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3 years ago

Melissa is saving $25 that she earned for washing her mom's car. She earns $10 every week

Mathematics

1 answer:

bogdanovich [222]3 years ago

4 0

Answer:

25+ 10n= t

Step-by-step explanation:

you already have 25. Then you get $10 each week, which is 10n.

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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

Uh please answer correctly I’ll mark you brainlist after I see if it’s correct!

zepelin [54]

Answer:

The equation for the volume of a rectangular prism is V=whl or width x height x length. The width is 7/2, the height is 7/5, and the length is 5. Multiply all of these numbers to get 49/2 or 24.5.

6 0

3 years ago

Read 2 more answers

What is the greatest common factor of 8 and 24?

Andrew [12]

Answer:

The G.C.F of (8,24) is 8

Step-by-step explanation:

look at the attachment above ☝️

7 0

2 years ago

- 24 divided by 6 = ?

liubo4ka [24]

Answer:

-4

Step-by-step explanation:

-24÷6 =-4

to double check you can do 6×-4 and you would get -24

hope this helps!

7 0

2 years ago

Read 2 more answers

Andy is rewarded each time his current grade of C improves his goal of an A grade. What term describes successive close approxim

Dmitry [639]

Answer:

Shaping

Explanation:

Shaping is the method of successively reinforcing closer and closer approximations to a target ultimate behavior. Shaping includes reinforcing responses which are approximating to the one needed. The shaping method is important because an individual is not always able to naturally exhibit the exact target behavior but by reinforcing the behavior closer to the target behavior, the terminal behavior can be learnt and taught. Shaping is carried out by successive approximation. This is done when the the target response is so rare that it will take a lot of time waiting for the response to take place so that it can be reinforced. In the given question Andy is rewarded every time his current grade C closely approximated to the desired goal that is A grade. So through these subtle reinforcements Andy's behavior is shaped.

The steps in shaping are:

Reinforce any response which is close to the target terminal behavior in some way.
Reinforce the response more closely associated and resembled with desired behavior. The previous behavior should no longer be reinforced.
Continue reinforcing the responses that more closely approximates target behavior.
Continue to reinforce the subsequent approximations until the desired behavior is met.
Reinforce only the ultimate desired behavior once the desired behavior is reached.

Example:

Suppose a teacher wants her student give presentation in front of the whole class but the student is shy so he would not be capable of giving a presentation at once. So the student should be given rewards for his behaviors/responses that approximates closely to the desired goal. Such as giving him reward when he stands in front of class. Next, giving him reward when he introduces himself to the class at the start of presentation. Rewarding him when he tells the topic of the presentation. Ultimately he can give a presentation. Shaping can be achieved this way.

7 0

3 years ago