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Lena [83]
4 years ago
12

Find the domain of the function

Mathematics
1 answer:
aev [14]4 years ago
8 0
Domain are those values of x for which function is defined. f(x) is not defined at x = 3. Therefore Domain is

x∈ (⁻∞, +∞) - {3}
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A serviceman charges ₱450 per hour for doing mechanical jobs. How much does he get for 6.75 hours of doing the job?
vredina [299]

Answer:

450×6.75 = 2 956.5

So the serviceman will get 2 956.5 after 6.75 hours of doing the job

7 0
2 years ago
A cube has a length of 8 in. What is the width and the height of the cube?​
Triss [41]

Answer:

8 in

Step-by-step explanation:

In a cube, the length, width, and height are all the same.

7 0
3 years ago
Read 2 more answers
Determine the number of x-intercepts that appear on a graph of each function.f (x) = (x - 6)2(x + 2)2
Orlov [11]
Short answer There are 2 places where the graph touches the x axis just as you said.
I take this to mean f(x) = (x - 6)^2*(x + 2)^2

 
There are two places where the graph touches the x axis but that is not an intercept. 
When a graph just touches the x axis, that counts as a root. As the graph shows, these 2 places are
x = 6 and
x = - 2 is where they touch.

If you have a graphing calculator, you can see what this looks like for yourself. The window settings are 
x min = - 3
x max = 7
y min = 0
y max = 250. 

There is a local maximum between x = -2 and x = 6.
So the answer, once again is at x = -2 and x = 6 and that makes 2 roots for the equation.














7 0
4 years ago
Read 2 more answers
10% of 5000 is...... <br> And 5% of 5000 is.....
stealth61 [152]

Step-by-step explanation:

• 10% of 5000

= 10\% \times 5000 \\  =  \frac{10}{100}  \times 5000 \\  = 10 \times 50 \\  = 500

So, 10% of 5000 is 500

• 5% of 5000

= 5\% \times 5000 \\  =  \frac{5}{100}  \times 5000 \\  = 5 \times 50 \\  = 250

So, 5% of 5000 is 250

5 0
3 years ago
You buy lunch for gas at a party. You can spend no more than $100. You will spend $20 on beverages and $10 per guest and sandwic
azamat

70$ used

i hop dis helps

3 0
3 years ago
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