Find all points of intersection of the given curves. (Assume 0 ≤ θ < 2π and r ≥ 0. Order your answers from smallest to largest θ. If an intersection occurs at the pole, enter POLE in the first answer blank.) r = 1 + cos(θ), r = 1 − sin(θ)

The points are (3π/4, ) and (7π/4, )

Step-by-step explanation:

Since the two equations r = 1 + cos(θ) and r = 1 − sin(θ) intersect, then

1 + cos(θ) = 1 − sin(θ)

collecting like terms

cos(θ) + sin(θ) = 1 - 1

cos(θ) + sin(θ) = 0

dividing through by cos(θ), we have

cos(θ)/cos(θ) + sin(θ)/cos(θ) = 0/cos(θ)

1 + tan(θ) = 0

tan(θ) = -1

Since tanθ is negative in both the second and fourth quadrant, we have

tan(π - θ) = 1 or tan(2π - θ) = 1

(π - θ) = tan⁻¹1 or (2π - θ) = tan⁻¹1

(π - θ) = π/4 or (2π - θ) = π/4

θ = π - π/4 = 3π/4 or θ = 2π - π/4 = 7π/4

Substituting these values into r = 1 + cos(θ), we have

r = 1 - cos(3π/4) or r = 1 - cos(7π/4)

r = 1 - (-1/√2) or r = 1 - 1/√2

r = 1 + √2/2 or r = 1 - √2/2

r = or r =

So, the points are (3π/4, ) and (7π/4, )

see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -

25/6

step-by-step explanation:

you multiply 4 by the denominator (4*6=24) and add it to the numerator which makes it 25/6.