They are congruent. Distance formula works.
Answer:
-1.15
Step-by-step explanation:
The tangent of the angle can be found as ...
tan(t) = (y-coordinate)/(x-coordinate) = -4/7
The secant of the angle is related by ...
sec(t)² = tan(t)² +1 = (-4/7)² +1 = 65/49
Then the secant of this 2nd-quadrant angle is ...
sec(t) = -√(65/49) = -(√65)/7 ≈ -1.15
Answer:
a=2.5
Step-by-step explanation:
3a-2.4=5.1 Add the like terms ( plus 2.4 cancels out -2.4)
+2.4 +2.4
3a=7.5 Now divide by 3
/3 /3
a=2.5
These are two questions and two answers.
Question 1) Which of the following polar equations is equivalent to the parametric equations below?
<span>
x=t²
y=2t</span>
Answer: option <span>A.) r = 4cot(theta)csc(theta)
</span>
Explanation:
1) Polar coordinates ⇒ x = r cosθ and y = r sinθ
2) replace x and y in the parametric equations:
r cosθ = t²
r sinθ = 2t
3) work r sinθ = 2t
r sinθ/2 = t
(r sinθ / 2)² = t²
4) equal both expressions for t²
r cos θ = (r sin θ / 2 )²
5) simplify
r cos θ = r² (sin θ)² / 4
4 = r (sinθ)² / cos θ
r = 4 cosθ / (sinθ)²
r = 4 cot θ csc θ ↔ which is the option A.
Question 2) Which polar equation is equivalent to the parametric equations below?
<span>
x=sin(theta)cos(theta)+cos(theta)
y=sin^2(theta)+sin(theta)</span>
Answer: option B) r = sinθ + 1
Explanation:
1) Polar coordinates ⇒ x = r cosθ, and y = r sinθ
2) replace x and y in the parametric equations:
a) r cosθ = sin(θ)cos(θ)+cos(θ)
<span>
b) r sinθ =sin²(θ)+sin(θ)</span>
3) work both equations
a) r cosθ = sin(θ)cos(θ)+cos(θ) ⇒ r cosθ = cosθ [ sin θ + 1] ⇒ r = sinθ + 1
<span>
b) r sinθ =sin²(θ)+sin(θ) ⇒ r sinθ = sinθ [sinθ + 1] ⇒ r = sinθ + 1
</span><span>
</span><span>
</span>Therefore, the answer is r = sinθ + 1 which is the option B.
