The equation given in the question has two unknown variables in the form of "x" and "y". The exact value of "x" and "y" cannot be determined as two equations are needed to get to the exact values of "x" and "y". This equation can definitely be used to show the way for determining the values of "x" in terms of "y"and the value of "y" in terms of "x". Now let us check the equation given.
2x - 5y = - 15
2x = 5y - 15
2x = 5(y - 3)
x = [5(y - 3)]/2
Similarly the way the value of y can be determined in terms of "x" can also be shown.
2x - 5y = - 15
-5y = - 2x - 15
-5y = -(2x + 15)
5y = 2x + 15
y = (2x +15)/5
= (2x/5) + (15/5)
= (2x/5) + 3
So the final value of x is [5(y -3)]/2 and the value of y is (2x/5) + 3.
Answer:
Yes, an arrow can be drawn from 10.3 so the relation is a function.
Step-by-step explanation:
Assuming the diagram on the left is the domain(the inputs) and the diagram on the right is the range(the outputs), yes, an arrow can be drawn from 10.3 and the relation will be a function.
The only time something isn't a function is if two different outputs had the same input. However, it's okay for two different inputs to have the same output.
In this problem, 10.3 is an input. If you drew an arrow from 10.3 to one of the values on the right, 10.3 would end up sharing an output with another input. This is allowed, and the relation would be classified as a function.
However, if you drew multiple arrows from 10.3 to different values on the right, then the relation would no longer be a function because 10.3, a single input, would have multiple outputs.
In distributive property, the answer would be 12(5 + 4j).
Answer:
B
Step-by-step explanation:
29 + (-37) = 29 -37
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1 cup would be the answer
