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3 years ago

E

Mathematics

1 answer:

Troyanec [42]3 years ago

7 0

Answer: The information in that "9" is the atomic number of the atom.

Step-by-step explanation:

For an atom of fluorine we have:

A = atomic weight = 18.99 amu.

Z = atomic number = 9.

(A - Z) = number of netrons = 10.

Then the "9" in the information of the atom of fluorine gives information about the atomic number.

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Solve for X then find the measure of DEF. Write your answers in decimal form.

Fittoniya [83]

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Since EF bisects ∠DEF, we know that EF splits ∠DEF in 2 halves from the word 'bisects', which can be written as 'bi- sects' , where bi means 2 and sects means sections

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4 years ago

Does this diverge or converge?

olga2289 [7]

Answer:

L = 7/6 , and the correct option is B, the series diverges.

Step-by-step explanation:

Here we have that:

$an = \frac{(3n + 3)*7^n}{6^{n+4}}$

Now we want to see the quotient test, then we need to do find the value of the quotient as n tends to infinity. Then we first need to find the quotient:

$a(n + 1)/a(n) = \frac{(3(n + 1) + 3)*7^{n + 1}}{6^{n + 1 + 4}}*\frac{6^{n + 4}}{(3*(n) + 3)*7^{n}}$

We can simplify the right part to get:

$\frac{(3(n + 1) + 3)*7^{n + 1}}{6^{n + 1 + 4}}*\frac{6^{n + 4}}{(3*(n) + 3)*7^{n}}= \frac{(3n + 3 + 3)*7^{n + 1}}{6^{n + 5}}*\frac{6^{n + 4}}{(3*n + 3)*7^{n}}$

$\frac{(3n + 3 + 3)*7^{n + 1}}{6^{n + 5}}*\frac{6^{n + 4}}{(3*n + 3)*7^{n}}= \frac{(3n + 3 + 3)*7}{(3*n + 3)*6}$

Now we want to find the limit of this as n goes to infinity, we can see that both parts will tend to infinity, then we need to use the Lhopital rule and find the limit of the quotient of the first derivatives.

The first derivatives are:

For the denominator: (3*6)

For the numerator: (3*7)

Then we need to find the limit as n goes to infinity to:

$\frac{3*7}{3*6} = \frac{7}{6}$

But this does not depend on the value of n, then the limit is just 6/7.

L = 7/6

And 7/6 > 1.

Then we can conclude that our series diverges.

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3 years ago