All categories

3 years ago

Which of the following valued of x is not in the domain of f(x) = x+3 / x-7

Mathematics

1 answer:

Rudiy273 years ago

8 0

Answer:

x = 7 is not there in the domain of the function.

Step-by-step explanation:

The given function is f(x) = $\frac{x+3}{x-7}$

We have to find out which all values are not there in the domain of the function.For the function to become not defined the function should have infinite value and this happens when the denominator becomes zero.

In the given function for the denominator to become zero , the value of x must be 7.

For all other values of x , the function is defined.

Hence , the domain of x is all real numbers except 7.

You might be interested in

4/5 gallon of water divided by 4 bottles what fraction of a gallon is in each bottle

Vinvika [58]

Answer:

1/5 gallon of water in each bottle

Step-by-step explain:

4/5 = 0.8

0.8/4 = 0.2 or 1/5 gallons

3 0

3 years ago

A candle factory made 534,449 candles one month. To the nearest ten thousand, about how many candles did the factory make?

olganol [36]

530,000 candles were produced by the factory (rounded of course!) in that month.

Hope this helps!

4 0

3 years ago

Read 2 more answers

Simplify completely the quantity 3 times x to the 4th power plus 5 times x to the 2nd power plus 2 times x all over x. (2 points

BartSMP [9]

3x^3+5x+2 is the answer

5 0

3 years ago

Read 2 more answers

Perform the indicated row operations, then write the new matrix.

Studentka2010 [4]

The matrix is not properly formatted.

However, I'm able to rearrange the question as:

$\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]$

Operations:

$2R_1 + R_2 ->R_2$

$-3R_1 +R_3 ->R_3$

Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.

Answer:

$\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]$

Step-by-step explanation:

The first operation:

$2R_1 + R_2 ->R_2$

This means that the new second row (R2) is derived by:

Multiplying the first row (R1) by 2; add this to the second row

The row 1 elements are:

$\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]$

Multiply by 2

$2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]$

Add to row 2 elements are: $\left[\begin{array}{ccc}-2&3&5|3\end{array}\right]$

$\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]$

$\left[\begin{array}{ccc}0&5&7|1\end{array}\right]$

The second operation:

$-3R_1 +R_3 ->R_3$

This means that the new third row (R3) is derived by:

Multiplying the first row (R1) by -3; add this to the third row

The row 1 elements are:

$\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]$

Multiply by -3

$-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]$

Add to row 2 elements are: $\left[\begin{array}{ccc}3&2&4|1\end{array}\right]$

$\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]$

$\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]$

Hence, the new matrix is:

$\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]$

3 0

3 years ago

Read 2 more answers

I don’t understand this question, what’s the answer?

Goryan [66]

Depending on how far it goes up and over up/run is the m and the y is were it touches the y axis

5 0

3 years ago