Answer:
the rate from the plane to the radar station increasing 3 minutes later is 19.81 km/min
Step-by-step explanation:
A plane flying with a constant speed of 14 km/min passes over a ground radar station at an altitude of 4 km and climbs at an angle of 45 degrees. At what rate is the distance from the plane to the radar station increasing 3 minutes later?
speed (Δs₁)= change in distance(Δd₁) /(Δt₁)change in time
mathematically,
= Δs₁
= 14 km/min = 
and will are to find the speed after passes the radar
=Δs₂ km/min = 
to find d₁ and d₂ with respect to angle 45 degree at 4km
sin 45 = opp/hyp = 4/d₁
d₁ = 4/sin45 = 5.66km
tan 45 == opp/adj = 4/d₂
d₂ =4/tan45 = 4km
for three minute increment,
=3 x d₁ x Δs₁ = 3 x d₂ Δs₂
= 3x5.66x14 = 3x 4 xΔs₂
= 237.72 = 12Δs₂
=Δs₂ = 19.81km/min
<u>Your answer is 28x^2-37x+12</u>
<u>1) Use the FOIL method: (a+b)(c+d)=ac+ad+bc+bd</u>
28x^2-21x-16x+12
<u>2) Collect like terms</u>
28x^2+(-21-16x)+12
<u>3) Simplify</u>
28x^2-37x+12
S = 10*m^(2/3)
S = 10*(450000)^(2/3)
S = 58,723.014617533
S = 58,723
<h3>Answer: Approximately 58,723 square cm.</h3>
Answer:
32.5 feet long and 20 feet wide
Step-by-step explanation:
To solve for the length first, you have it measured as 26 in long and each 4 in is 5 ft so you can go ahead and divide 26 in by 4 in and that gives you 6.5 in so what we did was figure out how many 4 in were in 26 in and therefore we have a 6.5 and so just to match that to the 5 ft per each 4 in (in this case we have 6.5) so we multiply 6.5*5 and that equals 32.5 ft which is the length of the garden bed.
Next we are going to solve for the width which is 16 in wide in the scale and so again each 4 in is 5 ft so here again we're going to divide 16 in by 4 in and that equals 4 in and so again each 4 in is 5 ft so we're going to go ahead and multiply 4 * 5 and that gets you 20 ft which is the width of the garden bed.
