If you multiplied a 2x3 matrix and a 3x5 matrix, what size would the resulting matrix

Mathematics

1 answer:

podryga [215]3 years ago

7 0

Answer:

Broooooowhere

N s tayt

Dwyane

Step-by-step explanation:

D ihalesinde merhaba hocam nasılsınız nasılsın bir yöntem olup olmadığını düşünelim sırasıyla görüşmek için randevu talep ediyoruz

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Which statement can be used to prove that a given parallelogram is a rectangle?

RSB [31]

The <em>correct answer</em> is:

The diagonals of the parallelogram are congruent.

Explanation:

In every parallelogram, opposite angles are congruent. This would not mean it is a rectangle.

Consecutive sides of a parallelogram are only congruent if the parallelogram is a rhombus or a square; this would not be a rectangle.

The diagonals of every parallelogram bisect each other. This would not mean it is a rectangle.

The diagonals of a rectangle bisect each other. If we know this is true about our parallelogram, this means our parallelogram is a rectangle.

7 0

3 years ago

Read 2 more answers

Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?

cluponka [151]

Under the given transformation, the Jacobian and its determinant are

$\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2$

so that

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv$

where $D'$ is the region $D$ transformed into the $u$ - $v$ plane. The remaining integral is the twice the area of $D'$ .

Now, the integral over $D$ is

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y$

but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

We require that $u>0$ , and the last equation tells us that we would also need $v>0$ . This means $1\le v\le\sqrt2$ and $0$ , so that the integral over $D'$ is