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3 years ago

Select the correct answer.

Mathematics

2 answers:

Naddika [18.5K]3 years ago

7 0

Answer:

y = -8

Step-by-step explanation:

y = 3

is a "flat" horizontal line, which means every point on that line has the same y value no matter what x value we would consider.

any line parallel to that is then also a horizontal line with a constant y value. which y value ? well, the y coordinate of the given point tells us : (-2, -8)

so, the line equation is

y = -8

aniked [119]3 years ago

5 0

Answer:

C, y=-8

Step-by-step explanation:

Which equation represents the line that is parallel to y= 3 and passes through (-2,-8)?

y=3 means the y value is always 3, so the line is horizontal and parallel to the y axis.

A, x= -2 this is a vertical line

B. x= 3 this is a vertical line

C, y=-8 this is a horizontal line, where y is always -8 so it is parallel and will pass through -2

D, y= 3x-2 this line will be at an angle

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Step-by-step explanation:

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3 years ago

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2 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.

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4 years ago

What is the amount of money p that will generate $40 in interest at a 10$ interest over 5 years

-BARSIC- [3]

If i read this correctly it should be $90. 10x5= $50 and that plus $40 equals $90 altogether

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4 years ago