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3 years ago

How to find x in a triangle with only two known lengths

Mathematics

2 answers:

kow [346]3 years ago

8 0

A triangle is 180 degrees. X+first length+second length=180. Then solve

Sidana [21]3 years ago

7 0

The whole triangle equals 180 so take the other two lengths and subtract it from 180 and get the last length

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Dimetri says that a function that is made of terms where the variables is raised only to an odd power will be an odd function.Do

lana [24]

Here we want to study the relation between the exponents in a polynomial function and the "odd" property of functions.

We will see that Dimetri is correct.

-------------------------------

We start by defining an odd function as a function such that:

f(-x) = -f(x).

Now, let's see a simple function with only odd exponents.

f(x) = a*x (the exponent is 1, so it is odd).

if we evaluate this in x = 1, we get:

f(1) = a

If we evaluate this in x = -1, we get:

f(-1) = a*-1 = -a

Then this is odd.

Now let's see a more general case and why Dimetri is correct.

For the rule of signs, when we multiply 2 negative numbers, the product is positive.

So always that we have an even exponent on a negative number, the outcome will be positive.

This does not happen for odd exponents, because we can't separate all the products in groups of 2, thus if we take the odd exponent of a negative number, the outcome is negative.

Then for functions with only odd exponents, the outcome for evaluating the function in a given input is exactly the opposite of evaluating the same function with the opposite input.

This means that the function is odd.

Again, note that this only happens if the function only has odd exponents.

If you want to learn more, you can read:

brainly.com/question/15775372

6 0

3 years ago

Work out 60 × 2/3 (2 over 3)

Taya2010 [7]

Solution:
60 x 2/3
60/1 x 2/3
120/3
40
Answer:
40

hope this helps!

3 0

3 years ago

Please help!posted picture of question

ale4655 [162]

We have the following arithmetic series:
4, 13, 22, 31, 40
The common difference is given by:
13-4 = 9
22-13 = 9
31-22 = 9
40-31 = 9
Since it is an arithmetic series, the difference is constant.
Answer:
The common difference is:
9
option 4

7 0

3 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

Can someone please answer. There is one problem. There's a picture. Thank you!

Over [174]

I believe the answer is A, 6,720.

4 0

3 years ago

Read 2 more answers