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3 years ago

Compare4/12 and 4/6 please help

Mathematics

2 answers:

Pachacha [2.7K]3 years ago

8 0

Compare them by making them like terms.
4/6 x2/2=8/12
4/6=8/12 so when compared to 4/12,
4/6 > 4-12

Leni [432]3 years ago

8 0

To compare, you must make the denominator common. In this case you find the great common factor which is 12. So you make 4/6 denominator equal to 12. So what can 6 times to, to equal to 12? (6x_=12)? 2.
So you times the denominator by 2 and the numerator by 2. And you’ll get: 8/12.
Now compare:
4/12 < 8/12
To make it easy to understand:
(4 slices out of 12 is less than 8 slices out of 12).

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Answer:

$3584 \div 28 \\ \\ = 128$

our post which explains the division of three thousand, five hundred and eighty-four by twenty-eight to you.

The number 3584 is called the numerator or dividend, and the number 28 is called the denominator or divisor.

The quotient of 3584 and 28, the ratio of 3584 and 28, as well as the fraction of 3584 and 28 all mean (almost) the same:

3584 divided by 28, often written as 3584/28.

Read on to find the result of 3584 divided by 28 in decimal notation, along with its properties

Here we provide you with the result of the division with remainder, also known as Euclidean division, including the terms in a nutshell:

The quotient and remainder of 3584 divided by 28 = 128 R 0

The quotient (integer division) of 3584/28 equals 128; the remainder (“left over”) is 0.

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HELP WILL GIV YOU 30 POINTS BUT HELP ME

MrRa [10]

Answer:

3/4

Step-by-step explanation:

Constant of Proportionality is just a fancy way of saying slope (for a line passing through (0, 0), at least).

You can find it by looking at y/x, which will be equal at any point on the line (which is why it is a "constant").

So, for multiple points on the line:

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Step-by-step explanation:

The value of $i$ is $i=\sqrt{-1}[tex] or [tex]i^{2}=-1[\tex].Now [tex]i^{4}$ in term of $i^{2}[\tex] can be written as, [tex]i^{4}=i^{2}\times i^{2}$

Substituting the value,

$i^{4}=\left(-1\right)\times \left(-1\right)$

Product of two negative numbers is always positive.

$\therefore i^{4}=1$

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Substituting the value,

$i^{3}=\left(-1\right)\times i$

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Now $\sqrt{-81}$ can be written as follows,

$\sqrt{-81}=\sqrt{\left(81\right)\times\left(-1\right)}$

Applying radical multiplication rule,

$\sqrt{ab}={\sqrt{a}}\sqrt{b}$

$\sqrt{\left(81\right)\times\left(-1\right)}={\sqrt{81}}\sqrt{-1}$

Now, $\sqrt{\left(81\right)=9$ and $\sqrt{-1}}=i$

$\therefore \sqrt{\left(81\right)\times\left(-1\right)}=9i$

Now substituting the above values in given expression,

$2i^4-5i^3+3i^2+\sqrt{-81}=2\left(1\right)-5\left(-i\right)+3\left(-1\right)+9i$

Simplifying,

$2+5i-3+9i$

Collecting similar terms,

$2-3+5i+9i$

Combining similar terms,

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The above expression is in the form of a+bi which is the required expression.

Hence, option number 4 is correct.

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