Add one term to the polynomial expression 14x^19 - 9x^15 + 11x^4 + 5x^2 + 3 to make it into a 22nd degree polynomial

Answers

Answer 1
Answer:

The required polynomial to a 22nd degree is P(x)=x^(22)+14x^(19) - 9x^(15) + 11x^4 + 5x^2

Given the polynomial function, 14x^(19) - 9x^(15) + 11x^4 + 5x^2,, we are to add one term to the polynomial to make it into a 22nd-degree polynomial.

Note the highest and leading power of the variable of any function is the degree of such function.

To convert the given polynomial to a 22nd-degree function, we will simply add a variable term x with a degree of 22 to have:

P(x)=x^(22)+14x^(19) - 9x^(15) + 11x^4 + 5x^2

Hence the required polynomial to a 22nd degree is P(x)=x^(22)+14x^(19) - 9x^(15) + 11x^4 + 5x^2

Learn more here: brainly.com/question/2706981

Answer 2
Answer:

Given:

The polynomial is

14x^(19)-9x^(15)+11x^4+5x^2+3

To find:

One term which is used to add in given polynomial to make it into a 22nd degree polynomial.

Solution:

Degree of a polynomial is the highest power of the variable.

Let,

P(x)=14x^(19)-9x^(15)+11x^4+5x^2+3

Here, the highest power of x is 19, so degree of polynomial is 19.

To make it into a 22nd degree polynomial, we need to need a term having 22 as power of x.

We can add kx^(22), where k is constant.

So add x^(22) in the given polynomial.

P(x)=x^22+14x^(19)-9x^(15)+11x^4+5x^2+3

Now, the degree of polynomial is 22.

Therefore, the required term is x^(22).


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Answers

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the coordinates of the endpoints of line AB and line CD are A(x1, y1), B(x2, y2), C(X3,Y3), and D(x4, y4). which condition proves that line AB is parallel the CD?

Answers

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Answers

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What is the 25th term of this arithmetic sequence? 3, 9, 15, 21, 27, …

Answers

Answer:   147

Step-by-step explanation:

The given arithmetic sequence : 3, 9, 15, 21, 27, ….....................

From the above sequence, it can be seen that the first term a= 3

The common difference = d=9-15=21-15=6

We know that in arithmetic sequence, the nth term is given by :-

a_n=a+d(n-1)

Then for, n=25, the 25th term will be :-

a_(25)=3+6(25-1)\n\n\Rightarrow\ a_(25)=147

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+6 +6  +6  +6

a(n) = a₁ + d(n - 1)
a(n) = 3 + 6(n - 1)
a(n) = 3 + 6(n) - 6(1)
a(n) = 3 + 6n - 6
a(n) = 6n + 3 - 6
a(n) = 6n - 3

  a(n) = 6n - 3
a(25) = 6(25) - 3
a(25) = 150 - 3
a(25) = 147