1. Rational or irrational? 11/5

2 answers:

8 0

For something to be rational, it has to be able to be put in fraction form. Since 11/5 is already a fraction, it is rational

nignag [31]3 years ago

4 0

The answer is irrational

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I need help!

Naddik [55]

Answer: Estimate: 36 ? units ? miles per hour ? feet per second?

Exact: 35.32142857 but if you are supposed to use significant digits, 35.32 is probably good.

Step-by-step explanation: For the estimate, round off the fractions to whole numbers, so multiply 6 x 6 to get 36

For exact I converted the mixed numbers to improper fractions, 5 3/4 = 23/4 and 6 1/7 = 43/7. Then multiplied the numerators 23 x 43= 989. And denominators, 4x7=28.

Divide numerator by denominator 989/28= 35.321428571

7 0

3 years ago

A distribution of values is normal with a mean of 99.5 and a standard deviation of 83.2. find p26, which is the score separating

Lynna [10]

A suitable calculator shows the score of 46.0 separates the bottom 26% from the top 74%.

5 0

4 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