Question 3 Find the distance from the vertices of AABC to the corresponding vertices of the other three triangles, and enter
them in the table. For AJKL you'll need to use the distance formula d= V(21 – 22) + (y1 - y2)?. Verify your
calculations using the tools available in GeoGebra.

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.

nswer: a   y-axis

step-by-step explanation:

answer

a) the midrange of the data set shown is 85.5

step by step explanation

let's find the midrange.

the midrange is nothing but add the greatest value and the least value, then divide it by 2.

here the greatest value = 99

the least value = 72

midrange = (99 + 72)/2

= 171/2

midrange = 85.5

you.

secant-secant formula:

(whole secant) x (external part) =           (whole secant) x (external part)

(8+y-2) *8=                                           (y+4+6) *6

(y+6)*8 = (y+10) *6

distribute

8y+48 = 6y +60

subtract 6y from each side

2y +48 = 60

subtract 48 from each side

2y = 12

divide by 2

y = 6

sv =     (y+4+6)

sv   = 6+4+6

sv = 16