Answer:
x = 9
Step-by-step explanation:
The product of the external part and the entire part of one secant is equal to the product of the external part and the entire part of the other secant, that is
5(x - 6 + 5) = 4(x - 3 + 4)
5(x - 1) = 4(x + 1) ← distribute parenthesis on both sides
5x - 5 = 4x + 4 ( subtract 4x from both sides )
x - 5 = 4 ( add 5 to both sides )
x = 9
Answer:
y = 2x - 3
Step-by-step explanation:
y = mx + b
m = (y2-y1)/(x2-x1)
m = (3-(-3))/(3-0)
m = 6/3
m = 2
y = mx + b
y = 2x + b
3 = 2(3) + b
3 = 6 + b
b = -3
therefore
y = 2x - 3
Both methods start out with a factor<span> tree. A </span>factor<span> tree is a diagram that is used to break down a number into its </span>factors until<span> all the numbers left are </span>prime. The first way you can use a factor<span> tree to find the </span>factorization of a number is to divide out prime<span> numbers only. Let's </span>factor<span> 24 using this method.</span>
Answer:
We have to prove,
(A \ B) ∪ ( B \ A ) = (A U B) \ (B ∩ A).
Suppose,
x ∈ (A \ B) ∪ ( B \ A ), where x is an arbitrary,
⇒ x ∈ A \ B or x ∈ B \ A
⇒ x ∈ A and x ∉ B or x ∈ B and x ∉ A
⇒ x ∈ A or x ∈ B and x ∉ B and x ∉ A
⇒ x ∈ A ∪ B and x ∉ B ∩ A
⇒ x ∈ ( A ∪ B ) \ ( B ∩ A )
Conversely,
Suppose,
y ∈ ( A ∪ B ) \ ( B ∩ A ), where, y is an arbitrary.
⇒ y ∈ A ∪ B and x ∉ B ∩ A
⇒ y ∈ A or y ∈ B and y ∉ B or y ∉ A
⇒ y ∈ A and y ∉ B or y ∈ B and y ∉ A
⇒ y ∈ A \ B or y ∈ B \ A
⇒ y ∈ ( A \ B ) ∪ ( B \ A )
Hence, proved......
