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tekilochka [14]

2 years ago

7

A coffe table is on sale for 20% off the orgingial price of $115. If there is 7.5% sales tax, what is the original price of the

coffee table?

1 answer:

Anuta_ua [19.1K]2 years ago

3 0

Answer: 98.90

Step-by-step explanation:

115*.2=23

115-23=92

0.075*92=6.9

6.90+92=$98.90

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Please help with this postulate! :)

atroni [7]

Answer:

C. ASA postulate

Step-by-step explanation:

<u>We observe two angles and the included side of ABC are congruent to two angles and the included side of DBE:</u>

∠A ≅ ∠D
∠C ≅ ∠ E
AC ≅ DE

This is called ASA

7 0

2 years ago

Read 2 more answers

Which of the following graphs shows the<br> solution set to 2x < 6 and 3x + 2 ><br> -4?

Ilya [14]

Answer:

The third picture

Step-by-step explanation:

Solve for x in both equations

2x<6

Divide both sides by 2:

x<3

3x+2>-4

Subtract 2 from both sides:

3x>-6

Divide both sides by 3:

x>-2

There is this trick you can use when x is on the left side of the equation to find out which way to shade in you graph. Keep in mind this is only for the left side, it will not work if your variable is on the right.

When the symbol is facing left < then shade left, imagine it is pointing which way to shade. x<3 is represented by the 3 picture on the left. When the symbol is facing right > then shade right, again it is pointing which way to shade. x>-2 is represented by the 3 picture on the right.

The circles are not filled in because the symbol is < and > rather than $\leq and \geq$ . When it is greater than or equal to or less than and equal to (represented by the line under the symbol), then the circle is shaded in.

5 0

1 year ago

Help me please I beg you

Dmitrij [34]

Answer:

Neither

Step-by-step explanation:

We need to transform one of the equations into its slope-intercept form:

y + 2/5x = 8

Subtract 2/5x from both sides:

y = -2/5x + 8

y = -5/2x - 7

Graphing both lines, they do intersect at point (-7.14, 10.86), but they are not parallel, nor are they perpendicular from each other. By definition, perpendicular lines must appear to be perpendicular (that is, they intersect at a 90° angle).

Looking at the attached graph (where the purple line is y = -2/5x + 8), their lines do not form a 90° angle.

Therefore, the correct answer is neither.

Please mark my answers as the Brainliest, if you find my explanations helpful :)

5 0

2 years ago

Read 2 more answers

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

2 years ago

Please help me on this

Travka [436]

You have to graph the points they are asking for it. Create the histogram..

5 0

2 years ago