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julsineya [31]
3 years ago
14

A square rug has an area 80ft which is the best approximation of the length of one side of the rug.....Explain...If you can show

work
A ) 6.4ft
B ) 7.5ft
C ) 8.9ft
D ) 20ft
Mathematics
1 answer:
kow [346]3 years ago
5 0
So to find the answer, first we need to understand how to approximate square roots. 80 is an imperfect square, so we're going to have to round the answer. To find that, we need to find the multiples 80.

72, 80, 81.
8, x, 9

So the distance apart between 72 and 81 is 9. The answer, x, is obviously between 8 and 9, and since the distance between these to is 9, your answer is going to be 8.9 .
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2,10,18,26 find the 61st term
saw5 [17]

Answer:

482

Step-by-step explanation:

We can see that the numbers shown resemble an arithmetic sequence because they have a common difference. The formula for the nth term of an arithmetic sequence is:

a_{n} =a_{1} +(n-1)d

Where a_{1} is the first term, a_{n} is the nth term, and d is the common difference. To find the 61st term, all we need is the first term and the common difference. By looking at what given, we can say the first term is 2. Now, to find the common difference, we find the difference of a term from the term before it. In this case we can do 10-2, which is 8 , or the common difference. Since we have everything we need, it can be plugged into the equation:

a_{61} =2 + (61 - 1)*8\\a_{61} = 2 + 60*8\\a_{61} = 2 + 480\\a_{61} = 482

So, the 61st term is 482.

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2 years ago
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Pepsi [2]

Answer:

A

Step-by-step explanation:

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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

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3 years ago
The probability that a randomly selected 40-year-old female will live to be 41 years old is 0.99855 according to the National Vi
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