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3 years ago

Solve: 3(2x-1) = 9(x+3)

Mathematics

2 answers:

Alja [10]3 years ago

4 0

Answer:

= − 1 0

Step-by-step explanation:

Hope this helps! :)

lions [1.4K]3 years ago

3 0

Answer:

x=-10

Explanation:

6x-3=9x+27(distributive property)

6x-9x=27+3(move like terms to one side)

-3x=30(combine like terms)

x=-10(divide both sides by -3)

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Write the equation of a line with that passes through the<br> points (-3, -2) and (6,1)?

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: Y = 1/3x - 1

M = y2 - y1/ x2-x1

M = 6 - (-3) / 1 - (-2)

Y = 1/3x + b

Then substitute one of the points into the equation

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3 years ago

How to represent lengths in line plot math

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3 years ago

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SCORPION-xisa [38]

Answer look it up learn to add and subtract radical expansions

Step-by-step explanation:

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3 years ago

Read 2 more answers

Please help! 7th one!

denis23 [38]

The method we need to use here is very particular to this type of a situation. The way we will find that polynomial, or the divisor, is to follow this formula: $g(x)= \frac{dividend-remainder}{quotient}$ . For us that will look like this: $g(x)= \frac{4x^4-5x^3-39x^2-46x-2-(-5x+8)}{x^2-3x-5}$ . First we will simplify as much as possible that very long numerator there. It simplifies to $g(x)= \frac{4x^4-5x^3-39x^2-41x-10}{x^2-3x-5}$ . What you do now is use long division of polynomials, which, unfortunately, is impossible to show in this forum. However, get familiar with the long division process if you are not already, and you will find that your polynomial g(x) is $4x^2+7x+2$ .

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4 years ago

The ground-state wave function for a particle confined to a one-dimensional box of length L is Ψ=(2/L)^1/2 Sin(πx/L). Suppose th

Hitman42 [59]

Answer:

(a) 4.98x10⁻⁵

(b) 7.89x10⁻⁶

(c) 1.89x10⁻⁴

(d) 0.5

(e) 2.9x10⁻²

Step-by-step explanation:

The probability (P) to find the particle is given by:

$P=\int_{x_{1}}^{x_{2}}(\Psi\cdot \Psi) dx = \int_{x_{1}}^{x_{2}} ((2/L)^{1/2} Sin(\pi x/L))^{2}dx$

$P = \int_{x_{1}}^{x_{2}} (2/L) Sin^{2}(\pi x/L)dx$ (1)

The solution of the intregral of equation (1) is:

$P=\frac{2}{L} [\frac{X}{2} - \frac{Sin(2\pi x/L)}{4\pi /L}]|_{x_{1}}^{x_{2}}$

(a) The probability to find the particle between x = 4.95 nm and 5.05 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{4.95}^{5.05} = 4.98 \cdot 10^{-5}$

(b) The probability to find the particle between x = 1.95 nm and 2.05 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{1.95}^{2.05} = 7.89 \cdot 10^{-6}$

(c) The probability to find the particle between x = 9.90 nm and 10.00 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{9.90}^{10.00} = 1.89 \cdot 10^{-4}$

(d) The probability to find the particle in the right half of the box, that is to say, between x = 0 nm and 50 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{50.00} = 0.5$

(e) The probability to find the particle in the central third of the box, that is to say, between x = 0 nm and 100/6 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{16.7} = 2.9 \cdot 10^{-2}$

I hope it helps you!

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4 years ago