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3 years ago

Simplify the expression.

Mathematics

1 answer:

Yuri [45]3 years ago

3 0

Answer:

The answer to your question is letter A) cot x

Step-by-step explanation:

$\frac{1 - cot x}{tan x - 1}$

Remember that cot x = $\frac{1}{tan x}$

Then

$\frac{1 - \frac{1}{tan x} }{tanx -1}$

Simplify

$\frac{\frac{tan x - 1}{tan x} }{tan x - 1}$

$\frac{tan x - 1}{tan x (tan x - 1)}$

$\frac{1}{tan x}$

Result

$\frac{1}{tan x} = cot x$

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Find the slope of the line from two points (9, 6) (13,11) Use the equation of the line

mafiozo [28]

Answer:

Step-by-step explanation:

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2 years ago

Y=6x 2x+3y=20 solving equation using substitution

RUDIKE [14]

Answer:

x = 1, y = 6

Step-by-step explanation:

y = 6x

2x + 3y = 20

Plug y as 6x in the second equation and solve for x.

2x + 3(6x) = 20

2x + 18x = 20

20x = 20

$\frac{20x}{20}= \frac{20}{20}$

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Plug x as 1 in the first equation and solve for y.

y = 6(1)

y = 6

6 0

3 years ago

Read 2 more answers

Can you please answer the question?

Roman55 [17]

The trigonometric expression $\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$ is equivalent to the trigonometric expression $\sec \alpha \cdot \csc \alpha + 1$ .

<h3>How to prove a trigonometric equivalence</h3>

In this problem we must prove that one side of the equality is equal to the expression of the other side, requiring the use of algebraic and trigonometric properties. Now we proceed to present the corresponding procedure:

$\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} + \frac{\frac{1}{\tan^{2}\alpha} }{\frac{1}{\tan \alpha} - 1 }$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} - \frac{\frac{1 }{\tan \alpha} }{\tan \alpha - 1}$

$\frac{\frac{\tan^{3}\alpha - 1}{\tan \alpha} }{\tan \alpha - 1}$

$\frac{\tan^{3}\alpha - 1}{\tan \alpha \cdot (\tan \alpha - 1)}$

$\frac{(\tan \alpha - 1)\cdot (\tan^{2} \alpha + \tan \alpha + 1)}{\tan \alpha\cdot (\tan \alpha - 1)}$

$\frac{\tan^{2}\alpha + \tan \alpha + 1}{\tan \alpha}$

$\tan \alpha + 1 + \cot \alpha$

$\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} + 1$

$\frac{\sin^{2}\alpha + \cos^{2}\alpha}{\cos \alpha \cdot \sin \alpha} + 1$

$\frac{1}{\cos \alpha \cdot \sin \alpha} + 1$

$\sec \alpha \cdot \csc \alpha + 1$

To learn more on trigonometric expressions: brainly.com/question/10083069

#SPJ1

6 0

2 years ago

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gladu [14]

Answer:

A sample size of 392 is required.

Step-by-step explanation:

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