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Readme [11.4K]
2 years ago
9

Find the domain for each expression. (1/x)+5 1/x X+5/x+5 X^2/x 7/(x-2)

Mathematics
1 answer:
wel2 years ago
6 0

Using it's concept, it is found that the domain for the expressions is, respectively, given by:

x \neq 0, x \neq 0, (-\infty, \infty), (-\infty, \infty), x \neq 2

<h3>What is the domain of a function?</h3>

It is the <u>set that contains all possible input values</u>.

In a fraction, the denominator cannot be zero, hence:

  • The domain of the first two expressions is of x \neq 0.
  • The domain of the last expression is of x \neq 2.

The third expression can be simplified, as:

(x + 5)/(x + 5) = 1.

The same is true for the fourth, as:

x²/x = 1.

Neither has any restriction, hence their domain is all real numbers, represented by (-\infty, \infty).

More can be learned about the domain of a function at brainly.com/question/25897115

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I need help with this problem please if you can help I would really appreciate it and the problem is number 10
aev [14]

Answer:

g(-5) = -127

Step-by-step explanation:

Plug in -5 for x:

g(-5) = (-5)^3 - 2

= -125 - 2

= -127

4 0
3 years ago
The table shows the relationship between r and s, where s is the independent variable. Which equation represents the relationshi
Harrizon [31]

Answer:

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Step-by-step explanation:

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6 0
3 years ago
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Alcona has a radius of 5 inches and a height of 5 inches what is the volume up and come to the nearest 10 inch cube :-)
Nookie1986 [14]
Pi * 5 squared = 25 Pi
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4 0
3 years ago
Recall that m(t) = (1/2)^t/h for radioactive decay, where h is the half-life. Suppose that a 500g sample of phosphorus-32 decays
katrin2010 [14]

The question is incomplete, here is the complete question:

Recall that m(t) = m.(1/2)^t/h for radioactive decay, where h is the half-life. Suppose that a 500 g sample of phosphorus-32 decays to 356 g over 7 days. Calculate the half life of the sample.

<u>Answer:</u> The half life of the sample of phosphorus-32 is 14.28days^{-1}

<u>Step-by-step explanation:</u>

The equation used to calculate the half life of the sample is given as:

m(t)=m_o(1/2)^{t/h}

where,

m(t) =  amount of sample after time 't' = 356 g

m_o = initial amount of the sample = 500 g

t = time period = 7 days

h = half life of the sample = ?

Putting values in above equation, we get:

356=500\times (\frac{1}{2})^{7/h}\\\\h=14.28days^{-1}

Hence, the half life of the sample of phosphorus-32 is 14.28days^{-1}

7 0
4 years ago
HELP FASTTTT<br> explain how to determine the end behavior of a polynomial
Crazy boy [7]

Answer:

Step-by-step explanation:

Given  : Statement.

To find : explain how to determine the end behavior of a polynomial.

Explanation :

End behavior of the polynomial is depending on the leading coefficient and the degree of the polynomial.

If the leading coefficient is positive and degree is even then the both end will be up .

If the leading coefficient is negative and degree is even then the both end will be down .

If the leading coefficient is positive and degree is odd then the right end will be up and left end will be down .

If the leading coefficient is negative  and degree is odd then the right end will be down and left end will be up .

5 0
3 years ago
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