Plz me but only the top two and plz show ur work and sorry if its blurry and nu, ber one says write an expression for this array using the distributibe property. number 2 says draw an array to match the expression

up

step-by-step explanation:

the orientation of the parabolas can easily be found by analysing the power of x and y and their signs. if x is squared , parabola opens up and also if its coefficient is negative it opens down. if y is squared , the parabola opens right and also if its coefficient is negative it opens left. however in order to do the above analysis our equation must be in standard form .

our parabola is

which is not standard form of any parabola . hence we square on both hand side and convert it into standard form.

hence we see that x is squared and its coefficient is non negative . there for it opens up.

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.