A plane traveled 580 miles to its destination and then returned home. The trip there was into the wind. The trip back took 10 hours. Find the speed of the plane in still air and the speed of the wind

This is the complete question;

A plane traveled 580 miles to its destination and then returned home. The trip there was with the wind. It took 5 hours. The trip back was into the wind. The trip took 10 hours. Find the speed of the plane in still air and the speed of the wind

Speed of plane in still air = 116 miles/hr

Speed of wind = 87 miles/hr

Step-by-step explanation:

Let the speed of the plane in still air be p

Let the speed of the wind be w

Now, we know that speed = distance/time.

Thus, on the trip going, the time was spent was 5 hours while the distance is 580 miles.

Thus, the total speed on this trip going is; 580/5 = 116 miles/hr

Now, the question says this trip was with the wind. Thus it means the plane speed and wind speed aided the movement.

So, p + w = 116 miles/hr - - - - - (eq1)

Now, on the return trip of 10 hours , it says plane was into the wind which means plane speed minus wind speed.

Thus,

p - w = 580/10

p - w = 58 miles/hr - - - - - (eq2)

Add eq 1 to eq 2 to get ;

2p = 116 + 58

2p = 174

p = 87 miles/hr

Plug in this into eq 2 to get

87 - w = 58

w = 87 - 58 = 29 miles/hr

idk

step-by-step explanation:

i don't know

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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