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STatiana [176]
2 years ago
9

If y varies directly with x, find the value of y when x = 27 for the direct variation with the given information. y = 7 when x =

-9 y =​
Mathematics
1 answer:
ElenaW [278]2 years ago
3 0

Given:

y varies directly with x,

It is given that y = 7 when x = -9.

To find:

The value of y when x = 27 for the direct variation.

Solution:

It is given that y varies directly with x, so

y\propto x

y=kx             ...(i)

Where, k is the constant of proportionality.

It is given that y = 7 when x = -9. Putting these values in (i), we get

7=k(-9)

-\dfrac{7}{9}=k

Putting k=-\dfrac{7}{9} in (i), we get

y=-\dfrac{7}{9}x

Putting x=27,  we get

y=-\dfrac{7}{9}(27)

y=-21

Therefore, the required value is y=-21.

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12. In the given figure, RS is parallel to PQ, If RS = 3 cm, PQ = 6 cm and ar(∆TRS) = 15cm³, then ar (∆TPQ) = ? (a) 70 cm² (b) 5
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\large\underline{\sf{Solution-}}

Given that,

In <u>triangle TPQ, </u>

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As it is given that, <u>RS || PQ</u>

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⇛∠TRS = ∠TPQ [ Corresponding angles ]

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\rm\implies \: \triangle TPQ \:  \sim \: \triangle TRS \:  \:  \:  \:  \:  \:  \{AA \}

<u>Now, We know </u>

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\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{ar( \triangle \: TRS)}  = \dfrac{ {PQ}^{2} }{ {RS}^{2} }

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15}  = \dfrac{ {6}^{2} }{ {3}^{2} }

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15}  = \dfrac{36 }{9}

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15}  = 4

\rm\implies \:ar( \triangle \: TPQ)  = 60 \:  {cm}^{2}

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