The equations are

,
The graphs of the solutions (x, y) of these equations are 2 parabolas, since the right hand side expressions are polynomials of degree 2.
The solution/s of the system are the x-coordinates of the point/s of intersection of the parabolas.
The solutions of the first equation form a parabola looking downwards (since the coefficient of x^2 is -), and the second, a parabola opening upwards (since the coefficient of x^2 is +).
We can draw both parabolas, but to find the solution we still need to solve the system algebraically.
The algebraic solution of the system is:

, so
the solutions are x=-1 and x=1.
The graph of the system is drawn using desmos.com
If we are allowed to use a graphic calculator, we can draw both graphs and point at the solution.
<span>the highest point; the top or apex </span>
Answer:
1/4, 2/6, 4/10
Step-by-step explanation:
1/4 = .25
2/6 = .333...
4/10 = .4
Answer:

Assuming that there are only 2 flavors of the sweet ⇻ strawberry & orange.
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P (strawberry flavored sweet) = 0.8
P (orange flavored sweet) = ?
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Now, we know that ➳ the value of probability ≤ 1. So, the probability of orange flavored sweets is less than 1.
____________________
So,
P (orange flavored sweet) = 1 - P (strawberry flavored sweet)
P (orange flavored sweet) = 
P (orange flavored sweet) = 
____________________
✐ The probability that the sweet is orange flavored is <u>0</u><u>.</u><u>2</u>
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# ꧁❣ RainbowSalt2²2² ࿐