What's 99 divided by 9,801

Mathematics

2 answers:

Akimi4 [234]2 years ago

7 0

__
99 divided by 9801 = 0.01

ra1l [238]2 years ago

3 0

Answer:

The answer is 0.01.

Step-by-step explanation:

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Rewrite in simplest rational exponent form √x • 4√x. Show each step of your process.

Lisa [10]

$First\ step:you\ must\ have\ the\ domain\ (D)\\x\geq0\to D:x\in[0;\ \infty)\\\\Second\ step:use\ \sqrt{a}\cdot\sqrt{a}=\left(\sqrt{a}\right)^2=a\ for\ a\geq0\\\\Solution:\\\sqrt{x}\cdot4\sqrt{x}=4(\sqrt{x})^2=\huge\boxed{4x}$

$or\ other\ method\\\\First\ step:\ use\ \sqrt{x}=x^{\frac{1}{2}}\\\\\sqrt{x}\cdot4\sqrt{x}=x^\frac{1}{2}\cdot4x^{\frac{1}{2}}\\\\Second\ step:\ use\ a^n\cdot a^m=a^{n+m}\\\\x^\frac{1}{2}\cdot4x^{\frac{1}{2}}=4\cdot x^\frac{1}{2}\cdot x^{\frac{1}{2}}=4x^{\frac{1}{2}+\frac{1}{2}}=4x^1=\huge\boxed{4x}$

7 0

2 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

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