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andriy [413]
2 years ago
7

The following graph shows a proportional relationship.

Mathematics
1 answer:
LenKa [72]2 years ago
4 0

Answer:

2

Step-by-step explanation:

If we look at the graph this is an increasing graph where y increases with respect to x so in other words we can say that y is directly proportional to x.

y\ \alpha\ x\\

and when we remove the proportionality sign we replace it with an equal to sign and a constant. So it becomes ,

y = kx\\

now to find the Constant of proportionality which is k we can use any ordered pair(coordinate/point) from the graph,

Lets take ( 2 , 4) where 2 represents the value of x on the x-axis and 4 represents the value of y on the y-axis so now,

y=kx\\(4)=k(2)\\4=2k\\4/2=k\\k=2\\

so the constant of proportionality is 2

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We know that,

\dag\bf\:sin^2A=\dfrac{1-cos2A}{2}

\dag\bf\:sin2A=2sinA\:cosA

<u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u>

<u>Now, Let's solve</u> !

\leadsto\:\bf\dfrac{sin^2A-sin^2B}{sinA\:cosA-sinB\:cosB}

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\leadsto\:\sf\dfrac{1-cos2A-1+cos2B}{sin2A-sin2B}

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\leadsto\:\sf\dfrac{sin(A+B)}{cos(A+B)}

\leadsto\:\bf{tan(A+B)}

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