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3 years ago

12x2 = 18x i need someone to explain this

Mathematics

1 answer:

masya89 [10]3 years ago

8 0

We have that
12x² = 18x------> 12x²-18x=0-----> divided by 12----> x²-(3/2)*x=0
this is the equation of a vertical parabola
x²-(3/2)*x=0-----> x*(x-1.5)=0
the x-intercepts of the parabola are
x=0
x=1.5

using a graph tool
see the attached figure

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3 years ago

Find the inverse matrix or type "none". (use decimals) {11 -5 3 -1}

alexira [117]

The inverse of the matrix $\left[\begin{array}{cc}11&-5\\3&-1\end{array}\right]$ is the matrix $\left[\begin{array}{cc}\frac{-1}{4} &\frac{5}{4} \\\\\frac{-3}{4} &\frac{11}{4} \end{array}\right]$ .

The inverse of a matrix A is calculated by the formula:

A' = (1/|A|)(Adj A),

where A' represents the inverse of matrix A,

|A| represents the determinant value of matrix A, and

Adj A represents the Adjoin matrix of matrix A.

So, to calculate the inverse of the matrix $\left[\begin{array}{cc}11&-5\\3&-1\end{array}\right]$ , we will first calculate its Adj.

Adj = $\left[\begin{array}{cc}-1&-3\\5&11\end{array}\right] ^{T}$ = $\left[\begin{array}{cc}-1&5\\-3&11\end{array}\right]$ .

Now, we calculate |A| = 11*(-1) - (-5)*3 = -11 + 15 = 4.

Therefore A' = (1/|A|)(Adj A),

or, A' = $(1/4)*\left[\begin{array}{cc}-1&5\\-3&11\end{array}\right]$ ,

or, A' = $\left[\begin{array}{cc}\frac{-1}{4} &\frac{5}{4} \\\\\frac{-3}{4} &\frac{11}{4} \end{array}\right]$ .

Therefore, the inverse of the matrix $\left[\begin{array}{cc}11&-5\\3&-1\end{array}\right]$ is the matrix $\left[\begin{array}{cc}\frac{-1}{4} &\frac{5}{4} \\\\\frac{-3}{4} &\frac{11}{4} \end{array}\right]$ .

Learn more about the inverse of a matrix at

brainly.com/question/16862241

#SPJ10

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