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mrs_skeptik [129]

3 years ago

13

Sal and two of his friends rent an apartment together. Their total cost to move in included first month's rent, last month's ren

t and a security deposit of $1,200. If Sal paid a total of $1,000 to move in, how much is his share of the rent each month? a. $300 b. $450 c. $600 d. $900

2 answers:

umka2103 [35]3 years ago

6 0

Answer:

A. $300

Step-by-step explanation:

Test on Edge - Sarah Robinsen <3

VMariaS [17]3 years ago

4 0

Answer:

Option a. $300

Step-by-step explanation:

Let

x ----> the cost of the rent monthly per person

we know that

The number of persons is 3 (Sal and two friends)

The cost of the security deposit per person is equal to

$\$1,200/3=\$400$

If Sal paid a total of $1,000 to move in

then

That means that the cost per person of two months of rent (first and last month) plus $400 (security deposit per person) must be equal to $1,000

$2x+400=1,000$

solve for x

$2x=1,000-400\\2x=600\\x=\$300$

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Select the assumption necessary to start an indirect proof of the statement. If 3a+7 < 28, then a<7.

pashok25 [27]

When applying indirect proofs, we assume the negation of the conclusion is true, and show that this assumption would lead to nonsense, or contradiction.

In our case we assume a is not smaller than 7, that is we assume a≥7.

a≥7 then, multiplying both sides by 3:
3a≥21, then, adding both sides 7:

3a+7≥28,

which is a contradiction because 3a+7 is smaller than 28.

So our assumption is wrong, which means the opposite of it is correct.

Answer: assume a≥7

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3 years ago

(9+8)+6/3-7×2 how to slove

klemol [59]

Answer:

Step-by-step explanation:

There is a specific order to do these kind of problems..

first, we do everything in parenthesis...

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17 + 6/3 - 7 * 2

now, starting from the LEFT side, do all multiplication and division in the order it comes

17 + 2 - 14

now, starting from the LEFT side, do all addition and subtraction in the order it comes

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PEMDAS

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3 years ago

Read 2 more answers

What is the percentage change in total number of players from Tram A 2021 to 2022 when more than 6 but less 20 matches where won

babunello [35]

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Step-by-step explanation:

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8 0

2 years ago

What is the volume of this figure? <br> Plz answer soon! <br> (iready)

Lena [83]

Answer:

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Step-by-step explanation:

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2 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