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3 years ago

Can someone help me this is due tonight

Mathematics

2 answers:

kobusy [5.1K]3 years ago

4 0

Answer:

Step-by-step explanation:

x% of y equals to 0.01*x*y

Just put the numbers in the formula

GalinKa [24]3 years ago

4 0

33% of 507 = 167.31

48% of 375 = 180

76% of 285 = 216.6

60% of 398 = 238.8

89% of 150 = 133.5

26% of 430 = 111.8

81% of 216 = 174.96

5% of 584 = 29.2

18% of 725 = 130.5

2% of 115 = 2.3

90% of 152 = 136.8

12% of 649 = 77.88

55% of 216 = 118.8

43% of 108 = 46.44

97% of 235 = 227.95

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So, first off it would help if you could define the terms irrational number and integer. Well, an irrational number is basically any real number that can't be expressed as a ratio of integers. They also cannot be represented by terminating or continuing decimals. And an integer is pretty much any number that cannot be written as a fraction or decimal, such as -2, 13, 257. It is not 2 and 1/2, or 4.75. Those would not be integers. Do you think you can figure out the difference?

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3 years ago

Can you simplify 14x+11 please??

ad-work [718]

Answer:

It cannot be further simplified

Step-by-step explanation:

11 is a prime number, so there are no numbers that can go into 14 and 11 (besides 1 and 1 would not simplify the problem)

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3 years ago

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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