Find the distance between (3,−2) and (2,−4)

Find the missing angle.

Sphinxa [80]

Answer:

$m\angle XYV=135^{\circ}$

Step-by-step explanation:

Notice that $m\angle XYV=m\angle XYW+m\angle VYW$ .

The diagram already marks $m\angle VYW=80^{\circ}$ , so we just need to find $m\angle XYW$ .

$\angle TYU$ and $\angle XYW$ are vertical angles, which means they are equal in measure.

Therefore,

$m\angle TYU=m\angle XYW=6x+7$

Since there are $180^{\circ}$ on each side of a line, we have the following equation:

$m\angle XYW+m\angle VYW+m\angle UYV=180^{\circ},\\6x+7+13+4x+80=180,\\10x+20=100,\\10x=80,\\x=8$ .

Plugging in $x=8$ , we have:

$m\angle XYW=6(8)+7=55^{\circ}$ .

Therefore,

$m\angle XYV=55^{\circ}+80^{\circ},\\m\angle XYV=\fbox{$135^{\circ}$}$ .

5 0

4 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 reductio ad absurdum. 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 reductio ad absurdum 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

Who can help and explain these 2 questions

Dominik [7]

Coefficients are added together because they are like terms, this can be proven with the distributive property. For example, x(2x+x)=2x^2+x^2=3x^2.

The commutative property of addition and the associative property demonstrate this.

The word "commutative" comes from "commute" or "move around", so the Commutative Property is the one that refers to moving values around.

The associative property states that you can add or multiply regardless of how the numbers are grouped.

6 0

4 years ago

The price of a technology stock has dropped to $9.79 yesterday. Today. the price fell to $9.68. Find the percentage decrease.

Citrus2011 [14]

The answer is 1.12% decrease
Explanation in image