Pleas help me out here.

Mathematics

1 answer:

Amiraneli [1.4K]3 years ago

4 0

Answer:

A and E

Step-by-step explanation:

The given problem shows a graph of a vertical line with an undefined slope, in which its equation is x = 3 (which matches Option A).

It is not considered a function because it has the same input for every given output. This means that regardless of its y-coordinate, its corresponding x-coordinate will always be x = 3.

The Vertical Line Test allows us to know whether or not a graph is actually a function. Remember that a function can only take on one output for each input. We cannot plug in a value and produce two output values. Since the graph represents a vertical line, it automatically fails the vertical line test. Therefore, it is not a function.

Because it is not a function, Options B, C, and D are invalid answers.

Hence, the correct answers are Options A and E.

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Sara can type 90 words in 4 minutes.About how many words would you expect her to type in 10 minutes?

Alexus [3.1K]

90 words in 4 minutes is the same as 45 words in 2 minutes because you divide both sides by 2. 2 goes into 10, 5 times, so 45x5=225.
You can expect her to type 225 words in 10 minutes.

3 0

3 years ago

How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?

BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

$\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}$

Where n = 8 and r = 0,1....8

When r = 0

$\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1$

When r = 1

$\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 2

$\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8$

When r = 3

$\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 4

$\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70$

When r = 5

$\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$