All categories

3 years ago

19. Write th three fractions that are greater than ; but less than 1.

Mathematics

1 answer:

Rainbow [258]3 years ago

3 0

Questions:

Write any 3 fractions greater than 0 but less than 1.

Answer:

2/3, 3/5, 7/8

Step-by-step explanation:

All proper fractions are greater than 0 but less than 1.

You might be interested in

HELP ASAP<br> 6+ x - y + 10=

Katena32 [7]

Answer:

x-y+16

Step-by-step explanation:

add the like terms, 6 and 10 which is 16. since the variables are not like terms, thats your final answer. hope it helped :)

7 0

3 years ago

Why can i keep giving brainlyest? they should make it like every day you can give 1

defon

That’s true!!!! They totally should do that!

6 0

3 years ago

Seven less than the quotient of x and 9

bulgar [2K]

X/9 - 7 is the answer. note x/9 is in fraction form

8 0

3 years ago

Which of the following is not a way to represent the solution of the inequality 2(x − 2) less than or equal to 2?

Nezavi [6.7K]

The answer is A A number line with a closed circle on 3-

8 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

$\displaystyle2\iint_{D'}\mathrm du\,\mathrm dv=2\int_1^{\sqrt2}\int_0^{6v}\mathrm du\,\mathrm dv=\boxed6$

4 0

2 years ago