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4 years ago

Find the area of a square with sides of length 12

Mathematics

1 answer:

Maru [420]4 years ago

5 0

Answer:

144

Step-by-step explanation:

12x12=144

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A company sells phone cases for $15 each. At the end of the week, they have sold $570 worth of phone cases. How many phone cases

Over [174]

Answer:

They sold 38 phone cases.

Step-by-step explanation:

The equation I used to solve is

$570 \div 15$

3 0

3 years ago

How to reduce 10/4 i don’t know how to

vredina [299]

answer: 10/4 reduced is 2.5/1 or 2.5.

work: what we are doing is trying to divide ten into groups of four. how many times does 10 fit into four? well, look at the multiples of four, 4, 8, 12... we can see that 8 fits into ten, but 12 doesn't. that means two whole fours can fit into ten (since 4 x 2 = 8) so we still have a 2 left over from the ten. since 2 is one half of four, we can fit in another .5. so, ten can be fit into 2.5 groups. hope this helps, have a great day!

3 0

3 years ago

Read 2 more answers

50 multiplied by 100

ra1l [238]

When multiplying by 100 just add 2 zero's behind your second number. ANSWER:5,000

3 0

3 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!

3 0

4 years ago

Please help STUCK !! Which of the following points lie on the graph of the function,

masha68 [24]

Answer: a. ( 1, 6 )

Step-by-step explanation:

If you go into desmos and plot your function you can go to each of the points in your answers and see just where the point is touching the line

7 0

3 years ago