Eliminate the parameter.
x = 3 cos t, y = 3 sin t

x^2+y^2 = 3^2

Step-by-step explanation:

We need to eliminate the parameter t

Given:

x = 3 cos t

y = 3 sin t

Squaring the above both equations

(x)^2=(3 cos t)^2

(y)^2 =(3 sin t)^2

x^2 = 3^2 cos^2t

y^2=3^2 sin^2t

Now adding both equations

x^2+y^2=3^2 cos^2t+3^2 sin^2t

Taking 3^2 common

x^2+y^2=3^2 (cos^2t+sin^2t)

We know that cos^2t+sin^2t = 1

so, putting the value

x^2+y^2=3^2(1)

x^2+y^2 = 3^2

Hence the parameter t is eliminated.

region b will make the inequalities y < 2x + 1 and y > -x - 2 true ⇒ answer a

step-by-step explanation:

* lets study the graph and the inequality to solve the question

- the inequality y > -x - 2

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = -2 that means the line intersect the y-axis at point (0 , -2)

∵ the sign of the inequality is >

∴ we shad the part over the line

  region a and b

- the inequality y < 2x + 1

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = 1 that means the line intersect the y-axis at point (0 , 1)

∵ the sign of the inequality is <

∴ we shad the part under the line

  region b and c

∵ the common region between the two lines is region b

∴ region b will make the inequalities y < 2x + 1 and y > -x - 2 true