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posledela
3 years ago
11

Simplify (square root 2)( square root of 2^3)

Mathematics
1 answer:
steposvetlana [31]3 years ago
8 0

Answer:

\large\boxed{(\sqrt2)(\sqrt{2^3})=4}

Step-by-step explanation:

(\sqrt2)(\sqrt{2^3})\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\sqrt{2\cdot2^3}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=\sqrt{2^{1+3}}=\sqrt{2^4}\\\\(1)=\sqrt{2^{2\cdot2}}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=\sqrt{(2^2)^2}\qquad\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=2^2=4\\\\(2)=\sqrt{2\cdot2\cdot2\cdot2}=\sqrt{16}=6\ \text{because}\ 4^2=16

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4m+2(m+1)=9m +5
4m+2m+2=9m+5
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Solve the matrix and prove that it is equal 0​
Art [367]

Step-by-step explanation:

\underline{ \underline{ \text{Given}}}  :

  • \tt{ {A}^{T}  = \begin{bmatrix} 2 &  - 4 \\ 4 & 3 \\ \end{bmatrix}}

\underline{ \underline { \text{To \: Find}}} :

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\underline{ \underline{ \text{Solution}}} :

The new matrix obtained from a given matrix by interchanging it's rows and columns is called the transposition of matrix. It is denoted by \sf{ {A}^{T}}. Again , Interchange it's rows and columns in order to find ' A '.

\tt{A = \begin{bmatrix} 2 &  4 \\  - 4 & 3 \\ \end{bmatrix}}

Now , LEFT HAND SIDE ( L.H.S )

\tt{ {A}^{2}  - 5A+ 22I}

Here, I refers to identity matrix. A diagonal matrix in which all the elements of leading diagonal are 1 ( unit ) is called unit or identity matrix.

⟼ \begin{bmatrix} 2 &   4 \\  - 4 & 3 \\ \end{bmatrix} \times \begin{bmatrix} 2 &  4 \\ -  4 & 3 \\ \end{bmatrix} - 5 \times \begin{bmatrix} 2 &   4 \\  - 4 & 3 \\ \end{bmatrix} + 22 \times \begin{bmatrix} 1 &   0 \\  0 & 1\\ \end{bmatrix}

⟼ \begin{bmatrix} 2  \times 2 + 4 \times ( - 4)&   2  \times 4 + 4 \times 3 \\  - 4 \times 2 + 3 \times ( - 4) &  - 4  \times 4 + 3 \times 3 \\ \end{bmatrix} - \begin{bmatrix} 10 &   20 \\   - 20& 15 \\ \end{bmatrix} + \begin{bmatrix} 22 &   0 \\  0 & 22 \\ \end{bmatrix}

⟼ \begin{bmatrix} 4 + ( - 16) &   8 + 12 \\   - 8 + ( - 12) &  - 16 + 9 \\ \end{bmatrix} - \begin{bmatrix} 10 &   20 \\   - 20 & 15 \\ \end{bmatrix} + \begin{bmatrix} 22 &   0 \\  0 & 22 \\ \end{bmatrix}

⟼ \begin{bmatrix} - 12 &   20\\  - 20&  - 7 \\ \end{bmatrix} - \begin{bmatrix} 10 &   20 \\   - 20 & 15 \\ \end{bmatrix} + \begin{bmatrix} 22 &   0 \\  0 & 22 \\ \end{bmatrix}

⟼ \begin{bmatrix}  - 22 &   0 \\  0&  - 22 \\ \end{bmatrix}  + \begin{bmatrix} 22 &   0 \\  0 & 22 \\ \end{bmatrix}

⟼ \begin{bmatrix}  - 22 + 22 &   0 + 0 \\  0 + 0 &  - 22  + 22 \\ \end{bmatrix}

⟼ \begin{bmatrix} 0 &   0\\  0 & 0 \\ \end{bmatrix}

⟼ \sf{0}

RIGHT HAND SIDE ( R.H.S ) : 0

L.H.S = R.H.S [ Hence , proved ! ]

Hope I helped ! ♡

Have a wonderful day / night ! ツ

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<span>The correct result would </span>be: <span>multiply equation a by −3.</span>
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