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3 years ago

What is the answer to 3-3X6+2=

1 answer:

5 0

<span>So, what you need to take into consideration here is the order of operations. First, you have to multiply 3 by 6 and then do the rest of the question in the order it is written. So, here is how to calculate this: 3 - 3x6 +2 = 3 - 18 + 2 = -15 + 2 = -13. The correct answer is -13.Hope this helps. Let me know if you need additional help!</span><span />

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ANSWER QUICKLY!!!!!!

SVEN [57.7K]

Answer:

It's A

Step-by-step explanation:

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3 years ago

Please help its due soon! :(

mariarad [96]

Answer:

So, the correct answer is that as x gets bigger, f(x) moves towards positive infinity.

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2 years ago

Brandon buys a radio for 43.99 in a state where sales tax is 7%.What is the total brandon pays for the radio

guajiro [1.7K]

Answer: $47.06 if you do not round or $47.07 if you are required to round up.

Step-by-step explanation:

$43.99× .07 = 3.0793

$43.99+ $3.07 = $47.06

(If you round up then follow below)

$43.99× .07= 3.0793 (the 9 would cause you to round the 7 up to 8)

$43.99+ $3.08= $47.07

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4 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

1. Triangle ABC is similar to triangle JKL.

Fittoniya [83]

1. It's all about pattern matching, as a lot of math is.
Letter A corresponds to letter J, as both are first in the names of their respective triangles.
Letter B corresponds to letter K, as both are second in the triangle names. Likewise, letter C corresponds to letter L, as both are last.

Realizing this, it should not be too much of a stretch to see
∠B ⇒ ∠K
∠C ⇒ ∠L
AC ⇒ JL
BC ⇒ KL

2. Same deal. Match the patterns. Here, you're counting rings in the angle marks.
1 ⇒ 1, so A ⇒ R
2 ⇒ 2, so B ⇒ Q
since the figures are reportedly similar, you can continue in the same order to finish.
ABCD ~ RQPS

3. The marked triangles cannot be similar. There are a number of ways to figure this. Basically, you want the ratios of sides to be the same for any similar triangles.

Here, you can eliminate the marked ones because the short side is too short relative to the others. (The average of the other two sides is double the short side in the similar triangles.)

8 0

4 years ago