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jasenka [17]
3 years ago
12

I will mark you as Brainpist if you answer this.

Mathematics
1 answer:
Zepler [3.9K]3 years ago
8 0

Answer:

Mark as a Brainpist , no need

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"Two-thirds of a number is<br> negative 20.
ExtremeBDS [4]
The number is -30
Exp
2/3=-20
1/3=-10
3/3=-30
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Question 6 (1 point)
nekit [7.7K]

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yes i agree

Step-by-step explanation:

easier way is .25x60

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I need help pls pls pls pls
nignag [31]

Answer:

9(w+8)  factor 9

Step-by-step explanation:

factor expression

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9/9w+72/9

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7 0
3 years ago
Read 2 more answers
Can you please answer the question?
Roman55 [17]

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

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

In this problem we must prove that <em>one</em> side of the equality is equal to the expression of the <em>other</em> side, requiring the use of <em>algebraic</em> and <em>trigonometric</em> 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

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

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

#SPJ1

6 0
2 years ago
Math help me please you will be mark as the brainiest I do not have much time left
Black_prince [1.1K]
A little tip: you should put the question in next time :)!
7 0
3 years ago
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