Acollection of dimes and quarters is worth $19.85. there are 128 coins in all. how many of each type of coin are in the collection?

Number of dimes were 81 and number of quarters were 47

Solution:

Let "d" be the number of dimes

Let "q" be the number of quarters

We know that,

value of 1 dime = $ 0.10

value of 1 quarter = $ 0.25

Given that There are 128 coins in all

number of dimes + number of quarters = 128

d + q = 128 ------ eqn 1

Also given that collection of dimes and quarters is worth $19.85

number of dimes x value of 1 dime + number of quarters x value of 1 quarter = 19.85

0.1d + 0.25q = 19.85  -------- eqn 2

Let us solve eqn 1 and eqn 2

From eqn 1,

d = 128 - q -------- eqn 3

Substitute eqn 3 in eqn 2

0.1(128 - q) + 0.25q = 19.85

12.8 - 0.1q + 0.25q = 19.85

12.8 + 0.15q = 19.85

0.15q = 7.05

Therefore from eqn 3,

d = 128 - q

d = 128 - 47

Thus number of dimes were 81 and number of quarters were 47

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.

54.864 meters

step-by-step explanation: