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raketka [301]
3 years ago
9

Please help!!

Mathematics
2 answers:
Romashka [77]3 years ago
4 0
22.       steps
          
         6w^2 + - 3w + -18
         -18 + -3w + 6w^2
         3(-6 + -1w + 2w^2
         3((-3 + -2w)(2 + -1w))
         3(-3 + -2w)(2 + -1w)

final answer :   3(-3 + -2w)(2 + -1w)
Nookie1986 [14]3 years ago
4 0
22)~6w^2-3w-18=3(2w^2-w-6)=3(2w+3)(w-2)\\\\
23)~a^3b^2-a^2b+a=a(a^2b^2-ab+1)\\\\
24)~6x^3+12x^2y-x-2y=6x^2(x+2y)-(x+2y)=(6x^2-1)(x+2y)\\\\
25)~2y^2-30y+72=2(y^2-15y+36)=2(y-3)(y-12)
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A fourth-order polynomial passes through the five points for which the independent and dependent variables, x and y, respectivel
Sladkaya [172]

Answer:

The fourth-order polynomial that passes through the points is P(x)=2x^4+x^3+3x^2+2x+5

Step-by-step explanation:

A fourth-order polynomial has this general form:

P(x)=ax^4+bx^3+cx^2+dx+e

We need to replace the given points one by one, which leads to a system of 5 equations and 5 variables (namely a,b,c,d,e)

37=a(-2)^4+b(-2)^3+c(-2)^2+d(-2)+e\\7=a(-1)^4+b(-1)^3+c(-1)^2+d(-1)+e\\5=a(0)^4+b(0)^3+c(0)^2+d(0)+e\\13=a(1)^4+b(1)^3+c(1)^2+d(1)+e\\61=a(2)^4+b(2)^3+c(2)^2+d(2)+e

We can rewrite this system as follows:

16a-8b+4c-2d+e=37\\a-b+c-d+e=7\\e=5\\a+b+c+d+e=13\\16a+8b+4c+2d+e=61

We can use the Gaussian Elimination to solve this system of equations. To use the Gaussian Elimination we need to express the system of linear equations in matrix form (the matrix equation Ax = b).

The coefficient matrix (A) for the above system is

\left[\begin{array}{ccccc}16&-8&4&-2&1\\1&-1&1&-1&1\\0&0&0&0&1\\1&1&1&1&1\\16&8&4&2&1\end{array}\right]

the variable matrix (x) is

\left[\begin{array}{c}a&b&c&d&e\end{array}\right]

and the constant matrix (b) is

\left[\begin{array}{c}37&7&5&13&61\end{array}\right]

The augmented matrix for this system, it is obtained by appending the columns of the coefficient matrix and the constant matrix.

\left[\begin{array}{ccccc|c}16&-8&4&-2&1&37\\1&-1&1&-1&1&7\\0&0&0&0&1&5\\1&1&1&1&1&13\\16&8&4&2&1&61\end{array}\right]

The augmented matrix can be transformed by a sequence of elementary row operations to the reduced row echelon form.

\left[\begin{array}{ccccc|c}1&0&0&0&0&2\\0&1&0&0&0&1\\0&0&1&0&0&3\\0&0&0&1&0&2\\0&0&0&0&1&5\end{array}\right]

Therefore the solutions are:

\left\begin{array}{c}a=2&b=1&c=3&d=2&e=5\end{array}\right

And the polynomial P(x) is:

P(x)=2x^4+x^3+3x^2+2x+5

We can check our solution plotting the polynomial and checking that it passes through the points.

4 0
3 years ago
Which one<br><br> A.<br><br> B.<br><br> C.<br><br> D.
Usimov [2.4K]

B. is correct! Hope this helps!

4 0
3 years ago
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Plzzzz help right answer gets a brainly!
masha68 [24]

Answer: A, The line starts at 9 and ends at 29

5 0
3 years ago
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Simplify (5x^3-x+14)-(3x^2-9x+4)
Mariana [72]
5x^3 -3x^2 +8x +10 Just combine all the like terms and multiply everything in the second parentheses by -1
5 0
3 years ago
Anyone please answer me ,<br> Please solve this question
Mamont248 [21]

Answer:

67

Step-by-step explanation:

Multi-Inscribed Quadrilateral: 15

Bananas: 4

Clock: 3

Therefore, we can substitute for the last equation the values, so it would become:

3 + 4 + 4 x 15 = ?

First, we solve the multiplication according to PEMDAS.

4 x 15 = 60

Then, we can add from left to right.

3 + 4 + 60 = ?

3 + 4  = 7

7 + 60 = 67

Therefore, our answer would be 67.

3 0
3 years ago
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