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3 years ago

What’s the property of 4+9=9+4

Mathematics

1 answer:

Ilya [14]3 years ago

5 0

Answer:

From change of places of summands the sum does not change.

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The figure is a square. Find the length of side x in simplest radical form with a rational denominator.

nika2105 [10]

Answer:

$10\sqrt{2}$

Step-by-step explanation:

The diagonal of any square with side length $s$ is equal to $s\sqrt{2}$ . Since the side length of the square is 10, the diagonal must be $\boxed{10\sqrt{2}}$ .

To support and prove this:

The diagonal creates two 45-45-90 triangles, with the diagonal being the hypotenuse of both of these triangles. The Pythagorean Theorem states that in any triangle, the sum of the squares of both legs is equal to the square of the hypotenuse ( $a^2+b^2=c^2$ ).

Call the diagonal $d$ . In these two triangles, both legs are equal to 10 (the side lengths of the square), and the hypotenuse is the diagonal. Thus, we have:

$10^2+10^2=d^2,\\100+100=d^2,\\d^2=200,\\d=\sqrt{200}=\sqrt{100}\cdot \sqrt{2}=\boxed{10\sqrt{2}}$

4 0

3 years ago

Read 2 more answers

Solve the matrix and prove that it is equal 0

Art [367]

Step-by-step explanation:

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

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

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

$\sf{ {A}^{2} - 5A+ 22I= 0}$

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

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.

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

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

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

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