Eva and Dulcina have 82 pins together. Dulcina has 34 more pins than Eva. How many pins are in Dulcina's collection? How many pins are in Eva's collection?

Dulcina's collection has 57 pins

Eva's collection has 25 pins

Step-by-step explanation:

first we are going to represent the relation of pins by means of an equation

d = Dulcina

e = Eva

e + d = 82

and now we represent the relation of the collections of each one

d = e + 32

having the 2 equations we can replace the d by (e + 32) and place it in the other equation

e + d = 82

e + (e + 32) = 82

2e + 32 = 82

2e = 82 - 32

2e = 50

e = 25

Now we replace e with 25 in the second equation and get the value of d

d = e + 32

d = 25 + 32

d = 57

there are 58 pins in Dulcina's collection and 24 pins in Eva's collection

Step-by-step explanation:

Let x represent the number of pins in Dulcina's collection.

Let y represent the number of pins in Eva's collection.

Eva and Dulcina have 82 pins together. This means that

x + y = 82- - - - - - - - - - - - - - - 1

Dulcina has 34 more pins than Eva. This means that

x = y + 34

Substituting x = y + 34 into equation 1, it becomes

y + 34 + y = 82

2y + 34 = 82

2y = 82 - 34

2y = 48

y = 48/2

y = 24

x = y + 34 = 24 + 34

x = 58

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.