Enlarge the triangle by scale factor 1.5 using (4,4) as the centre of enlargment

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.

y=-2.5x+12.5

step-by-step explanation:

5y=2x+10

y=0.4x+2

slope=-2.5

5=-2.5*3+b

b=12.5

the equation of the parallel line would be y = 2x + 9

step-by-step explanation:

to start we need to find the slope of the original line. we can do that by looking at the coefficient of x, which is 2

since the slope of the original line is 2 and parallel lines have the same slope, we know the new line will also have 2 slope. now we can use that along with the point given in point-slope form to get the find equation.

y - y1 = m(x - x1)

y - 5 = 2(x + 2)

y - 5 = 2x + 4

y = 2x + 9

hint: which coefficient is a multiple of 360°?

step-by-step explanation: