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julia-pushkina [17]
3 years ago
13

Solve the following inequality: n + 6 is less than or equal to 3

Mathematics
1 answer:
Rainbow [258]3 years ago
4 0

Answer:

n+6 is less than or equal to 3

Step-by-step explanation:

this answer is n is less than or equal to -3

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Solve for x if log 9 base x + log 3 base x^2 = 2.5​
Y_Kistochka [10]

Not sure if the equation is

\log_9x+\log_3(x^2)=\dfrac52

or

\log_x9+\log_{x^2}3=\dfrac52

  • If it's the first one:

9^{\log_9x+\log_3(x^2)}=9^{\log_9x}\cdot9^{\log_3(x^2)}

9^{\log_9x+\log_3(x^2)}=9^{\log_9x}\cdot(3^2)^{\log_3(x^2)}

9^{\log_9x+\log_3(x^2)}=9^{\log_9x}\cdot3^{2\log_3(x^2)}

9^{\log_9x+\log_3(x^2)}=9^{\log_9x}\cdot3^{\log_3(x^2)^2}

9^{\log_9x+\log_3(x^2)}=9^{\log_9x}\cdot3^{\log_3(x^4)}

9^{\log_9x+\log_3(x^2)}=x\cdot x^4

9^{\log_9x+\log_3(x^2)}=x^5

On the other side of the equation, we'd get

9^{5/2}=(3^2)^{5/2}=3^{2\cdot(5/2)}=3^5

Then

x^5=3^5\implies\boxed{x=3}

  • If it's the second one instead, you can use the same strategy as above:

x^{\log_x9+\log_{x^2}3}=x^{\log_x9}\cdot x^{\log_{x^2}3}

x^{\log_x9+\log_{x^2}3}=x^{\log_x9}\cdot\left((x^2)^{1/2}\right)^{\log_{x^2}3}

(Note that this step assume x>0)

x^{\log_x9+\log_{x^2}3}=x^{\log_x9}\cdot(x^2)^{(1/2)\log_{x^2}3}

x^{\log_x9+\log_{x^2}3}=x^{\log_x9}\cdot(x^2)^{\log_{x^2}\sqrt3}

x^{\log_x9+\log_{x^2}3}=9\sqrt3

Then we get

9\sqrt3=x^{5/2}\implies x=(9\sqrt3)^{2/5}\implies\boxed{x=3}

6 0
3 years ago
Eight rooks are placed randomly on a chess board. What is the probability that none of the rooks can capture any of the other ro
Brums [2.3K]

Answer:

9.11*10^-4 %

Step-by-step explanation:

To find the probability, you simply need to find the possible outcomes that allows no rooks to be in danger, and the possible amount of ways to place the rooks.

For the first outcome, you start by putting 1 rook in the first columns, you have 8 possible rows to do this. The next rook in the next column will only have 7 possible rows, as you have to exclude the one where the previous rook is located. The next rook, 6 possibilities, the next 5, and so on. So we conclude that the total amount of ways so that none of the rooks can capture any of the other rooks is 8*7*6*5*4*3*2*1 = 8! = 40320

In order to find the total amount of ways to place the rooks, you can just use a combinatoric:

\left[\begin{array}{ccc}64\\8\end{array}\right]= \frac{64!}{8!(64-8)!} = 4.43*10^9

Then:

P = \frac{40320}{4.43*10^9}*100\%=9.11*10^{-4}\%

5 0
4 years ago
Which is correct and explain
skad [1K]

Answer:

c

Step-by-step explanation:

7 0
2 years ago
I'll give brainlest if u can answer this question <br><br><br>1+1=?
yKpoI14uk [10]

Answer:

11

Step-by-step explanation:

LOL

3 0
3 years ago
Read 2 more answers
There are 50 students in the 5th grade at Clark School. One day, 20% of them were absent. How many fifth graders were in school
natita [175]
Well, 20% is equivalent to (1/5). So, you'd put 50 in the numerator and 5 in the denominator. (50/5) is equal to 10. Then you'd take that value and subtract it from your original value. [ 50 - 10 = 40 ]. 40 Students were at school that day.

5 0
4 years ago
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