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2 years ago

6

Point A is located at (-3, 9) and point B is located at (12,-10). Find the distance from point A to point B rounded to the neare

st tenth

1 answer:

Nezavi [6.7K]2 years ago

7 0

Answer:

20

Step-by-step explanation:

$\sqrt{(( - 10) - 9)^{2} + (12 - ( - 3)) ^{2} }$

$\sqrt{( - 19)^{2} + (15)^{2} }$

$\sqrt{361 + 225}$

$\sqrt{586}$

$24.2074$

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To make this a little clearer, let's give the pairs of inequalities the same denominator:

<span>Question 1:
</span> $\frac{11}{15}$ ? $\frac{5}{7}$
First, apply the common denominator to the first fraction:
$(\frac{11}{15})7 \\ \frac{11}{15} * \frac{7}{7} \\ \frac{11*7}{15*7} \\ \frac{77}{105}$
Do the same for the second: $15( \frac{5}{7}) \\ \frac{5}{7}* \frac{15}{15} \\ \frac{5*15}{7*15} \\ \frac{75}{105}$
Nest, compare the two fractions:
$\frac{77}{105} \ \textgreater \ \frac{75}{105}$
Therefore:
$\frac{11}{15}$ > $\frac{5}{7}$
<span>
Question Two:</span>
$\frac{5}{9}$ ? $\frac{7}{13}$
Apply the common denominator to fraction one:
$13( \frac{5}{9}) \\ \frac{5}{9} * \frac{13}{13} \\ \frac{5*13}{9*13} \\ \frac{65}{117}$
Fraction two:
$9(\frac{7}{13}) \\ \frac{7}{13} * \frac{9}{9} \\ \frac{7*9}{13*9} \\ \frac{63}{117}$
Evaluate:
$\frac{65}{117}$ > $\frac{63}{117}$
Therefore:
<span> $\frac{5}{9}$ > $\frac{7}{13}$
</span>
Hope this helps!

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