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3 years ago

8(x - 6) +10 = 8x - 37 pls help answer

Mathematics

2 answers:

Oduvanchick [21]3 years ago

7 0

Answer:

No solution for x

Step-by-step explanation:

Have a wonderful day!

Good luck!

Don't forget you are amazing, smart, and lovely!

kirill115 [55]3 years ago

3 0

Answer:

Its a no solution problem.

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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

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

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Answer:

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Step-by-step explanation:

240 x 3 = 720 pencils

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3 years ago

Read 2 more answers

Slav-nsk [51]

Let k be the scale factor relating two similar prisms $P_{1}$ and $P_{2}$ , such that for corresponding parts of prisms $P_{1}$ and $P_{2}$ (for heights, in particular) we have $k= \frac{height\ of \ P_{1} }{height \ of\ P_{2} }$ . In our case $k= \frac{4}{10} = \frac{2}{5}$ .
For surfaces area we have $\frac{Surface \ area \ of \ P_{1} }{Surface \ area \ of \ P_{2}} = k^{2} =( \frac{2}{5} )^{2}= \frac{4}{25}$ .
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