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3 years ago

1.6 x 10^9 in standard form

Mathematics

1 answer:

Dovator [93]3 years ago

8 0

Answer:

this is already in standard form

Step-by-step explanation:

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Marla scored 70% on her last unit exam in her statistics class. When Marla took the SAT exam, she scored at the 70th percentile

Lerok [7]

Answer:

The difference is that Marla's exam in her statistics class was graded by percent of correct answers, in her case 70%, while the SAT is graded into a curve, taking other students' grades also into account, and since she scored in the 70th percentile, Marla scored better than 70% of the students.

Step-by-step explanation:

Marla scored 70% on her last unit exam in her statistics class.

This means that in her statistics class, Marla got 70% of her test correct.

When Marla took the SAT exam, she scored at the 70th percentile in mathematics.

This means that on the SAT exam, graded on a curve, Marla scored better than 70% of the students.

Explain the difference in these two scores.

3 0

3 years ago

Write an expression that represents 6 less than a number time 5

den301095 [7]

5x-6 because 6 less than means after and a number times 5 would be 5x so it would be 5x-6

7 0

3 years ago

List the coordinates for the plotted points A, B, C, and D.

avanturin [10]

Answer:

A= (3,1) ; B=(-2,-1) ; C=(-4,1) ; D=(-4,0)

Step-by-step explanation:

Plotted points are expressed by (x,y) , where the x is equal to the value that you read on the x plot and y is equal to the value that you read on the y plot.

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Crm%5Cint_%7B0%7D%5E%7B%20%5Cinfty%20%7D%20%5Cleft%28%20%20%5Cfrac%7B%2

mylen [45]

Recall the geometric sum,

$\displaystyle \sum_{k=0}^{n-1} x^k = \frac{1-x^k}{1-x}$

It follows that

$1 - x + x^2 - x^3 + \cdots + x^{2020} = \dfrac{1 + x^{2021}}{1 + x}$

So, we can rewrite the integral as

$\displaystyle \int_0^\infty \frac{x^2 + 1}{x^4 + x^2 + 1} \frac{\ln(1 + x^{2021}) - \ln(1 + x)}{\ln(x)} \, dx$

Split up the integral at x = 1, and consider the latter integral,

$\displaystyle \int_1^\infty \frac{x^2 + 1}{x^4 + x^2 + 1} \frac{\ln(1 + x^{2021}) - \ln(1 + x)}{\ln(x)} \, dx$

Substitute $x\to\frac1x$ to get

$\displaystyle \int_0^1 \frac{\frac1{x^2} + 1}{\frac1{x^4} + \frac1{x^2} + 1} \frac{\ln\left(1 + \frac1{x^{2021}}\right) - \ln\left(1 + \frac1x\right)}{\ln\left(\frac1x\right)} \, \frac{dx}{x^2}$

Rewrite the logarithms to expand the integral as

$\displaystyle - \int_0^1 \frac{1+x^2}{1+x^2+x^4} \frac{\ln(x^{2021}+1) - \ln(x^{2021}) - \ln(x+1) + \ln(x)}{\ln(x)} \, dx$

Grouping together terms in the numerator, we can write

$\displaystyle -\int_0^1 \frac{1+x^2}{1+x^2+x^4} \frac{\ln(x^{2020}+1)-\ln(x+1)}{\ln(x)} \, dx + 2020 \int_0^1 \frac{1+x^2}{1+x^2+x^4} \, dx$

and the first term here will vanish with the other integral from the earlier split. So the original integral reduces to

$\displaystyle \int_0^\infty \frac{1+x^2}{1+x^2+x^4} \frac{\ln(1-x+\cdots+x^{2020})}{\ln(x)} \, dx = 2020 \int_0^1 \frac{1+x^2}{1+x^2+x^4} \, dx$

Substituting $x\to\frac1x$ again shows this integral is the same over (0, 1) as it is over (1, ∞), and since the integrand is even, we ultimately have

$\displaystyle \int_0^\infty \frac{1+x^2}{1+x^2+x^4} \frac{\ln(1-x+\cdots+x^{2020})}{\ln(x)} \, dx = 2020 \int_0^1 \frac{1+x^2}{1+x^2+x^4} \, dx \\\\ = 1010 \int_0^\infty \frac{1+x^2}{1+x^2+x^4} \, dx \\\\ = 505 \int_{-\infty}^\infty \frac{1+x^2}{1+x^2+x^4} \, dx$

We can neatly handle the remaining integral with complex residues. Consider the contour integral

$\displaystyle \int_\gamma \frac{1+z^2}{1+z^2+z^4} \, dz$

where γ is a semicircle with radius R centered at the origin, such that Im(z) ≥ 0, and the diameter corresponds to the interval [-R, R]. It's easy to show the integral over the semicircular arc vanishes as R → ∞. By the residue theorem,

$\displaystyle \int_{-\infty}^\infty \frac{1+x^2}{1+x^2+x^4}\, dx = 2\pi i \sum_\zeta \mathrm{Res}\left(\frac{1+z^2}{1+z^2+z^4}, z=\zeta\right)$

where $\zeta$ denotes the roots of $1+z^2+z^4$ that lie in the interior of γ; these are $\zeta=\pm\frac12+\frac{i\sqrt3}2$ . Compute the residues there, and we find