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WARRIOR [948]
3 years ago
5

What is 0.84 as a fraction in the lowest term​

Mathematics
2 answers:
Zarrin [17]3 years ago
5 0

Answer:

21/25

Step-by-step explanation:

jok3333 [9.3K]3 years ago
4 0

Answer:

well to start off its highest form is 84/100

Step-by-step explanation:

21/25 is the result of me trying to divide both top and bottom by numbers 1-10

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A highway noise barrier is 120 m long is constructed in 2 pieces. One piece is 45 m longer than the other one. Find the length o
bulgar [2K]

Answer:

37.5 and 82.5

Step-by-step explanation:

7 0
3 years ago
Circle any 2 factors of 400
Alexxandr [17]

A factor is a group of 2 or more numbers that can be multiplied to make a greater number.

The factors for 400 include:

1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80, 100, 200

When paired together:

(1, 400), (2, 200), (4, 100), (5, 80), (8, 50), (10,40), (16, 25), (20,20)

Prime factors:

1, 2, 5

6 0
3 years ago
Read 2 more answers
Taly's leading cultural center during Renaissance; important for trade and commerce;dominated by Medici's
ale4655 [162]

Answer:

Your question it is your answer

6 0
3 years ago
Please help me, I don’t know how to solve for x on this one
Brums [2.3K]

Use trigonometry to solve for x.

The tangent function is defined to be opposite side divided by the adjacent side.

tan(78) = x/10

tan(78)(10) = x

47.0463010948 = x

We now round off to two decimal places. In other words, we round off to the nearest tenths.

47.05 = x

Done!

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
2 years ago
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