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3 years ago

If 10x=1/0.001 find the value of x

Mathematics

1 answer:

DENIUS [597]3 years ago

4 0

$\\ \sf\longmapsto 10x=\dfrac{1}{0.001}$

Turn over the decimal

$\\ \sf\longmapsto 10x=\dfrac{1}{\dfrac{1}{1000}}$

$\\ \sf\longmapsto 10x=1000$

$\\ \sf\longmapsto x=\dfrac{1000}{10}$

$\\ \sf\longmapsto x=100$

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adell [148]

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2 years ago

Please respond in detail

Serggg [28]

Answer:

No.

Step-by-step explanation:

As x increases 2^x will grow faster than 5 x^2. This is because the x in 2^x is an exponent and its graph grows very steeply compared with 5x^2 , as x increases.

For example when x = 10

5x^2 = 5*10^2 = 500

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When x = 11:

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3 years ago

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Answer:

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Step-by-step explanation:

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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.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve $y=e^{2x-x^2}$ and the line $y=1$ , or $e^{2x-x^2}-1$ . The area of any such circle is $\pi$ times the square of its radius. Since the curve intersects the axis of revolution at $x=0$ and $x=2$ , the volume would be given by

$\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx$

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3 years ago

If 10x=1/0.001 find the value of x​

If 10x=1/0.001 find the value of x