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3 years ago

A jet moving at constant speed traveled 6,600 km in 3 hours. What was the speed of the jet?

Mathematics

2 answers:

puteri [66]3 years ago

7 0

Speed=distance ÷ time
speed=6600 ÷3
speed=2200km/h

DanielleElmas [232]3 years ago

3 0

Answer:

The speed of the jet is $2,200\frac{km}{h}$

Step-by-step explanation:

The speed traveled by the jet can be found using the relationship between distance and time. Because the problem gives the values for distance and time, they must be replaced in the equation and in this way the requested speed will be found.

1. Write the general equation for the speed:

$Speed=\frac{Distance}{Time}$

2. Replace the values of distance and time on the general equation for the speed and make the mathematical operations:

$Speed=\frac{6,600km}{3hours}$

$Speed=2,200\frac{km}{h}$

Therefore the speed of the jet is $2,200\frac{km}{h}$

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ryzh [129]

First find the critical points of f :

$f(x,y)=2x^2+3y^2-4x-5=2(x-1)^2+3y^2-7$

$\dfrac{\partial f}{\partial x}=2(x-1)=0\implies x=1$

$\dfrac{\partial f}{\partial y}=6y=0\implies y=0$

so the point (1, 0) is the only critical point, at which we have

$f(1,0)=-7$

Next check for critical points along the boundary, which can be found by converting to polar coordinates:

$f(x,y)=f(10\cos t,10\sin t)=g(t)=295-40\cos t-100\cos^2t$

Find the critical points of g :

$\dfrac{\mathrm dg}{\mathrm dt}=40\sin t+200\sin t\cos t=40\sin t(1+5\cos t)=0$

$\implies\sin t=0\text{ OR }1+5\cos t=0$

$\implies t=n\pi\text{ OR }t=\cos^{-1}\left(-\dfrac15\right)+2n\pi\text{ OR }t=-\cos^{-1}\left(-\dfrac15\right)+2n\pi$

where n is any integer. We get 4 critical points in the interval [0, 2π) at

$t=0\implies f(10,0)=155$

$t=\cos^{-1}\left(-\dfrac15\right)\implies f(-2,4\sqrt6)=299$

$t=\pi\implies f(-10,0)=235$

$t=2\pi-\cos^{-1}\left(-\dfrac15\right)\implies f(-2,-4\sqrt6)=299$

So f has a minimum of -7 and a maximum of 299.

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3 years ago

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Answer:

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Step-by-step explanation:

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ohaa [14]

Answer:

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