Apples lose 75% of their weight during drying. How many pounds of fresh apples do you need to pick in order to get 150 lb of dri
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Hello,
y=5+3*cos (2(x-π/3))
The function is periodic with periode=2π.
-1<=cos (2(x-π/3))<=1
==>-1*3<=3*cos (2(x-π/3))<=3*1
==>5-3<=5+3cos(2(x-π/3))<=5+3
==>2<= y<=8
y=5+3*cos (2(x-π/3))
The function is periodic with periode=2π.
-1<=cos (2(x-π/3))<=1
==>-1*3<=3*cos (2(x-π/3))<=3*1
==>5-3<=5+3cos(2(x-π/3))<=5+3
==>2<= y<=8
Distance traveled in clear weather = 50 miles
Distance traveled in thunderstorm = 15 miles
Let speed in clear weather = x
⇒ Speed in thunderstorm = x-20
Total time taken for trip = 1.5 hours
We need to determine average speed in clear weather (i.e. x) and average speed in the thunderstorm (i.e. x-20 ).
Total time taken for trip = Time taken in clear weather + Time taken in thunderstorm
⇒ Total time taken for trip = +
⇒ 1.5 = +
⇒ 1.5 =
⇒ 15*x*(x-20) = 10*[50*(x-20)+15*x]
⇒ 15x² - 300x = 500x - 10,000 + 150x
⇒ 15x² - 300x = 650x - 10,000
⇒ 15x² - 950x + 10,000 = 0
⇒ 3x² - 190x + 2,000 = 0
The above equation is in the format of ax² + bx + c = 0
To determine the roots of the equation, we will first determine 'D'
D = b² - 4ac
⇒ D = (-190)² - 4*3*2,000
⇒ D = 36,100 - 24,000
⇒ D = 12,100
Now using the D to determine the two roots of the equation
Roots are: x₁ = ; x₂ =
⇒ x₁ = and x₂ =
⇒ x₁ = and x₂ =
⇒ x₁ = and x₂ =
⇒ x₁ = 50 and x₂ = 13.33
So speed in clear weather can be 50 mph or 13.33 mph. However, we know that in thunderstorm was 20 mph less than speed in clear weather.
If speed in clear weather is 13.33 mph then speed in thunderstorm would be negative, which is not possible since speed can't be negative.
Hence, the speed in clear weather would be 50 mph, and in thunderstorm would be 20 mph less, i.e. 30 mph.