we are given
To find solution , we can factor it and then we can solve for x
step-1: Factoring
step-2: Solve for x
so, option-C and option-F...........Answer
<h2>Answer:
</h2><h2>14. 2z + 3yz - 6z +2yz + y - z - 4y
2z - 6z -4z - z
-5z + 3yz + 2yz + y - 4y 3yz + 2yz
-5z + 5yz + y - 4y
</h2><h2>Answer = -5z + 5yz - 3y </h2><h2 /><h2>15. -4ab + 3ax + ba - 6xa
-4ab + ba
Answer = -5ab - 3xa </h2><h2 /><h2>16. -c(3b - 7a)
-3bc - 7ac
Step-by-step explanation:
</h2><h2>Hopes this helps. Mark as brainlest plz!</h2>
Answer:
The relation is not a function
The domain is {1, 2, 3}
The range is {3, 4, 5}
Step-by-step explanation:
A relation of a set of ordered pairs x and y is a function if
- Every x has only one value of y
- x appears once in ordered pairs
<u><em>Examples:</em></u>
- The relation {(1, 2), (-2, 3), (4, 5)} is a function because every x has only one value of y (x = 1 has y = 2, x = -2 has y = 3, x = 4 has y = 5)
- The relation {(1, 2), (-2, 3), (1, 5)} is not a function because one x has two values of y (x = 1 has values of y = 2 and 5)
- The domain is the set of values of x
- The range is the set of values of y
Let us solve the question
∵ The relation = {(1, 3), (2, 3), (3, 4), (2, 5)}
∵ x = 1 has y = 3
∵ x = 2 has y = 3
∵ x = 3 has y = 4
∵ x = 2 has y = 5
→ One x appears twice in the ordered pairs
∵ x = 2 has y = 3 and 5
∴ The relation is not a function because one x has two values of y
∵ The domain is the set of values of x
∴ The domain = {1, 2, 3}
∵ The range is the set of values of y
∴ The range = {3, 4, 5}
Answer: If perpendicular lines intersect, then they form right angles.
Step-by-step explanation:
You take the first part, "perpendicular lines intersect", and add the "if" to it. Then, you take the second part, "to form right angles", and add the "then" in front of it to get your answer, "If perpendicular lines intersect, then they form right angles."
Hope it helps! :)
