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3 years ago

How do I know if the function is a linear function?

Mathematics

1 answer:

Lilit [14]3 years ago

3 0

Answer:

This is not a linear function.

Step-by-step explanation:

It is a quadratic function. If you were to make a table of the x and y values of the function it would look like this:

x y

0 -4

1 -1

2 8

Just by looking at these three values, we can see that they do not increase at a consistent rate. It increases by 3 from 0 to 1, but increase by 9 from 1 to 2. You can tell if a function is linear by seeing if it increases at a constant rate. (In this case it does not)

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Which values of a and b in the exponential function y = a times b Superscript x would result in the following graph? On a coordi

galina1969 [7]

Answer:

$a =-3$ $b =2$

Step-by-step explanation:

Given

$y = ab^x$

$(x_1,y_1) = (0,-3)$

Required

Find a and b

Substitute $(x_1,y_1) = (0,-3)$ for x and y in $y = ab^x$

$-3 = ab^0$

$-3 = a * 1$

$-3 = a$

Rewrite as:

$a =-3$

The above implies that (a) is correct;

Hence:

$a =-3$ $b =2$

So, the equation is:

$y = -3(2)^x$

6 0

2 years ago

Read 2 more answers

Please help me with this 8th grade math

VMariaS [17]

The answer is -21 I just took the test

7 0

3 years ago

Prove that if n is a perfect square then n + 2 is not a perfect square

notka56 [123]

Answer:

This statement can be proven by contradiction for $n \in \mathbb{N}$ (including the case where $n = 0$ .)

$\text{Let $n \in \mathbb{N}$ be a perfect square}$ .

$\textbf{Case 1.} ~ \text{n = 0}$ :

$\text{$n + 2 = 2$, which isn't a perfect square}$ .

$\text{Claim verified for $n = 0$}$ .

$\textbf{Case 2.} ~ \text{$n \in \mathbb{N}$ and $n \ne 0$. Hence $n \ge 1$}$ .

$\text{Assume that $n$ is a perfect square}$ .

$\text{$\iff$ $\exists$ $a \in \mathbb{N}$ s.t. $a^2 = n$}$ .

$\text{Assume $\textit{by contradiction}$ that $(n + 2)$ is a perfect square}$ .

$\text{$\iff$ $\exists$ $b \in \mathbb{N}$ s.t. $b^2 = n + 2$}$ .

$\text{$n + 2 > n > 0$ $\implies$ $b = \sqrt{n + 2} > \sqrt{n} = a$}$ .

$\text{$a,\, b \in \mathbb{N} \subset \mathbb{Z}$ $\implies b - a = b + (- a) \in \mathbb{Z}$}$ .

$\text{$b > a \implies b - a > 0$. Therefore, $b - a \ge 1$}$ .

$\text{$\implies b \ge a + 1$}$ .

$\text{$\implies n+ 2 = b^2 \ge (a + 1)^2= a^2 + 2\, a + 1 = n + 2\, a + 1$}$ .

$\text{$\iff 1 \ge 2\,a $}$ .

$\text{$\displaystyle \iff a \le \frac{1}{2}$}$ .

$\text{Contradiction (with the assumption that $a \ge 1$)}$ .

$\text{Hence the original claim is verified for $n \in \mathbb{N}\backslash\{0\}$}$ .

$\text{Hence the claim is true for all $n \in \mathbb{N}$}$ .

Step-by-step explanation:

Assume that the natural number $n \in \mathbb{N}$ is a perfect square. Then, (by the definition of perfect squares) there should exist a natural number $a$ ( $a \in \mathbb{N}$ ) such that $a^2 = n$ .

Assume by contradiction that $n + 2$ is indeed a perfect square. Then there should exist another natural number $b \in \mathbb{N}$ such that $b^2 = (n + 2)$ .

Note, that since $(n + 2) > n \ge 0$ , $\sqrt{n + 2} > \sqrt{n}$ . Since $b = \sqrt{n + 2}$ while $a = \sqrt{n}$ , one can conclude that $b > a$ .

Keep in mind that both $a$ and $b$ are natural numbers. The minimum separation between two natural numbers is $1$ . In other words, if $b > a$ , then it must be true that $b \ge a + 1$ .

Take the square of both sides, and the inequality should still be true. (To do so, start by multiplying both sides by $(a + 1)$ and use the fact that $b \ge a + 1$ to make the left-hand side $b^2$ .)

$b^2 \ge (a + 1)^2$ .

Expand the right-hand side using the binomial theorem:

$(a + 1)^2 = a^2 + 2\,a + 1$ .

$b^2 \ge a^2 + 2\,a + 1$ .

However, recall that it was assumed that $a^2 = n$ and $b^2 = n + 2$ . Therefore,

$\underbrace{b^2}_{=n + 2)} \ge \underbrace{a^2}_{=n} + 2\,a + 1$ .

$n + 2 \ge n + 2\, a + 1$ .

Subtract $n + 1$ from both sides of the inequality:

$1 \ge 2\, a$ .

$\displaystyle a \le \frac{1}{2} = 0.5$ .

Recall that $a$ was assumed to be a natural number. In other words, $a \ge 0$ and $a$ must be an integer. Hence, the only possible value of $a$ would be $0$ .

Since $a$ could be equal $0$ , there's not yet a valid contradiction. To produce the contradiction and complete the proof, it would be necessary to show that $a = 0$ just won't work as in the assumption.

If indeed $a = 0$ , then $n = a^2 = 0$ . $n + 2 = 2$ , which isn't a perfect square. That contradicts the assumption that if $n = 0$ is a perfect square, $n + 2 = 2$ would be a perfect square. Hence, by contradiction, one can conclude that

$\text{if $n$ is a perfect square, then $n + 2$ is not a perfect square.}$ .

Note that to produce a more well-rounded proof, it would likely be helpful to go back to the beginning of the proof, and show that $n \ne 0$ . Then one can assume without loss of generality that $n \ne 0$ . In that case, the fact that $\displaystyle a \le \frac{1}{2}$ is good enough to count as a contradiction.

7 0

3 years ago

Sara missed 7 of 184 school days this year. Approximately what percent of school days was Sara present?

gayaneshka [121]

Answer:

B. 96%

Step-by-step explanation:

Sara only missed 7 days of 184 days this year meaning 184 - 7 = 177 and

177 is how many days she attended school, so to figure out percentage take the number of days she attended school and divide by the total amount of days in this year which is 184: 177/184 = 0.96195652173

then multiply by 100 to get percentage so approximately 96%.

7 0

3 years ago

To estimate the average annual expenses of students on books and class materials a sample of size 36 is taken. The sample mean i

Elena L [17]

Answer:

C) $826.82 to $873.18.

Step-by-step explanation:

Sample mean (M) = $850

Standard deviation (s) = $54

sample size (n) = 36

Z for 99% confidence interval (Z) = 2.576

The confidence interval is determined by the following relationship:

$M \pm Z*\frac{s}{\sqrt{n}}$

Applying the given values, the lower (L) and upper (U) values are:

$850 \pm 2.576*\frac{54}{\sqrt{36}}\\L=\$826.82\\U=\$873.18$

The answer is C) $826.82 to $873.18.

6 0

3 years ago