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Flauer [41]
3 years ago
6

Help a girl out plsss.?

Mathematics
2 answers:
musickatia [10]3 years ago
8 0
The answer is the last one madam :)
Ludmilka [50]3 years ago
4 0
The last one and the first one 
hope you got it 

You might be interested in
What is the solution of 1/2 x= 5/6
Studentka2010 [4]

Answer:

5/3

Step-by-step explanation:

1/2 x = 5/6

Multiply by 2 on each side

(1/2 * 2)x = (5/6)2

x = 5/3

8 0
3 years ago
Read 2 more answers
Plz factor this completely
MrMuchimi
What am I supposed to be factoring?
3 0
3 years ago
Read 2 more answers
(110×110×110×110)2=?
sveticcg [70]

Answer:

2.9282 × 10^8

Step-by-step explanation:

(110×110×110×110)2

(146410000)2

292820000

292820000  can be simplified to 2.9282 × 10^8.

Hope this helps.

7 0
3 years ago
A house is worth $156,000
nordsb [41]

25 bahahahahahahahaha

5 0
3 years ago
HELLOOOO HELP PLEASE
MA_775_DIABLO [31]

Answer:

2*log(x)+log(y)

Step-by-step explanation:

So, there are two logarithmic identities you're going to need to know.

<em>Logarithm of a power</em>:

   log_ba^c=c*log_ba

   So to provide a quick proof and intuition as to why this works, let's consider the following logarithm: log_ba=x\implies b^x=a

   Now if we raise both sides to the power of c, we get the following equation: (b^x)^c=a^c

   Using the exponential identity: (x^a)^c=x^{a*c}

    We get the equation: b^{xc}=a^c

    If we convert this back into logarithmic form we get: log_ba^c=x*c

    Since x was the basic logarithm we started with, we substitute it back in, to get the equation: log_ba^c=c*log_ba

Now the second logarithmic property you need to know is

<em>The Logarithm of a Product</em>:

    log_b{ac}=log_ba+log_bc

    Now for a quick proof, let's just say: x=log_ba\text{ and }y=log_bc

    Now rewriting them both in exponential form, we get the equations:

    b^x=a\\b^y=c

    We can multiply a * c, and since b^x = a, and b^y = c, we can substitute that in for a * c, to get the following equation:

    b^x*b^y=a*c

   Using the exponential identity: x^{a}*x^b=x^{a+b}, we can rewrite the equation as:

 

   b^{x+y}=ac

   taking the logarithm of both sides, we get:

   log_bac=x+y

   Since x and y are just the logarithms we started with, we can substitute them back in to get: log_bac=log_ba+log_bc

Now let's use these identities to rewrite the equation you gave

log(x^2y)

As you can see, this is a log of products, so we can separate it into two logarithms (with the same base)

log(x^2)+log(y)

Now using the logarithm of a power to rewrite the log(x^2) we get:

2*log(x)+log(y)

3 0
1 year ago
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