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Debora [2.8K]
3 years ago
15

a. Determine whether the ratios 10/20 and 30/60 can form a proportion by finding a common multiplier. b. Show that the ratios in

part (a) are equal by writing them in simplest form. Please explain how you got the answers to these questions.
Mathematics
1 answer:
Flura [38]3 years ago
6 0
<span>To simplify one by one, step by step, proving that they are the same:
</span>
\frac{10}{20} \frac{\div2}{\div2}  =  \frac{5}{10}  \frac{\div5}{\div5} =  \boxed{\frac{1}{2} } \end{array}}\qquad\quad\checkmark
\frac{30}{60} \frac{\div2}{\div2} =  \frac{15}{30}  \frac{\div5}{\div5} =  \frac{3}{6}  \frac{\div3}{\div3} =  \boxed{\frac{1}{2} } \end{array}}\qquad\quad\checkmark


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Answer: (-4,-6) is the point that ALMOST satisfies both inequalities. IF they were equalities, this would be the solution.

The question is a bit confusing as it asks for "which points (x,y) satisfies both" It's ungrammatical, and many points (infinite within the shaded region) are solutions that SATISFY the system of inequalities!

Step-by-step explanation: Substitute the x and y-values and see if the inequalities are true.

y>x-2 -6> -4-2 -6= -6

That point (-4,-6) is on the dashed line, so not exactly a true solution; this is a question about inequalities. So y values have to be greater than-6 or x-values less than -4 for a true inequality.

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-6> -6 Again, equal, so for this y-values have to be greater than-6 and/or x-values less than -4 in order to have a true inequality.

If you have the graph to look at, you can select any points in the shaded region that satisfies both of the inequalities.

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