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mr Goodwill [35]
4 years ago
9

Put the following fractions in ascending order 5/12 1/3 3/4

Mathematics
1 answer:
mel-nik [20]4 years ago
5 0

Answer:

1/3,5/12,3/4

Step-by-step explanation:

Ascending order means from the smallest to the biggest. firstly you find the least common multiple of the denominators of the fractions

the l.c.m of 12,3,and 4= 12

you then divide 12 from the denominators the answer you get multiply it with the numerators

i.e 5/12÷12=1,1×5=5

1/3 ÷ 12=4, 4×1= 4

3/4 ÷ 12=3,3×3=9

so you use the solutions of the above to determine the order of magnitude

i.e 1/3, 5/12,3/4

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Answer:

The answer in the attached figure

Step-by-step explanation:

Let

x------> the abscissa

y-----> the ordinate

we know that

y=2x -----> given problem

The domain is [-1, 0, 1]

so

For x=-1

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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)

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y = 2x squared + 5x -3 can be written in the form y= 2( x+a) squared + b . Find the value of a and the value of b.
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y = 2x² + 5x  - 3

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y = 2(x²  +  \frac{5}{2}x + ( \frac{5}{4})²) - 3

- 2( \frac{5}{4})²

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Re-write to match signs of standard general form:

y = 2(x - ( -  \frac{5}{4}))² \:  + ( - 6 \frac{1}{8})

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