The equation representing the circle centered at (-27, 120) that passes through the origin is:
Solution:
The general equation of a circle is expressed as:
Where,
(a, b) denotes the center of the circle
r signifies the radius
Given the center as (-27, 120)
Thus;
a = -27
b = 120
Considering it intersects the origin, meaning (x, y) = (0, 0)
Substituting (a, b) = (-27, 120) and (x, y) = (0, 0) into the equation
Input 
= 15129 and (a, b) = (-27, 120) into the equation
Hence, the equation characterizing the circle is determined
The following table presents the conversion from degrees to gradients.
To calculate the slope, we take the difference between the two y-values (gradients) and divide it by the difference between the corresponding x-values (degrees).
For this purpose, we will use the initial and final points listed in the table. Therefore, the slope m is calculated as:
After rounding to two decimal places, the slope of the line converting degrees to gradients is 1.11
Response:
Step-by-step clarification:
Reply:
a) y-8 = (y₀-8), b) 2y -5 = (2y₀-5)
Clarification:
To address these equations, using direct integration is the simplest approach.
a) The equation provided is
dy / dt = -y + 8
dy / (y-8) = dt
We substitute variables
y-8 = u
dy = du
Substituting and integrating gives us
∫ du / u = ∫ dt
Ln (y-8) = t
Evaluating at the lower limits t = 0 for y = y₀
ln (y-8) - ln (y₀-8) = t-0
Simplifying the equation results in
ln (y-8 / y₀-8) = t
y-8 / y₀-8 =
y-8 = (y₀-8)
b) The equation here is
dy / dt = 2y -5
u = 2y -5
du = 2 dy
du / 2u = dt
Integrating now
½ Ln (2y-5) = t
Evaluating at limits gives
½ [ln (2y-5) - ln (2y₀-5)] = t
Ln (2y-5 / 2y₀-5) = 2t
2y -5 = (2y₀-5)
c) The equation here bears a strong resemblance to the previous one
u = 2y -10
du = 2 dy
∫ du / 2u = dt
ln (2y-10) = 2t
We evaluate this to get
ln (2y-10) –ln (2y₀-10) = 2t
2y-10 = (2y₀-10)
