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Alja [10]
3 years ago
13

Please help me with this!..

Mathematics
1 answer:
ch4aika [34]3 years ago
3 0

Answer:

I think it would be 12

Step-by-step explanation:

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Please help! <br> How do you know if the domain or the range of a graph are all real numbers?
RideAnS [48]

Answer:

Many reasons

Step-by-step explanation:

Is there an asymptote?

Is there a whole?

Is it a vertical or horizontal line?

What's the specific function?

4 0
3 years ago
Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A
iVinArrow [24]

Answer:

The answer is below

Step-by-step explanation:

We need to prove that:

(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.

Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.

Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

\frac{\sqrt{secA-1} }{\sqrt{secA+1} } +\frac{\sqrt{secA+1} }{\sqrt{secA-1} } =\frac{(\sqrt{secA-1)}(\sqrt{secA-1})+(\sqrt{secA+1)}(\sqrt{secA+1}) }{(\sqrt{secA+1})(\sqrt{secA-1}) } \\\\=\frac{secA-1+(secA+1)}{\sqrt{sec^2A-secA+secA-1} } \\\\=\frac{2secA}{\sqrt{sec^2A-1} } \\\\=\frac{2secA}{\sqrt{tan^2A} } \\\\=\frac{2secA}{tanA} \\\\=\frac{2*\frac{1}{cosA} }{\frac{sinA}{cosA} }\\\\= 2*\frac{1}{cosA}*\frac{cosA}{sinA}\\\\=2*\frac{1}{sinAA}\\\\=2cosecA

7 0
3 years ago
Isotope mass (amu) abundance (%) 1 203.97304 1.390 2 205.97447 24.11 3 206.97590 22.09 4 207.97665 52.41 find the atomic mass of
Nata [24]
Answer:
average atomic mass = 207.2172085 amu

Explanation:
To get the average atomic mass of an element using the abundance of its isotopes, all you have to do is multiply each isotope by its percentage of abundance and then sum up all the products.

For the question, we have:
203.97304 amu has an abundance of <span>1.390% (0.0139)
</span>205.97447 amu has an abundance of 24.11% (0.2411)
206.97590 amu has an abundance of 22.09% (0.2209)
207.97665 amu has an abundance of 52.41 % (0.5241)

The average atomic mass can be calculated as follows:
average atomic mass = 203.97304(0.0139) + 205.97447(0.2411)
                                     + 206.97590(0.2209) + 207.97665(0.5241)
average atomic mass = 207.2172085 amu

Hope this helps :)
7 0
3 years ago
Find the missing number of each unit rate.
UNO [17]
15/5 = 3/1 24/6 = 4/1
6 0
3 years ago
Read 2 more answers
As of April 2006, roughly 50 million web domain names were registered (e.g., ). How many domain names consisting of just two let
Furkat [3]

Answer:

a) 676, 1296

b) 17576, 46656

c) 456976, 1679616

Step-by-step explanation:

firstly we know that there are 26 letters in the alphabet.

there are 26 ways of choosing the first character and 26 ways choosing the second character,

Therefore, the possible number of domains with two leeter is;

n = 26²

= 26 × 26

= 676

now when the digits are allowed, there are 26 + 10 = 36 characters to be used in total

therefore the possible number of domains with two characters is;

n = 36²

= 36 × 36

= 1296

b)

there are 26 ways of choosing each character

therefore, the possible number of domains with three letters is

n = 26³

= 26 × 26 × 26

= 17576

when the digits are allowed, there are 26 + 10 = 36 characters to be used in total

therefore , the possible number of domains with three characters is;

n = 36³

= 36 × 36 × 36

= 46656

c)

there are 26 ways of choosing each character, therefore the possible number of domains with four letters is;

n = 26⁴

n = 26 × 26 × 26 × 26

n = 456976

when the digits are allowed, there are 26 + 10 = 36 characters to be used in total

therefore, the possible number of domains with four characters is;

n = 36⁴

n = 36 × 36 × 36 × 36

= 1679616

7 0
3 years ago
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