From the graph, we can see that the graph crosses the x-axis at the point (1.5, 0) the graph also passes through point (1, 1) and the graph crosses the y-axis at the point (0, 3).
Therefore, points (1.5, 0), (1, 1) and (0, 3) are some of the solutions of the graph.
Answer:
A) TA
Step-by-step explanation:
A radius goes from the center of a circle to the outside edge.
RP connects two points on the outside edge of the circle without going through the center. This is not a radius.
TR connects two opposite points on the outside edge of the circle while going through the center. This is a diameter, not a radius.
GP connects two opposite points on the outside edge of the circle while going through the center. This is a diameter, not a radius.
TA connects the center of the circle to the outside edge. This is a radius.
co-ordinates of point G are (1,2)
co-ordinates of point H are (3,-3)
so, by distance formula :

so, it is root(29) or 5.38 units
The Bernoulli equation is almost identical to the standard linear ODE.

Compare to the basic linear ODE,

Meanwhile, the Riccati equation takes the form

which in special cases is of Bernoulli type if

, and linear if

. But in general each type takes a different method to solve. From now on, I'll abbreviate the coefficient functions as

for brevity.
For Bernoulli equations, the standard approach is to write


and substitute

. This makes

, so the ODE is rewritten as

and the equation is now linear in

.
The Riccati equation, on the other hand, requires a different substitution. Set

, so that

. Then you have



Next, setting

, so that

, allows you to write this as a linear second-order equation. You have



where

and

.