All categories

3 years ago

Someone help me!!!Please no links

Mathematics

1 answer:

adell [148]3 years ago

5 0

Answer:

I will answer this for you and the links are so annyoing fr

Step-by-step explanation:

give me a couple of minutes to figure it out then I will post the answer!

You might be interested in

I NEED HELP ASAP!!!!!! I WILL GIVE PRAILIEST TO THE CORRECT ANSWER

Norma-Jean [14]

-13 (q + 4) = 7q + 16

I think to solve it you do this first:

-13q + -52 = 7q + 16

And then I think you do:

-6q + -36

And thats all I can remember sorry ):

That -6q + -36 might be the answer I think (:

Hope I could help (:

3 0

3 years ago

20) Chester bought a new car with a sticker price of $18,675. He paid 25% of the cost as a down payment. What

ELEN [110]

Answer:

B is correct

8 0

2 years ago

What’s the value of x in this equation???? Thank u!

andre [41]

Answer:

x = 9

Step-by-step explanation:

The exterior angle of a triangle is equal to the sum of the 2 opposite interior angles.

50° is an exterior angle of the triangle, thus

38 + 4x - 24 = 50 , that is

14 + 4x = 50 ( subtract 14 from both sides )

4x = 36 ( divide both sides by 4 )

x = 9

7 0

3 years ago

Read 2 more answers

If the total perimeter of the trail is 231ft, what is the length of the cave? (2 points)

valentinak56 [21]

Answer: yes it is indeed your very smart

8 0

3 years ago

Use a proof by contradiction to show that the square root of 3 is national You may use the following fact: For any integer kirke

Ierofanga [76]

Answer:

1. Let us proof that √3 is an irrational number, using <em>reductio ad absurdum</em>. Assume that $\sqrt{3}=\frac{m}{n}$ where $m$ and $n$ are non negative integers, and the fraction $\frac{m}{n}$ is irreducible, i.e., the numbers $m$ and $n$ have no common factors.

Now, squaring the equality at the beginning we get that

$3=\frac{m^2}{n^2}$ (1)

which is equivalent to $3n^2=m^2$ . From this we can deduce that 3 divides the number $m^2$ , and necessarily 3 must divide $m$ . Thus, $m=3p$ , where $p$ is a non negative integer.

Substituting $m=3p$ into (1), we get

$3= \frac{9p^2}{n^2}$

which is equivalent to

$n^2=3p^2$ .

Thus, 3 divides $n^2$ and necessarily 3 must divide $n$ . Hence, $n=3q$ where $q$ is a non negative integer.

Notice that

$\frac{m}{n} = \frac{3p}{3q} = \frac{p}{q}$ .

The above equality means that the fraction $\frac{m}{n}$ is reducible, what contradicts our initial assumption. So, $\sqrt{3}$ is irrational.

2. Let us prove now that the multiplication of an integer and a rational number is a rational number. So, $r\in\mathbb{Q}$ , which is equivalent to say that $r=\frac{m}{n}$ where $m$ and $n$ are non negative integers. Also, assume that $k\in\mathbb{Z}$ . So, we want to prove that $k\cdot r\in\mathbb{Z}$ . Recall that an integer $k$ can be written as

$k=\frac{k}{1}$ .

Then,

$k\cdot r = \frac{k}{1}\frac{m}{n} = \frac{mk}{n}$ .

Notice that the product $mk$ is an integer. Thus, the fraction $\frac{mk}{n}$ is a rational number. Therefore, $k\cdot r\in\mathbb{Q}$ .

3. Let us prove by <em>reductio ad absurdum</em> that the sum of a rational number and an irrational number is an irrational number. So, we have $x$ is irrational and $p\in\mathbb{Q}$ .

Write $q=x+p$ and let us suppose that $q$ is a rational number. So, we get that

$x=q-p$ .

But the subtraction or addition of two rational numbers is rational too. Then, the number $x$ must be rational too, which is a clear contradiction with our hypothesis. Therefore, $x+p$ is irrational.

7 0

4 years ago