Select all of the following that are quadratic functions. x = 3y2 – 6y + 5, y + 3 = –2 x 2+ 5 , y = 2 x 2 – 8 x + 6, y= –7 x – 4, y – 3 x = 4 x 3 – x2 + 9, y = 5 x(x + 9) – 8, y – 2 x 2 = 3 x – 2 x 2 + 4

The standard form is ax² + bx + c = 0 with a, b, and c being constants, or numerical coefficients, and x is an unknown variable and if the highest power in the quation is 2 then it is a quadratic equation.

----> Quadratic Equation

y +3 = -2x² +5----> Quadratic Equation

----> Quadratic Equation

----> Non Quadratic Equation

----> Non Quadratic Equation

y = 5 x(x + 9) - 8----> Quadratic Equation

----> Quadratic Equation

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what is the image?

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.