Answer:
(1,2021)
Step-by-step explanation:
P and q can vary subject to their sum being 2020.
Consider one parabola with p1 and q1 and another with p2 and q2.
y1=(x1)^2+(p1)(x1)+(q1)
y1=(x2)^2+(p2)(x2)+(q2)
At their intersection, the x and y coordinates are the same.
y1=y2=y
x1=x2=x
x^2+(p1)x+(q1)=x^2+(p2)x+(q2)
Solve for x
x(p1-p2)=q2-q1
x=(q2-q1)/(p1-p2)
Use the constraint that p+q=2020 to eliminate p1 and p2.
p1=2020-q1
p2=2020-q2
x=(q2-q1)/(2020-q1-2020+q2)
x=(q2-q1)/(q2-q1)
x=1
Substitute in the equation for y.
y=1^2+p(1)+q
y=2021
Answer:
57
Step-by-step explanation:
Substitute the given values into the expression
b² - 4ac
= 9² - (4 × - 2 × - 3)
= 81 - (24)
= 81 - 24
= 57
9 cakes and you would have 1/4 a cup of nuts left over. :)