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ivolga24 [154]
3 years ago
13

Which table shows a proportional relationship between x and y?

Mathematics
1 answer:
Semenov [28]3 years ago
3 0

Answer:

B

Step-by-step explanation:

A proportional relationship is a relationship which crosses through the origin (0,0) and which has a proportional constant. We can determine this either by finding (0,0) where x=0 and y=0 in the table or by dividing y/x. None of the tables contain (0,0) so we will divide y by x. We are looking for a table which when each y is divided by its x we have the same constant appearing.

<u>Table A</u>

\frac{12}{3} \neq \frac{15}{6} \neq \frac{18}{8} \neq \frac{20}{10}

These fractions are not equal. This is not proportional.

<u>Table B</u>

\frac{0.5}{1}=\frac{1}{2}  =\frac{3.5}{7} =\frac{4}{8}

These fractions are equal and each shows the numerator to be half of the denominator. This is proportional.

<u>Table C</u>

\frac{1}{3}\neq \frac{2.5}{7.5} \neq \frac{4}{15} \neq \frac{6}{20}

These fractions are not equal. This is not proportional.

<u>Table D</u>

\frac{7}{2} \neq \frac{9}{3}\neq  \frac{11}{4} \neq \frac{13}{5}

These fractions are not equal. This is not proportional.

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<u><em></em></u>sin(A) = \frac{opposite side}{Hypotenuse} = \frac{a}{c} = \frac{4}{5}<u><em></em></u>

<u><em></em></u>cos(A) = \frac{Adjacent side}{Hypotenuse} = \frac{b}{c} = \frac{3}{5}<u><em></em></u>

<u><em></em></u>tan(A) = \frac{opposite side}{Adjacent side} = \frac{a}{b} = \frac{4}{3}<u><em></em></u>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given  b = 3 and hypotenuse c = 5

Given ΔABC  is a right angle triangle

By using pythagoras theorem

        c² = a² + b²

  ⇒ a² = c² - b²

 ⇒  a² = 5²-3²

          =25 - 9

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<u><em>Step(ii):-</em></u>

<u><em></em></u>sin(A) = \frac{opposite side}{Hypotenuse} = \frac{a}{c} = \frac{4}{5}<u><em></em></u>

<u><em></em></u>cos(A) = \frac{Adjacent side}{Hypotenuse} = \frac{b}{c} = \frac{3}{5}<u><em></em></u>

<u><em></em></u>tan(A) = \frac{opposite side}{Adjacent side} = \frac{a}{b} = \frac{4}{3}<u><em></em></u>

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