What is an integer equivalent to 9 3/2

Mathematics

1 answer:

enyata [817]3 years ago

6 0

Answer:

No its a fraction (and decimal) so there is no integer equivalent

Step-by-step explanation:

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Find the? inverse, if it? exists, for the given matrix.<br><br> [4 3]<br><br> [3 6]

True [87]

Answer:

Therefore, the inverse of given matrix is

$=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}$

Step-by-step explanation:

The inverse of a square matrix $A$ is $A^{-1}$ such that

$A A^{-1}=I$ where I is the identity matrix.

Consider, $A = \left[\begin{array}{ccc}4&3\\3&6\end{array}\right]$

$\mathrm{Matrix\:can\:only\:be\:inverted\:if\:it\:is\:non-singular,\:that\:is:}$

$\det \begin{pmatrix}4&3 \\3&6\end{pmatrix}\ne 0$

$\mathrm{Find\:2x2\:matrix\:inverse\:according\:to\:the\:formula}:\quad \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}^{-1}=\frac{1}{\det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}}\begin{pmatrix}d\:&\:-b\:\\ -c\:&\:a\:\end{pmatrix}$

$=\frac{1}{\det \begin{pmatrix}4&3\\ 3&6\end{pmatrix}}\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}$

$\mathrm{Find\:the\:matrix\:determinant\:according\:to\:formula}:\quad \det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}\:=\:ad-bc$

$4\cdot \:6-3\cdot \:3=15$

$=\frac{1}{15}\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}$

$=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}$

Therefore, the inverse of given matrix is

$=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}$

4 0

3 years ago

If ex + e⁻¹= 2, then x =..

harkovskaia [24]

Answer:

$\rm x = \dfrac{2e - 1}{ {e}^{2} }$

Step-by-step explanation:

$\rm Solve \: for \: x: \\ \rm \longrightarrow ex + {e}^{ - 1} = 2 \\ \\ \rm \longrightarrow e x + \dfrac{1}{e} = 2 \\ \\ \rm Subtract \: \dfrac{1}{e} \: from \: both \: sides: \\ \rm \longrightarrow e x + \dfrac{1}{e} - \dfrac{1}{e} = 2 - \dfrac{1}{e} \\ \\ \rm \longrightarrow ex = \dfrac{2e}{e} - \dfrac{1}{e} \\ \\ \rm \longrightarrow ex = \dfrac{2e - 1}{e} \\ \\ \rm Divide \: both \: sides \: by \: e: \\ \rm \longrightarrow \dfrac{ex}{e} = \dfrac{2e - 1}{e \times e} \\ \\ \rm \longrightarrow x = \dfrac{2e - 1}{ {e}^{2} }$