Question

construct a parallelogram ABCD such that AC School to 6 CM BD is equal to 4.6 cm and angle between them is equal to 45 degree​

Answer 1

Answer:

Explanation:

★To Solve :

$\rm \: \blue{13(y+7) = 3(y-1)}$

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★SOLUTION :

$\rm\red\leadsto{13(y+7) = 3(y-1)}$

$\rm\red\leadsto{13\times{y}+13\times7= 3\times{y}-3\times1}$

$\rm\red\leadsto{13y+91=3y-3}$

$\rm\red\leadsto{13y-3y=-3-91}$

$\rm\red\leadsto{10y=-94}$

$\rm\red\leadsto{} y =\dfrac{ { \cancel{-94}}^{ \:-45 } }{ { \cancel{10}}^{ \: 5} }$

$\rm\red\leadsto{y=}\dfrac{-47}{5}$

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Answer 2

Answer:

13(y+7)=3(y−1)

\rm\red\leadsto{13\times{y}+13\times7= 3\times{y}-3\times1}⇝13×y+13×7=3×y−3×1

\rm\red\leadsto{13y+91=3y-3}⇝13y+91=3y−3

\rm\red\leadsto{13y-3y=-3-91}⇝13y−3y=−3−91

\rm\red\leadsto{10y=-94}⇝10y=−94

\rm\red\leadsto{} y =\dfrac{ { \cancel{-94}}^{ \:-45 } }{ { \cancel{10}}^{ \: 5} }⇝y=

10

5

−94

−45

\rm\red\leadsto{y=}\dfrac{-47}{5}⇝y=

5

−47

Step-by-step explanation:

y