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LenKa [72]
3 years ago
8

Can someone help me 7 + 2h = 3 + 4h

Mathematics
1 answer:
STALIN [3.7K]3 years ago
7 0

h = 2

7 + 2h = 3 + 4h

You must isolate the variable to solve using the order of operations, or PEMDAS.

4 = 2h

Subtract 3 from both sides of the equation, and subtract 2h from both sides of the equation.

\frac{4}{2}  = h

Divide both sides of the equation by 2.

h = 2

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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
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The steps that can be done to both sides of the equation is: Add 9.6, then divide by 3.2

Step-by-step explanation:

Solve the equation:

3.2x - 9.6 = 38.4

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3.2x -9.6 +9.6 = 38.4 +9.6

3.2x = 48

-Then, you divide both sides by 3.2:

\frac{3.2x}{3.2} = \frac{48}{3.2}

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7 0
3 years ago
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Zolol [24]

Answer:

Percentage increase in volume will be

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Step-by-step explanation:

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It's change of subject. :)

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I hope this helps you

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3 years ago
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