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Volgvan
3 years ago
14

100 POINTS !!!! Let f(x)=√3x and g(x)=x-6 What's the smallest number that is in the domain of fog?

Mathematics
1 answer:
Gennadij [26K]3 years ago
7 0

Answer:

6

Step-by-step explanation:

fog(x) = √3(x-6)

Domain => 3(x-6) ≥ 0

<=> x-6≥ 0

=> x ≥ 6

so, the smallest number is 6

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Each year for 4 years, a farmer increased the number of trees in a certain orchard by of the number of trees in the orchard the
Neko [114]

Answer:

The number of trees at the begging of the 4-year period was 2560.

Step-by-step explanation:

Let’s say that x is number of trees at the begging of the first year, we know that for four years the number of trees were incised by 1/4 of the number of trees of the preceding year, so at the end of the first year the number of trees wasx+\frac{1}{4} x=\frac{5}{4} x, and for the next three years we have that

                             Start                                          End

Second year     \frac{5}{4}x --------------   \frac{5}{4}x+\frac{1}{4}(\frac{5}{4}x) =\frac{5}{4}x+ \frac{5}{16}x=\frac{25}{16}x=(\frac{5}{4} )^{2}x

Third year    (\frac{5}{4} )^{2}x-------------(\frac{5}{4})^{2}x+\frac{1}{4}((\frac{5}{4})^{2}x) =(\frac{5}{4})^{2}x+\frac{5^{2} }{4^{3} } x=(\frac{5}{4})^{3}x

Fourth year (\frac{5}{4})^{3}x--------------(\frac{5}{4})^{3}x+\frac{1}{4}((\frac{5}{4})^{3}x) =(\frac{5}{4})^{3}x+\frac{5^{3} }{4^{4} } x=(\frac{5}{4})^{4}x.

So  the formula to calculate the number of trees in the fourth year  is  

(\frac{5}{4} )^{4} x, we know that all of the trees thrived and there were 6250 at the end of 4 year period, then  

6250=(\frac{5}{4} )^{4}x⇒x=\frac{6250*4^{4} }{5^{4} }= \frac{10*5^{4}*4^{4} }{5^{4} }=2560.

Therefore the number of trees at the begging of the 4-year period was 2560.  

7 0
3 years ago
The Knicks and Nets have a total 30 players.
Blababa [14]

Answer:

Step-by-step explanation:

From the problem statement, we can set up the following two equations:

K + N = 30

20K + 10N = 500

where K is the number of Knicks players, and N is the number of Nets players.

We can substitute the first equation into the second and solve for K

K + N = 30

N = 30 - K

20K + 10N = 500

20K + 10(30 - K) = 500

20K + 300 - 10K = 500

10K + 300 = 500

10K = 200

K = 20

5 0
3 years ago
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