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Dovator [93]
2 years ago
9

Round 47.219 to the nearest tenth.​

Mathematics
2 answers:
Aleksandr-060686 [28]2 years ago
8 0
47. 2 because 2 is greater than 1 so it won’t go a number up. Instead it stays the same
saveliy_v [14]2 years ago
7 0

Answer:

47.2

Step-by-step explanation:

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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
2 years ago
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