If the r-value, or correlation coefficient, of a data set is negative , the coefficient of determination is negative

A second important result is that electrons will fill the lowest energy states available. This would seem to indicate that every

Assoli18 [71]

Answer:

8

Explanation:

1) Principal quantum number, n = 2

n is the principal quantum number and indicates the main energy level.

2) Second quantum number, ℓ

The second quantum number, ℓ, is named, Azimuthal quantum number.

The possible values of ℓ are from 0 to n - 1.

Hence, since n = 2, there are two possible values for ℓ: 0, and 1.

This gives you two shapes for the orbitals: 0 corresponds to "s" orbitals, and 1 corresponds to "p" orbitals.

3) Third quantum number, mℓ

The third quantum number, mℓ, is named magnetic quantum number.

The possible values for mℓ are from - ℓ to + ℓ.

Hence, the poosible values for mℓ when n = 2 are:

for ℓ = 0: mℓ = 0

for ℓ = 1, mℓ = -1, 0, or +1.

4) Fourth quantum number, ms.

This is the spin number and it can be either +1/2 or -1/2.

Therfore the full set of possible states (different quantum number for a given atom) for n = 2 is:

(2, 0, 0 +1/2)
(2, 0, 0, -1/2)
(2, 1, - 1, + 1/2)
(2, 1, -1, -1/2)
(2, 1, 0, +1/2)
(2, 1, 0, -1/2)
(2, 1, 1, +1/2)
(2, 1, 1, -1/2)

That is a total of 8 different possible states, which is the answer for the question.

8 0

3 years ago

Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.

likoan [24]

Answer:

Thus, the two root of the given quadratic equation $x^2+4=6x$ is 5.24 and 0.76 .

Step-by-step explanation:

Consider, the given Quadratic equation, $x^2+4=6x$

This can be written as , $x^2-6x+4=0$

We have to solve using quadratic formula,

For a given quadratic equation $ax^2+bx+c=0$ we can find roots using,

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ ...........(1)

Where, $\sqrt{b^2-4ac}$ is the discriminant.

Here, a = 1 , b = -6 , c = 4

Substitute in (1) , we get,

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\Rightarrow x=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot 1 \cdot (4)}}{2 \cdot 1}$

$\Rightarrow x=\frac{6\pm\sqrt{20}}{2}$

$\Rightarrow x=\frac{6\pm 2\sqrt{5}}{2}$

$\Rightarrow x={3\pm \sqrt{5}}$

$\Rightarrow x_1={3+\sqrt{5}}$ and $\Rightarrow x_2={3-\sqrt{5}}$

We know $\sqrt{5}=2.23607$ (approx)

Substitute, we get,

$\Rightarrow x_1={3+2.23607}$ (approx) and $\Rightarrow x_2={3-2.23607}$ (approx)

$\Rightarrow x_1={5.23607}$ (approx) and $\Rightarrow x_2=0.76393}$ (approx)

Thus, the two root of the given quadratic equation $x^2+4=6x$ is 5.24 and 0.76 .

7 0

3 years ago

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Vertical Lines

noname [10]

Answer:

kk

Step-by-step explanation:

8 0

2 years ago

Use polynomial division to completely factor y=x^3+3x^2-13x-15 by x+5

Salsk061 [2.6K]

Answer:

(x - 3)(x + 1)(x + 5)

Step-by-step explanation:

I'd use synthetic division instead. If we were to find the roots of the given polynomial, we could from them write the factors as well.

The divisor x + 5 corresponds to root x = -5. Setting up synthetic div.,

-5 ) 1 3 -13 -15

-5 10 +15

-----------------------------

1 -2 -3 0

Since the remainder is 0, we know that -5 is a root and (x + 5) is a factor. Moreover, we know that the coefficients of the quotient are 1, -2 and -3.

1x² - 2x - 3 can be factored: the factors are (x - 3) and (x + 1).

So the end result for this problem is (x - 3)(x + 1)(x + 5).

7 0

3 years ago

Help ASAP please!!!

masya89 [10]

Total revenue the previous year would be $46,355

7 0

3 years ago

Read 2 more answers