All categories

3 years ago

How are integers and rational numbers different

Mathematics

1 answer:

irga5000 [103]3 years ago

5 0

Answer:

An integer is a whole number, positive or negative. A rational number is a number which can be expressed as a fraction between 2 whole numbers. An integer is a rational number, because every integer can be expressed as a fraction between itself and 1, both whole numbers.

Step-by-step explanation:

You might be interested in

You are given a line that has a slope of 4 and passed through the point (3/8, 1/2). Which statements about the question of the l

alexgriva [62]

Answer:

Statement 1, 2 and 4 are true where as statement 3 is not true.

Step-by-step explanation:

Statement 1: The y-intercept is -1

Point (3/8, 1/2)

Slope = m = 4

y = mx + c

1/2 = 4(3/8) + c

1/2 = 3/2 + c

1/2 = 3/2 + 2c/2

-2 = 2c

c = -1

This statement is true as the y-intercept is -1.

Statement 2: The slope intercept equation is y= 4x - 1

slope = m = 4

y-intercept = c = -1

y = mx + c

y = 4x - 1

This statement is true as the slope intercept equation is y= 4x -1

Statement 3: The point slope equation is y - 3/8 = 4 (x - 1/2)

Point slope equation: y - y1 = m (x - x1)

Points: (x, y) (3/8, 1/2)

y1 = 1/2

x1 = 3/8

Slope = m = 4

y - 1/2 = 4 (x - 3/8)

This statement is not true as the slope intercept equation is y - 1/2 = 4 (x - 3/8) instead of y - 3/8 = 4 (x - 1/2).

Statement 4: The point (3/8, 1/2) corresponds to (x1, y1) in the point slope form of the equation

This statement is true as shown in statement 3's explanation where x1 = 3/8 and y1 = 1/2

6 0

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