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igomit [66]
3 years ago
13

Classify the polynomial according to its degree and number of terms 3x^2+8x​

Mathematics
1 answer:
Sunny_sXe [5.5K]3 years ago
6 0

Answer:

Classifying Polynomials

Polynomials can be classified two different ways - by the number of terms and by their degree.

1. Number of terms.

A monomial has just one term. For example, 4x2 .Remember that a term contains both the variable(s) and its coefficient (the number in front of it.) So the is just one term.

A binomial has two terms. For example: 5x2 -4x

A trinomial has three terms. For example: 3y2+5y-2

Any polynomial with four or more terms is just called a polynomial. For example: 2y5+ 7y3- 5y2+9y-2

Practice classifying these polynomials by the number of terms:

1. 5y

2. 3x2-3x+1

3. 5y-10

4. 8xy

5. 3x4+x2-5x+9

Answers: 1) Monomial 2) Trinomial 3) Binomial 4) Monomial 5) Polynomial

2. Degree. The degree of the polynomial is found by looking at the term with the highest exponent on its variable(s).

Examples:

5x2-2x+1 The highest exponent is the 2 so this is a 2nd degree trinomial.

3x4+4x2The highest exponent is the 4 so this is a 4th degree binomial.

8x-1 While it appears there is no exponent, the x has an understood exponent of 1; therefore, this is a 1st degree binomial.

5 There is no variable at all. Therefore, this is a 0 degree monomial. It is 0 degree because x0=1. So technically, 5 could be written as 5x0.

3x2y5 Since both variables are part of the same term, we must add their exponents together to determine the degree. 2+5=7 so this is a 7th degree monomial.

Classify these polynomials by their degree.

1.7x3+52+1

2.6y5+9y2-3y+8

3.8x-4

4.9x2y+3

5.12x2

Answers 1) 3rd degree 2) 5th degree 3) 1st degree 4) 3rd degree 5) 2nd degree

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makvit [3.9K]

Step-by-step explanation:

The value of sin(2x) is \sin(2x) = - \frac{\sqrt{15}}{8}sin(2x)=−

8

15

How to determine the value of sin(2x)

The cosine ratio is given as:

\cos(x) = -\frac 14cos(x)=−

4

1

Calculate sine(x) using the following identity equation

\sin^2(x) + \cos^2(x) = 1sin

2

(x)+cos

2

(x)=1

So we have:

\sin^2(x) + (1/4)^2 = 1sin

2

(x)+(1/4)

2

=1

\sin^2(x) + 1/16= 1sin

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(x)+1/16=1

Subtract 1/16 from both sides

\sin^2(x) = 15/16sin

2

(x)=15/16

Take the square root of both sides

\sin(x) = \pm \sqrt{15/16

Given that

tan(x) < 0

It means that:

sin(x) < 0

So, we have:

\sin(x) = -\sqrt{15/16

Simplify

\sin(x) = \sqrt{15}/4sin(x)=

15

/4

sin(2x) is then calculated as:

\sin(2x) = 2\sin(x)\cos(x)sin(2x)=2sin(x)cos(x)

So, we have:

\sin(2x) = -2 * \frac{\sqrt{15}}{4} * \frac 14sin(2x)=−2∗

4

15

∗

4

1

This gives

\sin(2x) = - \frac{\sqrt{15}}{8}sin(2x)=−

8

15

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