\ \dfrac{dh}{dt}= \dfrac{20}{25\pi}\\ \dfrac{dh}{dt}=0.25$ feet per minute (to two decimal places.)

When the pile is 10 feet high, the height of the pile is increasing at a rate of 0.25 feet per minute

5 0
3 years ago
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