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uysha [10]
3 years ago
9

Please help! 3 Questions please show how you solved the problem

Mathematics
1 answer:
borishaifa [10]3 years ago
8 0

Answer:

1. 3y^2+2y-11

2. 8y^{7}x^{5}z^2

3. x^3-9x^2+14x+24

Step-by-step explanation:

1. We have been given an expression (-y^2-4y-8)-(-4y^2-6y+3) and we are asked to subtract and simplify our given expression.

Using order of operations (PEMDAS), first of all let us remove parenthesis.

-y^2-4y-8+4y^2+6y-3

Now let us group like terms.

(-y^2+4y^2)+(-4y+6y)+(-8-3)

Upon combining like terms our expression simplifies to:

(3y^2)+(2y)+(-11)

3y^2+2y-11

Therefore, our expression simplifies to 3y^2+2y-11.

2. We have been given an expression (2x^2y^3z^2)*(4xy^4x^2) and we are asked to multiply and simplify our given expression.

(2x^2y^3z^2)*(4y^4x*x^2)

(2x^2y^3z^2)*(4y^4x^3)

Using exponent property a^b*a^c=a^{b+c} we will get,

2*4x^{2+3}y^{3+4}z^2

8x^{5}y^{7}z^2

8y^{7}x^{5}z^2

Therefore, our given expression simplifies to 8y^{7}x^{5}z^2.

3. We are asked to multiply and simplify our expression (x-4)(x^2-5x-6).

Using distributive property we will get,

x*x^2-5x*x-6*x-4*x^2-5x*-4-6*-4

x^3-5x^2-6x-4x^2+20x+24

Upon combining like terms we will get,

x^3+(-5x^2-4x^2)+(-6x+20x)+24

x^3+(-9x^2)+(14x)+24

x^3-9x^2+14x+24

Therefore, our given expression simplifies to x^3-9x^2+14x+24.

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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)}

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\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

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