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NNADVOKAT [17]
3 years ago
11

Can somebody please help me with question b please?

Mathematics
1 answer:
snow_tiger [21]3 years ago
4 0
\bf A=
\begin{bmatrix}
-2&0\\1&3
\end{bmatrix}\qquad B=
\begin{bmatrix}
1&0&3\\-1&3&0
\end{bmatrix}\\\\
-----------------------------\\\\
AB=
\begin{bmatrix}
(-2\cdot 1)+(0\cdot -1)&(-2\cdot 0)+(0\cdot 3)&(-2\cdot 3)+(0\cdot 0)\\\\
(1\cdot 1)+(3\cdot -1)&(1\cdot 0)+(3\cdot 3)&(1\cdot 3)+(3\cdot 0)
\end{bmatrix}

as you notice above, is the first-row components from A, multiplying all the columns subsequently on B, and you add the products of that row, that gives you one component on the AB matrix

in the one above, we end up with a 2x3 AB matrix
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19) Given that f(x)x² - 8x+ 15x² - 25find the horizontal and vertical asymptotes using the limits of the function.A) No Vertical
Tems11 [23]

EXPLANATION

Since we have the function:

f(x)=\frac{x^2-8x+15}{x^2}

Vertical asymptotes:

For\:rational\:functions,\:the\:vertical\:asymptotes\:are\:the\:undefined\:points,\:also\:known\:as\:the\:zeros\:of\:the\:denominator,\:of\:the\:simplified\:function.

Taking the denominator and comparing to zero:

x+5=0

The following points are undefined:

x=-5

Therefore, the vertical asymptote is at x=-5

Horizontal asymptotes:

\mathrm{If\:denominator's\:degree\:>\:numerator's\:degree,\:the\:horizontal\:asymptote\:is\:the\:x-axis:}\:y=0.If\:numerator's\:degree\:=\:1\:+\:denominator's\:degree,\:the\:asymptote\:is\:a\:slant\:asymptote\:of\:the\:form:\:y=mx+b.If\:the\:degrees\:are\:equal,\:the\:asymptote\:is:\:y=\frac{numerator's\:leading\:coefficient}{denominator's\:leading\:coefficient}\mathrm{If\:numerator's\:degree\:>\:1\:+\:denominator's\:degree,\:there\:is\:no\:horizontal\:asymptote.}\mathrm{The\:degree\:of\:the\:numerator}=1.\:\mathrm{The\:degree\:of\:the\:denominator}=1\mathrm{The\:degrees\:are\:equal,\:the\:asymptote\:is:}\:y=\frac{\mathrm{numerator's\:leading\:coefficient}}{\mathrm{denominator's\:leading\:coefficient}}\mathrm{Numerator's\:leading\:coefficient}=1,\:\mathrm{Denominator's\:leading\:coefficient}=1y=\frac{1}{1}\mathrm{The\:horizontal\:asymptote\:is:}y=1

In conclusion:

\mathrm{Vertical}\text{ asymptotes}:\:x=-5,\:\mathrm{Horizontal}\text{ asymptotes}:\:y=1

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