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3 years ago

What is 8 plus (-2) gehehfbf

Mathematics

1 answer:

nalin [4]3 years ago

3 0

This is just another way of saying 8 - 2.

So, 8 + (-2) = 6.

Hope this helps.

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What is an integer in math

san4es73 [151]

Answer:

An integer is a whole number that is not a fraction.

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3 years ago

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Use Gaussian Elimination to find an equation of a polynomial that passes through points A(-5,-3), B(-2,3). C(3,3), D(6,19). Indi

Marrrta [24]

Answer:

The polynomial equation that passes through the points is $2-\frac{2}{3}x+\frac{1}{12}x^{2}+\frac{1}{12}x^{3}$

Step-by-step explanation:

Suppose you have a function y = f(x) which goes through these points

A(-5,-3), B(-2,3). C(3,3), D(6,19)

there is a polynomial P(x) of degree 3 which goes through these point.

We use the fact that four distinct points will determine a cubic function.

P(x) is the degree 3 polynomial through the 4 points, a standard way to write it is

$P(x) = a+bx+cx^2+dx^3$

Next replace the given points one by one, which leads to a system of 4 equations and 4 variables (namely a,b,c,d)

$-3=a+b\cdot-5+c\cdot -5^2+d\cdot -5^3\\3=a+b\cdot-2+c\cdot -2^2+d\cdot -2^3\\3=a+b\cdot 3+c\cdot 3^2+d\cdot 3^3\\19=a+b\cdot 6+c\cdot 6^2+d\cdot 6^3$

We can rewrite this system as follows:

$-3=a-5\cdot b+25\cdot c-125\cdot d\\3=a-2\cdot b+4\cdot c-8\cdot d\\3=a+3\cdot b+9\cdot c+27\cdot d\\19=a+6\cdot b+36\cdot c+216\cdot d$

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

$\left[\begin{array}{cccc}1&-5&25&-125\\1&-2&4&-8\\1&3&9&27\\1&6&36&216\end{array}\right]$

the variable matrix (x) is

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

and the constant matrix (b) is

$\left[\begin{array}{c}-3&3&3&19\end{array}\right]$

We also need the augmented matrix, it is obtained by appending the columns of the coefficient matrix and the constant matrix.

$\left[\begin{array}{cccc|c}1&-5&25&-125&-3\\1&-2&4&-8&3\\1&3&9&27&3\\1&6&36&216&19\end{array}\right]$

To transform the augmented matrix to the reduced row echelon form we need to follow these steps:

Subtract row 1 from row 2 $\left(R_2=R_2-R_1\right)$

$\left[\begin{array}{cccc|c}1&-5&25&-125&-3\\0&3&-21&117&6\\1&3&9&27&3\\1&6&36&216&19\end{array}\right]$

Subtract row 1 from row 3 $\left(R_3=R_3-R_1\right)$

$\left[\begin{array}{cccc|c}1&-5&25&-125&-3\\0&3&-21&117&6\\0&8&-16&152&6\\1&6&36&216&19\end{array}\right]$

Subtract row 1 from row 4 $\left(R_4=R_4-R_1\right)$

$\left[\begin{array}{cccc|c}1&-5&25&-125&-3\\0&3&-21&117&6\\0&8&-16&152&6\\0&11&11&341&22\end{array}\right]$

Divide row 2 by 3 $\left(R_2=\frac{R_2}{3}\right)$

$\left[\begin{array}{cccc|c}1&-5&25&-125&-3\\0&1&-7&39&2\\0&8&-16&152&6\\0&11&11&341&22\end{array}\right]$

Add row 2 multiplied by 5 to row 1 $\left(R_1=R_1+\left(5\right)R_2\right)$

$\left[\begin{array}{cccc|c}1&0&-10&-70&7\\0&1&-7&39&2\\0&8&-16&152&6\\0&11&11&341&22\end{array}\right]$

Subtract row 2 multiplied by 8 from row 3 $\left(R_3=R_3-\left(8\right)R_2\right)$

$\left[\begin{array}{cccc|c}1&0&-10&-70&7\\0&1&-7&39&2\\0&0&40&-160&-10\\0&11&11&341&22\end{array}\right]$

Subtract row 2 multiplied by 11 from row 4 $\left(R_4=R_4-\left(11\right)R_2\right)$

$\left[\begin{array}{cccc|c}1&0&-10&-70&7\\0&1&-7&39&2\\0&0&40&-160&-10\\0&0&88&-88&0\end{array}\right]$

Divide row 3 by 40 $\left(R_3=\frac{R_3}{40}\right)$

$\left[\begin{array}{cccc|c}1&0&-10&-70&7\\0&1&-7&39&2\\0&0&1&-4&-1/4\\0&0&88&-88&0\end{array}\right]$

Add row 3 multiplied by 10 to row 1 $\left(R_1=R_1+\left(10\right)R_3\right)$

$\left[\begin{array}{cccc|c}1&0&0&30&9/2\\0&1&-7&39&2\\0&0&1&-4&-1/4\\0&0&88&-88&0\end{array}\right]$

Add row 3 multiplied by 7 to row 2 $\left(R_2=R_2+\left(7\right)R_3\right)$

$\left[\begin{array}{cccc|c}1&0&0&30&9/2\\0&1&0&11&1/4\\0&0&1&-4&-1/4\\0&0&88&-88&0\end{array}\right]$

Subtract row 3 multiplied by 88 from row 4 $\left(R_4=R_4-\left(88\right)R_3\right)$

$\left[\begin{array}{cccc|c}1&0&0&30&9/2\\0&1&0&11&1/4\\0&0&1&-4&-1/4\\0&0&0&264&22\end{array}\right]$

Divide row 4 by 264 $\left(R_4=\frac{R_4}{264}\right)$

$\left[\begin{array}{cccc|c}1&0&0&30&9/2\\0&1&0&11&1/4\\0&0&1&-4&-1/4\\0&0&0&1&1/12\end{array}\right]$

Subtract row 4 multiplied by 30 from row 1 $\left(R_1=R_1-\left(30\right)R_4\right)$

$\left[\begin{array}{cccc|c}1&0&0&0&2\\0&1&0&11&1/4\\0&0&1&-4&-1/4\\0&0&0&1&1/12\end{array}\right]$

Subtract row 4 multiplied by 11 from row 2 $\left(R_2=R_2-\left(11\right)R_4\right)$

$\left[\begin{array}{cccc|c}1&0&0&0&2\\0&1&0&0&-2/3\\0&0&1&-4&-1/4\\0&0&0&1&1/12\end{array}\right]$

Add row 4 multiplied by 4 to row 3 $\left(R_3=R_3+\left(4\right)R_4\right)$

$\left[\begin{array}{cccc|c}1&0&0&0&2\\0&1&0&0&-2/3\\0&0&1&0&1/12\\0&0&0&1&1/12\end{array}\right]$

From the reduced row-echelon form the solutions are:

$\left[\begin{array}{c}a=2&b=-2/3&c=1/12&d=1/12\end{array}\right]$

The polynomial P(x) is:

$2-\frac{2}{3}x+\frac{1}{12}x^{2}+\frac{1}{12}x^{3}$

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

3 0

3 years ago

Find the domain and range of f(x). Please explain your answer. f(x) = √(x^2+5x+6)

timama [110]

Answer:

Domain:— [ x ≥ -2, x ≤ -3 ]

Range:— [ y ≥ 0 ]

Step-by-step explanation:

You may use graphing calculator to draw a graph and examine the graph’s domain and range. However, I’ll explain further about the graph of quadratic in a surd.

First, factor the quadratic expression in the surd:—

$\displaystyle \large{f(x)=\sqrt{(x+3)(x+2)}}$

We can find the x-intercepts by letting f(x) = 0.

I’ll be separating in two parts — one for finding x-intercept and one for finding y-intercept.

__________________________________________________________

Finding x-intercepts

Let f(x) = 0.

$\displaystyle \large{0=\sqrt{(x+3)(x+2)}}$

Solve for x, square both sides:—

$\displaystyle \large{0^2 = (\sqrt{(x+3)(x+2)})^2}\\\displaystyle \large{0=(x+3)(x+2)}$

Simply solve a quadratic equation:—

$\displaystyle \large{x=-3,-2}$

Therefore, x-intercepts are:—

$\displaystyle \large{\boxed{x=-3,-2}}$

__________________________________________________________

Finding y-intercept

Let x = 0.

$\displaystyle \large{f(x)=\sqrt{(0+3)(0+2)}}\\\displaystyle \large{f(x)=\sqrt{3 \cdot 2}}\\\displaystyle \large{f(x)=\sqrt{6}}$

Therefore, y-intercept is:—

$\displaystyle \large{\boxed{\sqrt{6}}}$

__________________________________________________________

However, I want you to focus on x-intercepts instead. We know that the square root only gives you a positive value. That means the range of function can only be y ≥ 0.

For domain, first, we have to know how or what the graph looks like. You can input the function in a graphing calculator as you’ll see that when x ≥ -2, the graph heads to the right while/when x ≤ -3, the graph heads to the left. This means that the lesser value of x-intercept gets left and more value get right.

See, between -3 < x < -2, there is no curve, point or anything between the interval. Therefore, -3 < x < -2 does not exist in function.

Hence, the domain is:—

x ≥ -2, x ≤ -3

4 0

2 years ago

CAN ANYONE HELP ME WITH THIS QUESTION...Plzz

Lelechka [254]

Answer:

AO and OC

x+8 and 16-x

x+8 = 16-x

x+x= 16-8

2x= 8

x= 8/2

x=4

DO and OB

2y+13 and 5y+4

2y+13 = 5y+4

2y-5y= 4-13

-3y = -9

y= 9/3

y= 3

Step-by-step explanation:

i hope this helps

if this helps let me know

5 0

3 years ago

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Can any one answer these questions !!! anyone

Murljashka [212]

Answer:

Step-by-step explanation:

Hope this helps u!!

7 0

3 years ago