2) When Janet put her marble collection in groups of 7 there were two marbles left over, when she put them in groups of 5 there were 3 marbles left over. What is the fewest number of marbles that Janet could have had in her collection?​

23 marbles

Step-by-step explanation:

3 groups of 7 = 21 +2 = 23

4 groups of 5 = 20 +3 = 23

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.

the answer is two. hope this !