The measure of angle A is 144.3 degrees and the angle to cut the molding is 54.3 degrees
<h3>How to solve for angle A?</h3>
Start by solving the acute part of angle A using the following sine function
sin(Ax) = (30 - 4)/32
Evaluate the quotient
sin(Ax) = 0.8125
Take the arc sin of both sides
Ax = 54.3
The measure of angle A is then calculated as:
A = 90 + Ax
This gives
A = 90 + 54.3
Evaluate
A = 144.3
Hence, the measure of angle A is 144.3 degrees
<h3>The angle to cut the molding</h3>
In (a), we have:
Ax = 54.3
This represents the angle where the molding would be cut
Hence, the angle to cut the molding is 54.3 degrees
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Answer:
ρ_air = 0.15544 kg/m^3
Step-by-step explanation:
Solution:-
- The deflated ball ( no air ) initially weighs:
m1 = 0.615 kg
- The air is pumped into the ball and weight again. The new reading of the ball's weight is:
m2 = 0.624 kg
- The amount of air ( mass of air ) pumped into the ball can be determined from simple arithmetic between inflated and deflated weights of the ball.
m_air = Δm = m2 - m1
m_air = 0.624 - 0.615
m_air = 0.009 kg
- We are to assume that the inflated ball takes a shape of a perfect sphere with radius r = 0.24 m. The volume of the inflated ( air filled ) ball can be determined using the volume of sphere formula:
V_air = 4*π*r^3 / 3
V_air = 4*π*0.24^3 / 3
V_air = 0.05790 m^3
- The density of air ( ρ_air ) is the ratio of mass of air and the volume occupied by air. Expressed as follows:
ρ_air = m_air / V_air
ρ_air = 0.009 / 0.05790
Answer: ρ_air = 0.15544 kg/m^3
Find the median first. The middle of all the numbers.
5, 7, 8, 10, 11, 13, 14, 18, 27
11 is the median.
Then find the median of the upper quartile.
The upper quartile consists of numbers...
5, 7, 8, 10
Since it is an even set of numbers add the two in the middle and divided by two. So 7 plus 8= 15/2.
7.5 is your upper quartile.
Procedure:
1) Integrate the function, from t =0 to t = 60 minutues to obtain the number of liters pumped out in the entire interval, and
2) Substract the result from the initial content of the tank (1000 liters).
Hands on:
Integral of (6 - 6e^-0.13t) dt ]from t =0 to t = 60 min =
= 6t + 6 e^-0.13t / 0.13 = 6t + 46.1538 e^-0.13t ] from t =0 to t = 60 min =
6*60 + 46.1538 e^(-0.13*60) - 0 - 46.1538 = 360 + 0.01891 - 46.1538 = 313.865 liters
2) 1000 liters - 313.865 liters = 613.135 liters
Answer: 613.135 liters
The answer to this question is false because it cannot still be both negative after you reflect over two positives.
