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2 years ago

Find the whole if the part is 69 and the percent is 12.

Mathematics

2 answers:

wlad13 [49]2 years ago

5 0

Answer:

$\frac{ (69 \times 12) }{100} \\ \\ = \frac{828}{100} \\ \\ = 8.28$

Lyrx [107]2 years ago

5 0

Answer:

575

Step-by-step explanation:

if 69 = 12% then find what is 1%

69 / 12 = 5.75 = 1%

To get 100% :

5.75 x 100

:. 575 is the whole number

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On a school trip, the ratio of the teachers to students is 2:21. The ratio boys to girls is 4:3. If there are 18 girls girls on

Alexandra [31]

Answer:

Step-by-step explanation:

Let's say B is the number of boys and T is the number of teachers.

T / (B + 18) = 2/21

B/18 = 4/3

Solve the second equation for B.

B = 24

Plug into the first equation to find T.

T / (24 + 18) = 2/21

T = 4

6 0

3 years ago

2xy , cu x + y = 9 divizibil cu 2. repede va roggg

Anna [14]

Answer:

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Step-by-step explanation:

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3 years ago

Which statement is true of triangles P and Q? Triangle P has side lengths of 6, 8, 10 and angle measures of 53.1 degrees, 90 deg

Wewaii [24]

Answer:

Triangle P and Triangle Q are mathematically similar shapes (?).

Step-by-step explanation:

Hi, so the question asks which statement is true, given the following information, but you haven't written what statements we can choose from.

After reading the information, we can see that Triangle Q is the same shape as Triangle P but just larger.

I'm assuming that one of the statements given is about Triangle P and Triangle Q being mathematically similar shapes?

If you need to show your working out, here it is:

18 ÷ 6 = 3

24 ÷ 8 = 3

30 ÷ 10 = 3

All the angles are the same.

This means that the length scale factor is +3 from Triangle P to Triangle Q, the area scale factor is +9 (because 3 x 3 = 9) from Triangle P to Triangle Q, and that the two shapes are mathematically similar.

*DISCLAIMER* The majority of question askers on Brainly seem to be from the US, and I'm not, so the way I work things out / the mathematical terms I use might be different. Sorry!

Hope this helped anyway!

Bluey :)

3 0

3 years ago

Read 2 more answers

The product is 180. one factor is a multiple of 10

AfilCa [17]

To figure this out you would have to divide 180 by 10(which equals 18))

7 0

3 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

3 years ago