On a 12 problem test how many ways can 2 problems correct and 10 questions wrong?

Basically we have to calculate how many ways 2 objects can be chosen from 12?

We use this formula

where n = 12 and r =2

12! / (2!) * (12 -2)! =

12! / (2) * 10! =

12 * 11 / 2 = 132 / 2

= 66 ways

Step-by-step explanation:

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.

i dont know nkw

step-by-step explanation:

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