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2 years ago

What is the difference between the smallest six-digit whole number and the greatest 4-digit number

1 answer:

8 0

What's the smallest six-digit whole number possible?
It is 100000

try counting, six digits, right? if you even take 1 off from it, it's no longer six digits, so it is the smallest six-digit whole number possible.

what's the greatest 4-digit number?
It should be 9999
try adding 1 to it, it becomes five-digit number.

So, 9999 is the greatest 4-digit whole number possible.

To get a difference, subtraction.
100000 - 9999 = 90001

Just as a tip, for six digit number, 000000 does not work, as it is same as 0

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A number and its absolute value are equal. If you subtract two from the number, the new number and its absolute value are not eq

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A number and its absolute value are not equal when the number is less than zero. Since subtracting 2 makes the number less than zero, we know ...

... the number is a non-negative number less than 2. (0 ≤ n < 2)

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2 years ago

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3 years ago

What is the answer for this problem 3-3×6+2=?

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3 years ago

Read 2 more answers

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

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

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1 year ago

A right prism has a triangular base. The length of the base is 6 meters and the height of the base is 4 meters. The height of th

NNADVOKAT [17]

The volume of this right prism that has a triangular base is equal to 108 cubic meters.

<h3>The volume of a right prism.</h3>

Mathematically, the volume of a right prism is given by this formula:

V = L x W x H

<u>Where:</u>

L is the length.

W is the width.

H is the height.

V is the volume.

<u>Given the following data:</u>

Length of base = 6 meters.

Width of base = 4 meters. cm.

Height of prism = 4.5 meters.

Substituting the given parameters into the formula, we have;

V = 6 x 4 x 5

V = 108 cubic meters.

Read more on volume of prism here: brainly.com/question/3867601

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1 year ago