How can you tell if a function is linear by looking at the graph or the equation

1 answer:

7 0

Every linear graph is just a straight line, so if there are any curves or unnatural shapes, than you know it is not linear. As for equations, if it can be shaped into y=mx+b where m and b are numbers, then it will be a linear equation.

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Pls help meeee Idk how to do this

34kurt

Distribute the x3 to each number. The coefficient stays the same but add exponent (3) onto each x. Final answer is: x to the seventh power plus 8X to the sixth power plus 2X to the fifth power +18 X to the fourth power +9x the third power

8 0

3 years ago

Read 2 more answers

Help, I need it by 9:00 am

olya-2409 [2.1K]

<h3>Answer:</h3>

Option B. because all rational numbers are integers.

Have a look at the above venn diagram showing the relationship between - Real, irrational, rational, integers, whole and natural numbers.

$\longrightarrow{\green{Option \: B.}}⟶Option \: B.$

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2 years ago

In a smoothie the ratio of ounces of blueberries to ounces of yogurt is 4 to 5.

KonstantinChe [14]

Answer:

12.5

Step-by-step explanation:

for people who don't get another quizzez assignment about ratios from their teachers, this question showed & the correct answer was 12.5 Hope this answer somehow helps.

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2 years ago

All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be

Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a student is proficient in reading but not mathematics and $A \cap B$ is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

$B = b + (A \cap B)$

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

$(A \cup B) + C = 1$