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Sauron [17]
3 years ago
8

3/5 + 1/2 and can you please explain how to do it.

Mathematics
2 answers:
Anna11 [10]3 years ago
5 0
The simple way is cross multiplication
I will attach a paper you check it
brainliest pls

raketka [301]3 years ago
3 0
When adding fractions, you have to make sure the bottom numbers are the same. Otherwise, you can't add them.

So currently, the bottom number for the first fraction is 5. The bottom number for the second fraction is 2.

As you can see, 5 is not the same as 2, so we need to find one number to make sure the bottom number is the same for both the fractions.

You can multiply the the first fraction by 2, and the second fraction by 5 to give you two fractions that have the same denominator (bottom number.)

So, 3/5 multiplied by 2 is the same as multiplying 3 by 2, and 5 by 2. That gives you 6/10.

1/2 multiplied by 5 is 1 multiplied by 5, and 2 multiplied by 5, which is 5/10.

Now that you have your converted fractions, you can add them and forget about the bottom number, as it's the same. When adding fractions, if the bottom numbers are equal, you leave it as it is, so the 10 stays as 10:

6/10 + 5/10 = 11/10 

Notice how I added 6 and 5 to give me 11, but I don't add 10 + 10, because I don't need to.

So the answer is 11/10.
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Consider the contrapositive of the statement you want to prove.

The contrapositive of the logical statement

<em>p</em> ⇒ <em>q</em>

is

¬<em>q</em> ⇒ ¬<em>p</em>

In this case, the contrapositive claims that

"If there are no scalars <em>α</em> and <em>β</em> such that <em>c</em> = <em>α</em><em>a</em> + <em>β</em><em>b</em>, then <em>a₁b₂</em> - <em>a₂b₁</em> = 0."

The first equation is captured by a system of linear equations,

\begin{cases}c_1 = \alpha a_1 + \beta b_1\\ c_2 = \alpha a_2 + \beta b_2\end{cases}

or in matrix form,

\begin{pmatrix}c_1\\c_2\end{pmatrix} = \begin{pmatrix}a_1&b_1\\a_2&b_2\end{pmatrix}\begin{pmatrix}\alpha\\\beta\end{pmatrix}

If this system has no solution, then the coefficient matrix on the right side must be singular and its determinant would be

\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix} = a_1b_2-a_2b_1 = 0

and this is what we wanted to prove. QED

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