How do you name a vector?

What are the solutions to the system of equations?

svet-max [94.6K]

Answer:the third option is correct

Step-by-step explanation:

The system of equations are

y = 2x^2 - 5x - 7 - - - - - - - - - - -1

y = 2x + 2 - - - - - - - - - - - - - 2

We would equate equation 1 and equation 2. It becomes

2x^2 - 5x - 7 = 2x + 2

2x^2 - 5x - 2x - 7 - 2 = 0

2x^2 - 7x - 9 = 0

We would find two numbers such that their sum or difference is -7x and their product is - 18x^2. The two numbers are 2x and - 9x. Therefore

2x^2 + 2x - 9x - 9 = 0

2x(x + 1) - 9(x + 1) = 0

2x - 9 = 0 or x + 1 = 0

2x = 9 or x = - 1

x = 9/2 = 4.5

Substituting x = 4.5 or x = -1 into equation 2, it becomes

y = 2 × 4.5 + 2 or y = 2 × - 1 + 2

y = 11 or y = 0

Therefore, the solutions are

(4.5, 11) (- 1, 0)

6 0

4 years ago

On december 1, jonas jewelry wholesale center started with an inventory of 1,000 watches. shipments of 168, 231, and 411 were se

denis-greek [22]

Answer: Third option is correct.

Step-by-step explanation:

Since we have given that

Total watches in an inventory = 1000

Number of shipments sent out = 168, 231, 411

So, Remaining inventory of watches would be :

$1000-(168+231+411)\\\\=1000-810\\\\=190$

New receipts = 21, 145 and 11

So, Number of watches on hand becomes :

$190+(21+145+11)\\\\=367$

Hence, Third option is correct.

5 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.