It’s 24 you can use a calculator
Answer:
c
Step-by-step explanation:
The limit exists because it's continuous since they equal each other
when x approaches zero.
Substitute both expressions with x=0 and both of them will equal 2, so the limit should be 2.
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.
1st one 17/100 2nd one 0.25
Because 0.17 would be like 17% out of 100% so the fraction simplified is 17/100
2. Since the area of a square is length×length
But according to the question we are asked to find the length and the area is given so we will have to solve this
L×L= area
L^2=0.25
Square root both sides
L=√0.25
L=0.5
Therefore the length of the square is 0.5
Step-by-step explanation:
the answer is the keyboard keys getting stuck, it's something you can touch and feel, thus it's an hardware problem
