Which of the following equations is of a parabola with a vertex at (3, 0)? Oy=x^2+3
o y=x^2-3
Oy= (x+3)^ 2
y=(x-3)^ 2

D. y = (x - 3)²

Step-by-step explanation:

The parabola formula in vertex form is:

y = a(x - h)² + k

where the vertex is located at (h, k). Replacing with vertex (3, 0) we get:

y = a(x - 3)² + 0

y = a(x - 3)²

In all possible options the leading coefficient a is equal to one. Therefore:

y = (x - 3)²

solution:

we have to construct

∠m nt ≅ ∠p qr

the steps to draw this construction are

step 1: use a compass to draw an arc from point q which intersects the side pq at point a and the side qr at point b.

step 2: draw a segment nt and use the same width of the compass to draw an arc from point n which intersects the segment nt at a point x.

step 3:

step 4: join points n and y using a straightedge.

step 3:

option (d).maintaining the same width of the compass as ab, draw an arc from point x such that it intersects the arc drawn from n in a point y.

to draw   congruent angles , 1. width of compasses while drawing two angles should be same i.e from point at which angle has to be drawn, 2.the distance between the two points where the compass intersects the two sides should be same while drawing any two congruent angles.

34.5 cm

step-by-step explanation:

generally for a circular segment of angle, θ° and radius r

arc length

= (θ/360) x circumference

= (θ/360) x 2πr

in our case, θ = 60°, π=3.142 and r = 33 cm

substitute into above formula

arc length

= (60/360) x 2(3.142)( 33) = 34.5 cm