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Allisa [31]
3 years ago
8

Which numbers are greater than 0.7? 0.37, 0.9, 0.08, 0.69, 0.71

Mathematics
1 answer:
Anna35 [415]3 years ago
7 0
Remember that that the further the the left a digit is, the bigger the number, for example 0.8 is bigger than 0.08 which is in turn bigger than 0.008

Using this we can see that;

0.9, 0.71

Are the only two numbers bigger than 0.7
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5.5hr

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3 years ago
Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )
DedPeter [7]
(a) First find the intersections of y=e^{2x-x^2} and y=2:

2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}

So the area of R is given by

\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx

If you're not familiar with the error function \mathrm{erf}(x), then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line y=1 with y=e^{2x-x^2}.

1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2

So the area of S is given by

\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx
\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx

which is approximately 1.546.

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\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx
5 0
3 years ago
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ra1l [238]
I believe you combine the first and last terms first because they have the same denominator
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