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3 years ago

Write the decimal expansion for 5/6.

Mathematics

2 answers:

SashulF [63]3 years ago

8 0

Answer:

≈ 0.8333333333333

Step-by-step explanation:

Since anything divided by 6 is a repeating decimal, we know it will go on forever.

Vikentia [17]3 years ago

4 0

Answer:

0.8333333333333

Step-by-step explanation:

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Which table represents an exponential function?

iogann1982 [59]

Answer:

The second table

Step-by-step explanation:

The second table you described represents an exponential function. It has a common ratio of 4, meaning that each y-value is multiplied by 4 to get to the next value. (every time the Xs increase by 1, the Ys increase by a multiple of 4)

I hope this helps :))

3 0

3 years ago

Write the number in standard form and expanded form. Three million four hundred eighty- seven thousand six hundred fifty-one

snow_tiger [21]

Answer:

down below

Step-by-step explanation:

3,487, 651

(3 x 1,000,000) + (4 x 100,000) + (8 x 10,000) + ( 7 x 1,000) + (6 x 100 ) +

(5 x 10) + (1 x 1)

7 0

2 years ago

Which of the following statements shows a characteristic of a statistical question? (4 points)

cestrela7 [59]

The answer is D I hope that helped

5 0

3 years ago

Read 2 more answers

What is the standard form of 5.58 x 10^-2

alukav5142 [94]

Answer:

0.0558

Step-by-step explanation:

3 0

2 years ago

Rewrite the following integral in spherical coordinates.

lora16 [44]

In cylindrical coordinates, we have $r^2=x^2+y^2$ , so that

$z = \pm \sqrt{2-r^2} = \pm \sqrt{2-x^2-y^2}$

correspond to the upper and lower halves of a sphere with radius $\sqrt2$ . In spherical coordinates, this sphere is $\rho=\sqrt2$ .

$1 \le r \le \sqrt2$ means our region is between two cylinders with radius 1 and $\sqrt2$ . In spherical coordinates, the inner cylinder has equation

$x^2+y^2 = 1 \implies \rho^2\cos^2(\theta) \sin^2(\phi) + \rho^2\sin^2(\theta) \sin^2(\phi) = \rho^2 \sin^2(\phi) = 1 \\\\ \implies \rho^2 = \csc^2(\phi) \\\\ \implies \rho = \csc(\phi)$

This cylinder meets the sphere when

$x^2 + y^2 + z^2 = 1 + z^2 = 2 \implies z^2 = 1 \\\\ \implies \rho^2 \cos^2(\phi) = 1 \\\\ \implies \rho^2 = \sec^2(\phi) \\\\ \implies \rho = \sec(\phi)$

which occurs at

$\csc(\phi) = \sec(\phi) \implies \tan(\phi) = 1 \implies \phi = \dfrac\pi4+n\pi$

where $n\in\Bbb Z$ . Then $\frac\pi4\le\phi\le\frac{3\pi}4$ .

The volume element transforms to

$dx\,dy\,dz = r\,dr\,d\theta\,dz = \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi$

Putting everything together, we have

$\displaystyle \int_0^{2\pi} \int_1^{\sqrt2} \int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta = \boxed{\int_0^{2\pi} \int_{\pi/4}^{3\pi/4} \int_{\csc(\phi)}^{\sqrt2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta} = \frac{4\pi}3$

4 0

2 years ago