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3 years ago

What is another way to write 42 + 63

Mathematics

2 answers:

satela [25.4K]3 years ago

7 0

You could write this by going 63+42 or add 63 and 42
hope that's what you are looking for ;-)

sleet_krkn [62]3 years ago

6 0

43 + 62 = 105
42 +63 = 105

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Estimate the sum or difference.

dangina [55]

1) $\frac{4}{6}+\frac{1}{8}=\frac{19}{24}$

2) $\frac{2}{6}+\frac{7}{8}=\frac{29}{24}$

3) $\frac{5}{6}-\frac{3}{8}=\frac{11}{24}$

4) $\frac{4}{6}+\frac{3}{8}=\frac{25}{24}$

5) $\frac{7}{8}-\frac{5}{6}=\frac{1}{24}$

6) $\frac{1}{6}+\frac{7}{8}=\frac{25}{24}$

Step-by-step explanation:

In order to calculate the sum of two fractions, we first have to find the lowest common denominator of the two fractions, then multiply the numerator of each fraction by the ratio between the lowest common denominator and the original denominator, and then add/subtract the two new numerators.

$\frac{4}{6}+\frac{1}{8}=$

Here the lowest common denominator between 6 and 8 is 24,

So we have to rewrite each fraction as having denominator 24: this means that we have to multiply both numerator and denominator of the 1st fraction by 4 (because 24/6=4), and both numerator and denominator of the 2nd fraction by 3 (because 24/8=3).

So the new numerators of the two fractions are:

$4\cdot 4 = 16\\1\cdot 3 = 3$

The expression then becomes:

$\frac{4}{6}+\frac{1}{8}=\frac{16}{24}+\frac{3}{24}=\frac{16+3}{24}=\frac{19}{24}$

$\frac{2}{6}+\frac{7}{8}=$

Here the lowest common denominator between 6 and 8 is again 24,

so we have again to multiply both numerator and denominator of the 1st fraction by 4, and both numerator and denominator of the 2nd fraction by 3.

So the new numerators of the two fractions are:

$2\cdot 4 = 8\\7\cdot 3 = 21$

And we get:

$\frac{2}{6}+\frac{7}{8}=\frac{8}{24}+\frac{21}{24}=\frac{8+21}{24}=\frac{29}{24}$

$\frac{5}{6}-\frac{3}{8}$

The denominators are the same, so the lowest common denominator is always 24. So we can adopt the same procedure, and new numerators are:

$5\cdot 4 = 20\\3\cdot 3 = 9$

And so:

$\frac{5}{6}-\frac{3}{8}=\frac{20}{24}-\frac{9}{24}=\frac{20-9}{24}=\frac{11}{24}$

$\frac{4}{6}+\frac{3}{8}=$

Using the same lowest common denominator, 24, the new numerators are:

$4\cdot 4 = 16\\3\cdot 3 = 9$

And so we can rewrite the expression as

$\frac{4}{6}+\frac{3}{8}=\frac{16}{24}+\frac{9}{24}=\frac{16+9}{24}=\frac{25}{24}$

$\frac{7}{8}-\frac{5}{6}=$

Again, the lowest common denominator is 24. This time the denominator of the 1st fraction is 8 while the denominator of the 2nd fraction is 6, so we have to multiply the numerator of the 1st fraction by 3 and the numerator of the 2nd fraction by 4.

We get:

$7\cdot 3 = 21\\5\cdot 4 = 20$

So the expression will be rewritten as:

$\frac{7}{8}-\frac{5}{6}=\frac{21}{24}-\frac{20}{24}=\frac{21-20}{24}=\frac{1}{24}$

$\frac{1}{6}+\frac{7}{8}=$

Here the situation is similar to the first 4 exercises: using 24 as lowest common denominator, the numerators become

$1\cdot 4 = 4\\7\cdot 3 = 21$

So the expression becomes

$\frac{1}{6}+\frac{7}{8}=\frac{4}{24}+\frac{21}{24}=\frac{4+21}{24}=\frac{25}{24}$

Learn more about fractions:

brainly.com/question/605571

brainly.com/question/1312102

#LearnwithBrainly

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Which of the following illustrates the truth value of the given statements? January is the first month of the year or December i

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First, the need to determine if the statements are true or false.

1) January is the first month of the year. (This statement is true)
2) December is the last month of the year. (This statement is also true)

With this in mind we can determine that what will illustrate the truth value would be:

T T -> T

In other words, since the first statement is true and the second statement is also true then conjunction of both statements would be true.

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