Answer:
#1 is 2
#2 is -2/3
#3 is -2/5
#4 is -1/2
#5 is 4
#6 is 1
Step-by-step explanation:
I hope this helps
Answer:
(a) false
(b) true
(c) true
(d) true
(e) false
(f) true
(g) false
(h) true
(i) true
Step-by-step explanation:
(a) 15 ⊂ A, since 15 is not a set, but an element, we cannot say of an element to be subset of a set. False
(b) {15} ⊂ A The subset {15} is a subset of A, since every element of {15}, that is 15, belongs to A.
15 ∈ {15} and 15 ∈ { x ∈ Z: x is an integer multiple of 3 } 15 is an integer multiple of 3. since 15/3=5. True
(c)∅ ⊂ A
∅ is a subset of any set. True
(d) A ⊆ A
A is a subset of itself. True
(e)∅ ∈ B
∅ is not an element, it is a subset, so it does not belong to any set. False
(f)A is an infinite set.
Yes, there are infinite integers multiple of 3. True
(g)B is a finite set.
No, there are infinite integers that are perfect squares. False
(h)|E| = 3
The number of elements that belong to E are 3. True
(i)|E| = |F|
The number of elements that belong to F are 3. So is the number of elements of E. True
The first one (2-x^3) is the correct one
The solutions of the equations of the situation can be:
z = 1 , y = 5, x = 4
z= 2 , y = 3, x = 5
z=3 , y = 1, x = 6
z = 0 , y = 7, x =3
The question can be expressed as a equation
6 x + 8 y + 10 z = 84
also, x + y + z = 10
⇒ x = 10- y - z
Putting it in first equation,
6(10 - y - z ) +8y + 10z = 84
⇒ 60 +2y + 4z = 84
⇒2(y + 2z ) = 14
⇒ y + 2z = 7
Now putting
z = 1 , we get y = 5,
z= 2 , y = 3
z=3 , y = 1
z = 0 , y = 7
So, only 4 possible solutions.
Therefore, there can only be 4 possible solutions for the equations.
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