Answer:
Linear
Step-by-step explanation:
A linear equation is an equation of a straight line, which means that the degree of a linear equation must be 0 or 1 for each of its variables. In this case, the degree of variable y is 1 and the degree of variable x is 1
.
Linear
Answer:

Below is the procedure that was used to find the answer.
Step-by-step explanation:
Let be "e" the weight in pounds of the elephant and "c" the weight in pounds of the cat.
According to the information provided in the exercise, we know that The weight of an elephant is
times the weight of a cat. Based on this we can write the following equation:

If the weight in pounds of the elephant is:

We must substitute this value into the equation and then solve for "c" in order to find the weight in pounds of the cat.
Then we get:

Question Two is D. $14.80
Answer:
I have no idea but I suggest making the rule to round the output to the nearest tenth because all the numbers on there have a 9 at the end.
Hope I helped, if I'm right please give brainliest because I cant level up until I get 2 more ;(
<3 Enjoy,
Dea
Answer:
Find the value of x and y in coordinate form, that'll be the point of intersection.
Question 1

Therefore, points of intersection are two
Answer: <u> </u><u>(</u><u>3</u><u>,</u><u> </u><u>4</u><u>)</u><u> </u><u>and</u><u> </u><u>(</u><u>1</u><u>,</u><u> </u><u>2</u><u>)</u>
Question 2:
Following the steps as in question 1

Answer: <u>(2, 0)</u>