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2 years ago

Divide 2/9 by 2/4. I need this answer asap.

Mathematics

1 answer:

il63 [147K]2 years ago

6 0

Answer:

The answer is 4/9

Step-by-step explanation:

On dividing 2/9 by 2/4, we multiply $\frac{2}{9}$ with the reciprocal of $\frac{2}{4}$ .

= $\frac{2}{9}$ × $\frac{4}{2}$ = 4/9

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Which of the following equations below is a linear inequality?

Ket [755]

Answer:

Step-by-step explanation:

A <u>linear inequality</u> is an inequality which involves a linear function. A linear inequality contains one of the symbols of inequality: <, >, ≤, ≥.

Consider all options:

A. The inequality $\dfrac{y}{x}>3\cdot x+2$ is not linear linear inequality, because x is in the left part denominator and in the right part numerator.

B. This option shows linear equation, not inequality.

C. This option shows quadratic equation, not linear inequality.

D. This option shows linear inequality $y\le 5x-5$

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2 years ago

Whats the area of a circle with a diameter of 12 cm?

Vedmedyk [2.9K]

Answer:

113.1cm

Step-by-step explanation:

area= $\pi$ r²

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3 years ago

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