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seropon [69]
3 years ago
12

Rewrite the expression in the form y^ny n y, start superscript, n, end superscript. \left(y^{^{\scriptsize -\dfrac12}}\right)^{4

}= ⎝ ⎜ ⎛ ​ y − 2 1 ​ ⎠ ⎟ ⎞ ​ 4

Mathematics
1 answer:
sladkih [1.3K]3 years ago
5 0

Answer

Today’s mathematicians would probably agree that the Riemann Hypothesis is the most significant open problem in all of math. It’s one of the seven Millennium Prize Problems, with a million dollar reward for its solution. It has implications deep into various branches of math, but it’s also simple enough that we can explain the basic idea right here.

There is a function, called the Riemann zeta function, written in the image above.

For each s, this function gives an infinite sum, which takes some basic calculus to approach for even the simplest values of s. For example, if s=2, then (s) is the well-known series 1 + 1/4 + 1/9 + 1/16 + …, which strangely adds up to exactly ²/6. When s is a complex number—one that looks like a+b, using the imaginary number —finding (s) gets tricky.

So tricky, in fact, that it’s become the ultimate math question. Specifically, the Riemann Hypothesis is about when (s)=0; the official statement is, “Every nontrivial zero of the Riemann zeta function has real part 1/2.” On the plane of complex numbers, this means the function has a certain behavior along a special vertical line. You can see this in the visualization of the function above—it’s along the boundary of the rainbow and the red. The hypothesis is that the behavior continues along that line infinitely.

The Hypothesis and the zeta function come from German mathematician Bernhard Riemann, who described them in 1859. Riemann developed them while studying prime numbers and their distribution. Our understanding of prime numbers has flourished in the 160 years since, and Riemann would never have imagined the power of supercomputers. But lacking a solution to the Riemann Hypothesis is a major setback.

If the Riemann Hypothesis were solved tomorrow, it would unlock an avalanche of further progress. It would be huge news throughout the subjects of Number Theory and Analysis. Until then, the Riemann Hypothesis remains one of the largest dams to the river of math research.

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52% or B.

Step-by-step explanation:

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3 years ago
Type the correct answer in each box
Luba_88 [7]

Answer:

We know that the equation of the circle in standard form is equal to <em>(x-h)² + (y-k)² = r²</em> where (h,k) is the center of the circle and r is the radius of the circle.

We have x² + y² + 8x + 22y + 37 = 0, let's get to the standard form :

1 - We first group terms with the same variable :

(x²+8x) + (y²+22y) + 37 = 0

2 - We then move the constant to the opposite side of the equation (don't forget to change the sign !)

(x²+8x) + (y²+22y) = - 37

3 - Do you recall the quadratic identities ? (a+b)² = a² + 2ab + b². Now that's what we are trying to find. We call this process <u><em>"completing the square"</em></u>.

x²+8x = (x²+8x + 4²) - 4² = (x+4)² - 4²

y²+22y = (y²+22y+11²)-11² = (y+11)²-11²

4 - We plug the new values inside our equation :

(x+4)² - 4² + (y+22)² - 11² = -37

(x+4)² + (y+22)² = -37+4²+11²

(x+4)²+(y+22)² = 100

5 - We re-write in standard form :

(x-(-4)²)² + (y - (-22))² = 10²

And now it is easy to identify h and k, h = -4 and k = - 22 and the radius r equal 10. You can now complete the sentence :)

5 0
3 years ago
The eccentricity e of an ellipse is defined as the number c/a, where a is the distance of a vertex from the center and c is the
Anna71 [15]

Answer:

Check below, please.

Step-by-step explanation:

Hi, there!

Since we can describe eccentricity as e=\frac{c}{a}

a) Eccentricity close to 0

An ellipsis with eccentricity whose value is 0, is in fact, a degenerate one almost a circle. An ellipse whose value is close to zero is almost a degenerate circle. The closer the eccentricity comes to zero, the more rounded gets the ellipse just like a circle. (Check picture, please)

\frac{x^2}{a^2} +\frac{y^2}{b^2} =1 \:(Ellipse \:formula)\\a^2=b^2+c^2 \: (Pythagorean\: Theorem)\:a=longer \:axis.\:b=shorter \:axis)\\a^2=b^2+(0)^2 \:(c\:is \:the\: distance \: the\: Foci)\\\\a^2=b^2 \\a=b\: (the \:halves \:of \:each\:axes \:measure \:the \:same)

b) Eccentricity =5

5=\frac{c}{a} \:c=5a

An eccentricity equal to 5 implies that the distance between the Foci has to be five (5) times larger than the half of its longer axis! In this case, there can't be an ellipse since the eccentricity must be between 0 and 1 in other words:

If\:e=\frac{c}{a} \:then\:c>0 , and\: c>0 \:then \:1>e>0

c) Eccentricity close to 1

In this case, the eccentricity close or equal to 1 We must conceive an ellipse whose measure for the half of the longer axis a and the distance between the Foci 'c' they both have the same size.

a=c\\\\a^2=b^2+c^2\:(In \:the\:Pythagorean\:Theorem\: we \:should\:conceive \:b=0)

Then:\\\\a=c\\e=\frac{c}{a}\therefore e=1

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