Suppose we want to make a quadratic polynomial y = f(x) = a0 + a1x + a2x
2
that passes through three specified points p1 = (p1, q1), p2 = (p2, q2), p3 = (p3, q3). Consider the
equation
0 = det




1 x x2 y
1 p1 p
2
1
q1
1 p2 p
2
2
q2
1 p3 p
2
3
q3




.
The determinant above implicitly gives an equation y = f(x) (it’s easy to solve for y since no y2, y3,etc terms appear).
(a) Write out the matrix above, using p1 = (0, 0), p2 = (1, 1), p3 = (3, 5) for the constants pi, qi, but leaving x and y as variables. Solve the equation det(A) = 0 to get y = a quadratic polynomial in x. Check directly that the parabola passes through p1, p2, p3.

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!

hello from mrbilldoesmath!

answer:   x^2 + y^2 = 1024

discussion:

the equation of a circle with center (h,k) and radius "r" is

(x-h)^2 + (y-k)^2 = r^2.

in our case

(x-0)^2 + (y-0) ^2 = 32^2 = 1024

you,

mrb

he answer is x=1x3 because x=1 to x=3

english maybe i can ..