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3 years ago

8

What is the solution of 3x+8/x-4 greater than or equal to 0

1 answer:

DochEvi [55]3 years ago

6 0

Answer:

x ≥ -2 2/3

Step-by-step explanation:

multiply each side by x-4, you get:

3x + 8 ≥ 0

3x ≥ -8

x ≥ -8/3

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i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

3 0

2 years ago

If we know the diameter of a circle is 10 cm, how long<br> will the circumference be?

mihalych1998 [28]

Answer:

C= π*d

c= π*10

c= 10 π

c= 10 * π

c= 10*3.14= 31.4 cm

7 0

2 years ago

Read 2 more answers

Which conditional statement is true?

Zigmanuir [339]

Answer:

if it is wednesday, it is a weekday.

Step-by-step explanation:

not all tables have 4 legs, not all printed words are in books, and not all months have 30 days.

5 0

2 years ago

Find the angle between u = (8.- 3) and v = (-3,- 8) Round to the nearest tenth of a degree.

Nimfa-mama [501]

Answer:

<h2>90°</h2>

Step-by-step explanation:

First you must calculate the module or the magnitude of both vectors

The module of u is:

$|u|=\sqrt{(8)^2 + (-3)^2} \\\\|u|=\sqrt{64 + 9}\\\\|u|=8.544$

The module of v is:

$|v|=\sqrt{(-3)^2 + (-8)^2} \\\\|u|=\sqrt{9 + 64}\\\\|u|=8.544$

Now we calculate the scalar product between both vectors

$u*v = 8*(-3) + (-3)*(-8)\\\\u*v = -24+ 24=0$

Finally we know that the scalar product of two vectors is equal to:

$u*v = |u||v|*cos(\theta)$

Where $\theta$ is the angle between the vectors u and v. Now we solve the equation for $\theta$

$0 = 8.544*8.544*cos(\theta)\\\\0 = cos(\theta)\\\\\theta= arcos(0)\\\\\theta=90\°$

the answer is 90°

Whenever the scalar product of two vectors is equals to zero it means that the angle between them is 90 °

5 0

3 years ago

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krok68 [10]

Answer:

there is 8 quadrilateral shapes

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3 years ago