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3 years ago

The graph attaches shows the function f(x) and the table represents the function g(x)

Mathematics

1 answer:

Fed [463]3 years ago

5 0

Answer:

Statement D is true

Step-by-step explanation:

We should know the following:

x-intercept is the value of x when y = 0

y-intercept is the value of y when x = 0

As shown on the graph, we can deduce the following:

For f(x) : at y = 0 ⇒ x = 1 And at x = 0 ⇒ y = -1

So, for f(x) ⇒ x-intercept = 1 & y-intercept = -1 

For g(x) : at g(x) = 0 ⇒ x = -1 And at x = 0 ⇒ g(x) = 1

So, for g(x) ⇒ x-intercept = -1 & y-intercept = 1

Comparing the results with the given options:

A) the x-intercept of f(x) is equal to the x-intercept of g(x) (Wrong)

B) the y-intercept of f(x) is equal to the y-intercept of g(x) (Wrong)

C) the y-intercept of f(x) Is greater than the y-intercept of g(x) (Wrong)

D) the x-intercept of f(x) is greater than the x-intercept of g(x) (true)

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2 years ago

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3 years ago

Read 2 more answers

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katen-ka-za [31]

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3 years ago

Which of the following correctly classifies the number -5 to all of the number sets in which it belongs?

valina [46]

<h3>Answer: Choice C) Real, Rational</h3>

=============================================

Explanation:

Any number you'll encounter so far in your math course will be a real number. This is until you get to something like $\sqrt{-1}$ which is a complex number, but your teacher may not have covered that chapter yet.

Anyways, -5 is a real number because we aren't taking the square root of a negative or anything like that.

-5 is also an integer, but that choice isn't listed. The set of integers is any positive or negative whole numbers, including 0 as well.

Furthermore, -5 is considered rational because -5 = -5/1. We can write -5 as a fraction of two integers where 0 is not in the denominator. With -5 being rational, it cannot be irrational. The term "irrational" means "not rational".

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