Answer:
<h3>44 ft</h3>
Step-by-step explanation:
The set up will give a right angled triangle where;
Base of the triangle = 90 ft (distance of the offshore from the base of the house)
theta = 26 degrees (angle of elevation)
Height of the house will be the opposite side = x
Using the TOA trig identity;
tan 26 = opp/hyp
tan 26 x/90
x = 90tan 26
x = 90(0.4877)
x = 43.89 ft
Hence the nearest height of the house is 44ft
Answer:
1. x = -6, -8, -90
2. solutions for include -5, while -5 is not a solution for
Step-by-step explanation:
I put it in my calculator and go 17/30
14400 = P(0.08)(30)
14400 = P(2.4)
P = 14400/2.4
P = 6000 - The initial principal
Answer:
The minimum volume of hot air needed to lift a hot air balloon carrying 800kg
Is 653.1m³
Step-by-step explanation:
This problem bothers on the density of materials in this problem air
Given data
Volume of air V=?
Mass of air m= 800kg
From tables the density of air is
ρ =1.225 kg/m3
Now to solve for the volume we have
ρ=m/v
v=m/ρ
v=800/1.225
Volume v = 653.1m³