3 = 4 - 4^0 * 4/4
4 = 4^0 + 4^0 + 4^0 + 4^0
5 = 4^2 / (4^0 * 4) + 4^0
6 = 4^2 / 4 + 4^0 + 4^0
If;
A = Adjacent
O = Opposite
H = Hypotenuse
Then,
Sin Ф = O/H
Cos Ф = A/H
Therefore,
(Sin Ф)/Cos Ф) = (O/H)/(A/H) = (O/H)*(H/A) = O/A
Now,
tan Ф = O/A ----
Therefore, it is true that
tan Ф = SinФ/CosФ
-3/4x + 5/6 y = 15
Multiply through by 6
-9/2x + 5y = 90
Add 9/2x to both sides
5y = 90 + 9/2x
Divide both sides by 5
y = 18 + 9/10x
y = 9/10x + 18
Comparing with the general equation of a straight line y = mx + c
Gradient (m) = 9/10 or 0.9
and the y - intercept (c) = 18
From the given question we come to know of certain number of facts and they are:
At 1:00 PM the water level of the pond was = 13 inches
At 1:30 PM the water level of the pond was = 18 inches
At 2:30 PM the water level of the pond was = 28 inches
From the above given facts we can easily find the amount of water changing every half an hour.
Amount of increase in water from 1:00PM to 1:30 PM = (18 - 13) inches
= 5 inches
Amount of increase in water level from 1:30PM to 2:30PM = (28 -18) inches
= 10 inches
From the above two deductions we can come to the conclusion the the constant rate of change in water level is 5 inches for every half an hour.
