Let the number of nickels found be n, the number of dimes be d and the number of quarters be q.
i) "you have twice as many quarters as dimes and 42 coins in all."
means that 2d=q, and n+d+q=42
we can reduce the number of unknowns by substituting q with 2d:
n+d+q=42
n+d+2d=42
n+3d=42
We can write all n, d and q in terms of d as follows:
there are n=42-3d nickels, d dimes and q=2d quarters.
ii) In total there are $6.60 dollars,
1 nickel = 5 cent = $0.05
1 dime = 10 cent = $0.1
1 quarter = 25 cent = $0.25
thus
(42-3d)*0.05 + d*0.1 +2d*0.25= $6.60
2.1 - 0.15d+0.1d+0.5d=6.60
2.1+0.45d=6.6
0.45d=6.6-2.1=4.5
d=4.5/0.45=10
iii)
so, there are 10 dimes, 2d=2*10=20 quarters and 42-3d=42-3*10=12 nickels.
Answer: 10 dimes, 20 quarters, 12 nickels
Answer:
See below
Step-by-step explanation:
Remember the notation and rules of quantifiers. ∀ is the universal quantifier and ∃ is the existential quantifier. To negate ∀x p(x) , write ∃x ¬p(x). To negate ∃x p(x) , write ∀x ¬p(x)
Part I:
A) None of life's problems have a simple solution.
B) All of life's problems have a simple solution.
C) Some of life's problems have a simple solution
D) All of life's problems have a simple solution (notice how the original statements in B and D mean exactly the same)
E) Some of life's problems do not have a simple solution.
Part II: Let x be a variable representing one of life's problems, y be a variable representing solutions, p(x):="x has a simple solution", and q(x,y):="y is a simple solution of x".
A) (∀x)(¬p(x)) or ¬(∃x)(p(x))
B) (∀x)(∃y)(q(x,y))
C) (∃y)(∀x)(q(x,y)). Note that the order of quantifiers is important. B) and C) have different meanings. In C) there is an universal solution of all problems, in B) each problem has its solution.
D) (∀x)(p(x))
E) Same as C)
4(c+2)<4c+10
4c+8<4c+10
-4c -4c
--------------------
8<10
Which gives you No Solution.
