1,459.75 - 200.25 - 359.45 - 125.00 - 299.35 = 475.70
475.70 + 375.00 = 850.70
Her account balance after the bills and deposit is $850.70
Let x- intercept represents time
Let y-intercept represents population
Let 1990 represent initial year
Points are (0,14.2)(10,12.4)
slope m= 
= 
General linear equation
y= mx+b --------------(i)
here m is slope and b is intercept
plug the 1st point in equation
14.2= -0.018(0) + b
b=14.2
y= 0.018x + 14.2
replace y with p(t) and x with t
p(t) = -0.018t + 14.2
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.
Step-by-step explanation:



HOPE IT HELPS Y
14-10=? less than indicates that a minus sign is needed.