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4 years ago

Y=-3x+2 I need help plssssssss

1 answer:

8 0

Graphing or solving for variable? I can solve for either. Of your looking for a graph just comment but I can help you when looking for a variable. So I don’t know if your solving for x or y but I will assume your solving for x. So,
1. Flip the equation; -3x+ 2 = y
2. Add -2 to both sides
3. Divide both sides by -3
Your answer: x= -1/3y + 2/3.
Feedback is appreciated along with brainliest :-)

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4 years ago

5m - 3n = 19; m - 6n = -7 /pls. help me solve this linear equation with two variables.

Allisa [31]

Answer:

m = 5, n = 2

Step-by-step explanation:

$5m - 3 n = 19\\m - 6n = - 7 \ => m = -7 + 6n\\\\Substitute \ m \ in \ 5m - 3n = 19\\\\5(-7 + 6n) - 3n = 19\\-35 + 30n -3n = 19\\27n = 19 + 35 \\27n = 54 \\n = 2\\\\Substitute \ n = 2 \ in \ m = -7 + 6n\\\\m = -7 + (6 \times 2) = -7 + 12 = 5$

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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 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.