I need some help pls :'(

hi, i dont undertand number 20 because i was absent in class today and i rerally need help, i will appraciate with the help, and

Mariulka [41]

Given:

The equation is,

$2\log _3x-\log _3(x-2)=2$

Explanation:

Simplify the equation by using logarthimic property.

$\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{ \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}$

Simplify further.

$\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}$

Solve the quadratic equation for x.

$\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}$

From the above equation (x - 6) = 0 or (x - 3) = 0.

For (x - 6) = 0,

$\begin{gathered} x-6=0 \\ x=6 \end{gathered}$

For (x - 3) = 0,

$\begin{gathered} x-3=0 \\ x=3 \end{gathered}$

The values of x from solving the equations are x = 3 and x = 6.

Substitute the values of x in the equation to check answers are valid or not.

For x = 3,

$\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}$

Equation satisfy for x = 3. So x = 3 is valid value of x.

For x = 6,

$\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}$

Equation satifies for x = 6.

Thus values of x for equation are x = 3 and x = 6.

6 0

10 months ago

An English teacher needs to pick 9 books to put on his reading list for the next school year, and he needs to plan the order in

Nikitich [7]

Answer: $4.37\times10^3$

Step-by-step explanation:

Given : Number of choices for novels = 4

Number of choices for plays = 6

Number of choices for poetry books = 5

Number of choices for nonfiction books = 5

Total books =4+6+5+5=20

If he wants to include all 4 novels, then the number of books left to select = 9-4=5

Remaining choices for books = 20-4=16

Number of combinations of n things taking r at a time : $\dfrac{n!}{r!(n-r)!}$

Then, the number of different reading schedules are possible :_

$^4C_4\times^{16}C_5\\\\=\dfrac{4!}{4!(4-4)!}\times\dfrac{16!}{5!(16-5)!}\\\\=(1)\times\dfrac{16\times15\times14\times13\times12\times11!}{120\times11!}\\\\=4368=4.368\times10^3\approx4.37\times10^3$

Hence, the required answer is $4.37\times10^3$ .

8 0

2 years ago

I need to know the answers to this problem

FinnZ [79.3K]

The amplitude of this function is 4/5. Be sure you understand why that is so.

7 0

3 years ago

A customer purchased a total of 23 hamburgers and chicken sandwiches. Each hamburger costs $3.50 and each chicken sandwich costs

Bezzdna [24]

Answer:

25 2/7

Step-by-step explanation:

88.50/3.50 divide 88.50 by 3:50

6 0

2 years ago

Simplify: (d - 3)(d + 9)

podryga [215]

<span>Simplify: (d - 3)(d + 9)

A) d^2 + 6
B) d^2 - 27
C) d^2 - 12d - 27
D) d^2 + 6d - 27</span>

8 0

3 years ago