Answer:
The horizontal distance between the parking lot and the launch site is 470 meters
Step-by-step explanation:
Let
x-----> the horizontal distance between the parking lot and the launch site
we know that
The tangent of angle of 12 degrees is equal to divide the altitude of 100 meters by the horizontal distance
so
tan(12°)=100/x
Solve for x
x=100/tan(12°)=470 meters
Answer:
Density is defined as:
Density = mass/volume.
We know that:
For liquid A:
Density = 70kg/m^3
Mass = 1400kg
Then the volume is:
Volume = mass/density = (1400kg)/(70kg/m^3) = 20 m^3
For liquid B:
Density = 280 kg/m^3
Volume = 30m^3
We can find the mass of liquid B as:
mass = density*volume = (280kg/m^3)*(30m^3) = 8400 kg
We know that liquid C is a mixture of liquid A and B.
Then the mass of liquid C will be equal to the sum of the masses of liquid A and B, then:
Mass of liquid C = 1400kg + 8400kg = 9800kg
The same happens for the volume, then:
Volume of liquid C = 30m^3 + 20m^3 = 50m^3
Then the density of liquid C is:
Density of liquid C = (9800kg)/(50m^3) = 196 kg/m^3
Answer:
Step-by-step explanation:
Y=-3x+15
Comparing to y = mx + c
Slope m = -3
Intercept c = 15
<span> I am assuming you want to prove:
csc(x)/[1 - cos(x)] = [1 + cos(x)]/sin^3(x).
</span>
<span>If we multiply the LHS by [1 + cos(x)]/[1 + cos(x)], we get:
LHS = csc(x)/[1 - cos(x)]
= {csc(x)[1 + cos(x)]/{[1 + cos(x)][1 - cos(x)]}
= {csc(x)[1 + cos(x)]}/[1 - cos^2(x)], via difference of squares
= {csc(x)[1 + cos(x)]}/sin^2(x), since sin^2(x) = 1 - cos^2(x).
</span>
<span>Then, since csc(x) = 1/sin(x):
LHS = {csc(x)[1 + cos(x)]}/sin^2(x)
= {[1 + cos(x)]/sin(x)}/sin^2(x)
= [1 + cos(x)]/sin^3(x)
= RHS.
</span>
<span>I hope this helps! </span>
