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3 years ago

Nine twentieths is equal

Mathematics

2 answers:

ruslelena [56]3 years ago

5 0

An equivalent fractions would be 18/40.

kondaur [170]3 years ago

4 0

Nine twentieths is equal to...

9/20 and 0.45.

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How do I factor 8x^2+12+48?

wolverine [178]

We can factor a 4 out of this expression.

4(2x^2 + 3x + 12)

Now that the expression is simplified, we can see if splitting the middle term is possible.

<em><u>Display factors of 24.</u></em>

1 * 24

-1 * - 24

2 * 12

-2 * -12

3 * 8

-3 * - 8

4 * 6

-4 * -6

All factors of 24 have been listed, and none of these digits satisfy the criteria.

Trinomial cannot be further factored.

8 0

2 years ago

Determine the commutators of the operators a and a+,where a = (x + ip)/2 ^1/2 and a+ = (x - ip)/ 2 ^1/2

Vsevolod [243]

Answer:

Given that:

$a = (\frac{x+ip}{2})^{\frac{1}{2}}$ and $a+= (\frac{x-ip}{2})^{\frac{1}{2}}$

if a , a+ commutator, it obeys $aa^+ = a^+a$

First find:

$aa^+ = (\frac{x+ip}{2})^{\frac{1}{2}} (\frac{x-ip}{2})^{\frac{1}{2}}$

= $(\frac{(x)^2-(ip)^2}{4})^{\frac{1}{2}}=(\frac{(x)^2+(p)^2}{4})^{\frac{1}{2}}$

Now;

$a^+a =(\frac{x-ip}{2})^{\frac{1}{2}} (\frac{x+ip}{2})^{\frac{1}{2}} = (\frac{(x)^2-(ip)^2}{4})^{\frac{1}{2}}$

= $(\frac{(x)^2-(ip)^2}{4})^{\frac{1}{2}}=(\frac{(x)^2+(p)^2}{4})^{\frac{1}{2}}$

therefore, $aa^+ = a^+a$ which implies the operators a and a+ are commutators.

7 0

2 years ago

Karl bought ? 3 1/2 ? gallons of milk that cost $2.70 per gallon. How much did he spend? Please I need it super fast I will mark

lisov135 [29]

He spent $9.45
3.5x2.7= 9.45

6 0

2 years ago

For a carnival game, containers are arranged in a triangular display. The top row has 1 container. The second row has 2 containe

docker41 [41]

26 are left. Its just addition then subtraction.

4 0

2 years ago

Read 2 more answers

3.8*10 to the 9th power divided by 4*10 to the 2nd power

user100 [1]

Answer:

$9.5\times 10^{6}$

Step-by-step explanation:

1. Divide the coefficients and the exponentials separately

$\dfrac{3.8 \times 10^{9}}{4 \times 10^{2}} = \dfrac{3.8}{4} \times \dfrac{10^{9}}{10^{2}}$

2. Divide the coefficients

$\dfrac{3.8}{4} = 0.95$

3. Divide the exponentials

Subtract the exponent in the denominator from the exponent in the numerator.

$\dfrac{10^{9}}{10^{2}} = 10^{(9 - 2)} = 10 ^{7}$

4. Re-join the new coefficient and the new exponential

$\dfrac{3.8 \times 10^{9}}{4 \times 10^{2}} = 0.95 \times 10^{7}$

5. Put the new number into standard form

The number before the power of 10 must be greater than or equal to one and less than 10.

Multiply the answer by 10/10.

$(0.95 \times 10^{7}) \times \dfrac{10}{10} = (0.95\times10) \times \dfrac{10^{7}}{10} = \mathbf{9.5\times 10^{6}}$

5 0

2 years ago