Rational exponent form of the square root of 10

Mathematics

1 answer:

sp2606 [1]3 years ago

3 0

To make the exponent rational, for square root use 1/2. To wit:
10^(1/2)

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Use the definition to find an expression for the area under the curve y = x3 from 0 to 1 as a limit. lim n→∞ n i = 1 R (b) The f

Harlamova29_29 [7]

The summand (R?) is missing, but we can always come up with another one.

Divide the interval [0, 1] into $n$ subintervals of equal length $\dfrac{1-0}n=\dfrac1n$ :

$[0,1]=\left[0,\dfrac1n\right]\cup\left[\dfrac1n,\dfrac2n\right]\cup\cdots\cup\left[1-\dfrac1n,1\right]$

Let's consider a left-endpoint sum, so that we take values of $f(\ell_i)={\ell_i}^3$ where $\ell_i$ is given by the sequence

$\ell_i=\dfrac{i-1}n$

with $1\le i\le n$ . Then the definite integral is equal to the Riemann sum

$\displaystyle\int_0^1x^3\,\mathrm dx=\lim_{n\to\infty}\sum_{i=1}^n\left(\frac{i-1}n\right)^3\frac{1-0}n$

$=\displaystyle\lim_{n\to\infty}\frac1{n^4}\sum_{i=1}^n(i-1)^3$

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$=\displaystyle\lim_{n\to\infty}\frac{n^2(n-1)^2}{4n^4}=\boxed{\frac14}$

8 0

3 years ago

Let ∠1, ∠2, and ∠3 have the following relationships.

Kaylis [27]

Answer:

sry im being Dumb im pretty sure its 180 because intersecting lines form vertical angles if those angles are acute the one in between is obtuve a like is 180 degrees

7 0

3 years ago

Help me guys <br> thx so much

navik [9.2K]

Answer:

Option D, $10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}$

Step-by-step explanation:

<u>Step 1: Multiply</u>

<u /> $(\sqrt{10x^4} -x\sqrt{5x^2} )*(2\sqrt{15x^4} + \sqrt{3x^3})\\ (\sqrt{10 * x^2 * x^2} -x\sqrt{5 * x^2} ) * (2\sqrt{15 * x^2 * x^2} +\sqrt{3 * x^2 * x})\\(x^2\sqrt{10} -x^2\sqrt{5} )*(2x^2\sqrt{15} +x\sqrt{3x}) \\\\$