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3 years ago

What is 4/3 times 3.14

7D%5E%7B3%7D%20" id="TexFormula1" title=" \frac{4}{3} \times 3.14 \times {6}^{3} " alt=" \frac{4}{3} \times 3.14 \times {6}^{3} " align="absmiddle" class="latex-formula">

Mathematics

1 answer:

kakasveta [241]3 years ago

6 0

$\dfrac{4}{3}\times3.14\times6^3=\dfrac{12.56}{3}\times216=\dfrac{12.56}{1}\times72=904.32$

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I need help with this question <br> v2–2v–24

Lelu [443]

Answer:

(v - 6)(v + 4)

Step-by-step explanation:

v² - 2v - 24

v² + 4v - 6v - 24

v(v + 4) - 6(v + 4)

(v - 6)(v + 4)

6 0

3 years ago

Calculate the probability that a particle in a one-dimensional box of length aa is found between 0.29a0.29a and 0.34a0.34a when

Helga [31]

Answer:

P(0.29a < X < 0.34a) = -(4a/π) [(cos 0.34πa²) - (cos 0.29πa²)]

Assuming the box is of unit length, a = 1,

0.29a = 0.29 and 0.34a = 0.34

P(0.29 < X < 0.34) = (-4/π) [(cos 0.34π) - (cos 0.29π)] = 0.167 = 0.17 to 2 s.f

Step-by-step explanation:

ψ(x) = 2a(√sin(πxa)

The probability of finding a particle in a specific position for a given energy level in a one-dimensional box is related to the square of the wavefunction

The probability of finding a particle between two points 0.29a and 0.34a is given mathematically as

P(0.29a < X < 0.34a) = ∫⁰•³⁴ᵃ₀.₂₉ₐ ψ²(x) dx

That is, integrating from 0.29a to 0.34a

ψ²(x) = [2a(√sin(πxa)]² = 4a² sin(πxa)

∫⁰•³⁴ᵃ₀.₂₉ₐ ψ²(x) dx = ∫⁰•³⁴ᵃ₀.₂₉ₐ (4a² sin(πxa)) dx

∫⁰•³⁴ᵃ₀.₂₉ₐ (4a² sin(πxa))

= - [(4a²/πa)cos(πxa)]⁰•³⁴ᵃ₀.₂₉ₐ

- [(4a/π)cos(πxa)]⁰•³⁴ᵃ₀.₂₉ₐ = -(4a/π) [(cos 0.34πa²) - (cos 0.29πa²)]

P(0.29a < X < 0.34a) = -(4a/π) [(cos 0.34πa²) - (cos 0.29πa²)]

Assuming the box is of unit length, a = 1

P(0.29 < X < 0.34) = (-4/π) [(cos 0.34π) - (cos 0.29π)] = (-1.273) [0.482 - 0.613] = 0.167.

7 0

4 years ago

Amie works 6 hours each saturday and 4 hours each monday she also makes $20 a week as a dog walker she makes $130 a week if h re

OLga [1]

Answer:

Step-by-step explanation:

4 0

3 years ago

Someone get me the answers ASAP!! thank you

Virty [35]

This relies on the distance formula like number two, which you answered correctly.

I think it's helpful to draw this out. I made a digital drawing on GeoGebra (free) and connected the vertices. Now, to find the area, we need the side lengths to start. This program actually gives you the side lengths as well, but pretending we don't already have that, you can plug in each vertex into the distance formula.

$a = \sqrt{(-4-3)^{2}+ (2-2)^{2}} = 7$

$b = \sqrt{(3-3)^{2}+(2-(-5))^{2}} = 7$

$c = \sqrt{(3-(-4))^{2}+(-5-(-2))^{2}} = \sqrt{58}$

$d = \sqrt{(-4-(-4))^{2}+(2-(-2))^{2}} = 4$

So, unfortunately, we don't have a regular polygon, so this isn't simply length times width. But we can easily cut this into a rectangle and a triangle, whose areas are much easier to find (see second image).

The area, A, of a triangle is $\frac{1}{2} B * h$ and the area of a rectangle is simply $l * w$ . So the base of our triangle is 7 units, and the height is 3 units, so the area is 10.5 units squared. The length of our rectangle is 7 units, and the width is 4 units, so the area is 28 units squared.

$A = A_{triangle} + A_{rectangle} = 10 \frac{1}{2}+28 = 38\frac{1}{2}$ units squared.

I would like to clarify that you do not actually need to find all those distances for this problem. Once it's drawn out, it's fairly easy to tell.

7 0

4 years ago

2783 and 7283. The value of 2 in _ is _ times the value of two in _. The 2 is the underlined digit.

NARA [144]

The value of 2 in 2783 (2000) is 10 times the value of two in 7283 (200).

Hope this helped!

Nate

3 0

3 years ago