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arsen [322]
3 years ago
9

Which compound inequality is represented by the graph?

Mathematics
1 answer:
tankabanditka [31]3 years ago
3 0

Answer: Choice C

x < -2 \ \text{ or } \ x \ge 4

========================================================

Explanation:

The blue portion on the left is represented by x < -2 as it describes values smaller than -2. Notice the open hole at -2 which indicates we do not include this value. So that's why we go with x < -2 instead of x \le -2

The blue portion on the right is represented as x \ge 4 describing values that are equal to 4, or larger than 4.

So overall, the entire graph visually represents x values such that x < -2 \ \text{ or } \ x \ge 4

Side note: the "or" means we pick one region or the other, but not both regions at the same time.

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Answer:

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Step-by-step explanation:


6 0
4 years ago
Transform x2 + 6x + 8 = 0 into the form (x − p)2 = q?
liraira [26]
So hmmm x²+6x+8=0

alrite.. let's do some grouping now
( x² + 6x + [?]²) + 8 = 0

notice above, we have a missing fellow in order to get a perfect square trinomial... hmm who would that be?

let's take a peek at the middle guy of the trinomial.. 6x.. hmmm let's factor it, 2*3*x, wait a minute!  2 * 3 * x... we already have x² on the left-side, since the middle term is just 2 * the square root of the other two terms, that means that the guy on the right, our missing guy must be "3"

alrite, let's add 3² then, however, bear in mind that, all we're doing is borrowing from our very good friend Mr Zero, 0

so if we add 3², we also have to subtract 3², let's do so

(x² + 6x +3² - 3²) + 8 = 0

(x² + 6x +3²) + 8 - 3² = 0

(x+3)²=3² - 8

(x+3)² = 1
5 0
4 years ago
Read 2 more answers
Calculate cos theta where theta is the angle between u and v
koban [17]
You need to upload the image if you want the answer.
However, I can explain how to solve the problem so that you can determine the angle.

cos theta = side opposite of theta divided by the hypotenuse.

Reminder that the hypotenuse is the side NOT touching the 90° angle.

Hope this helps!  If not, upload an image of the drawing.
3 0
3 years ago
If sin theta = 2/3 and sex theta &lt; 0 , find cos theta and tan theta
FromTheMoon [43]

Answer:

\displaystyle cos\theta=-\frac{\sqrt{5}}{3}

\displaystyle tan\theta=-\frac{2\sqrt{5}}{5}

Step-by-step explanation:

<u>Trigonometric Formulas</u>

To solve this problem, we must recall some basic relations and concepts.

The main trigonometric identity relates the sine to the cosine:

sin^2\theta+cos^2\theta=1

The tangent can be found by

\displaystyle tan\theta=\frac{sin\theta}{cos\theta}

The cosine and the secant are related by

\displaystyle cos\theta=\frac{1}{sec\theta}

They both have the same sign.

The sine is positive in the first and second quadrants, the cosine is positive in the first and fourth quadrants.

The sine is negative in the third and fourth quadrants, the cosine is negative in the second and third quadrants.

We are given

\displaystyle sin\theta=\frac{2}{3}

Find the cosine by solving

sin^2\theta+cos^2\theta=1

\displaystyle \left(\frac{2}{3}\right)^2+cos^2\theta=1

\displaystyle cos^2\theta=1-\left(\frac{2}{3}\right)^2=1-\frac{4}{9}=\frac{5}{9}

\displaystyle cos\theta=\sqrt{\frac{5}{9}}=-\frac{\sqrt{5}}{3}

\boxed{\displaystyle cos\theta=-\frac{\sqrt{5}}{3}}

We have placed the negative sign because we know the secant ('sex') is negative and they both have the same sign.

Now compute the tangent

\displaystyle tan\theta=\frac{sin\theta}{cos\theta}=\frac{\frac{2}{3}}{-\frac{\sqrt{5}}{3}}=-\frac{2}{\sqrt{5}}

Rationalizing

\displaystyle tan\theta=-\frac{2}{\sqrt{5}}=-\frac{2\sqrt{5}}{5}

\boxed{\displaystyle tan\theta=-\frac{2\sqrt{5}}{5}}

5 0
4 years ago
Solve the following system.
Brilliant_brown [7]
<h3>The solutions are:</h3>

(0, -1)\ and\ (\frac{1}{3}, 0)

<em><u>Solution:</u></em>

<em><u>Given that,</u></em>

y = 3x^2 + 2x-1-------- eqn 1\\\\\3x-y=1 ------- eqn 2

From eqn 2,

3x - y = 1

y = 3x - 1

<em><u>Substitute the above in eqn 1</u></em>

3x -1 = 3x^2 + 2x - 1\\\\3x^2 + 2x - 3x =0\\\\3x^2 -x = 0\\\\x(3x - 1) = 0\\\\Therefore,\\\\x = 0\\\\And\\\\3x - 1 = 0\\\\3x = 1\\\\x = \frac{1}{3}

<h3>When, x = 0 </h3>

Substitute x = 0 in eqn 2

3(0) - y = 1

y = -1

<h3>When x = 1/3</h3>

Substitute x = 1/3 in eqn 2

3\frac{1}{3} - y = 1\\\\ 1 - y = 1\\\\y = 0

Thus solutions are:

(0, -1)\ and\ (\frac{1}{3}, 0)

4 0
4 years ago
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