5(2y-4) - 3y = 1. 10y-20-3y=1. 7y-20=1. 7y=21. y =3. x = 2(3)-4. x = 2. x*y =6.
Answer:
A
Step-by-step explanation:
Choice A is right because there is no dot above the two.
Choice B is not right because there is 2 dots above 1 and not 3.
Choice C is not right because the data doesn't go passed 5 to get to 8
Choice D is not right because it was 5 games because there is 5 dots above 3
An angle between (not equal to) 0 and 90 degrees
Answer:
<h2>DNE</h2>
Step-by-step explanation:
Given the limit of the function
, to find the limit, the following steps must be taken.
Step 1: Substitute the limit at x = 0 and y = 0 into the function

Step 2: Substitute y = mx int o the function and simplify


<em>Since there are still variable 'm' in the resulting function, this shows that the limit of the function does not exist, Hence, the function DNE</em>
Answer:
the answer is B
Step-by-step explanation:
i just took the test, hope this can help?