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3 years ago

Find the area of a rectangle with a length of (8m³)² and a width of (4x²m⁴)

Mathematics

1 answer:

Alisiya [41]3 years ago

7 0

Answer:nhnc

add it mc

Step-by-step explanation:vgnhc

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${ \tt{ {(hypotenuse)}^{2} = {(adjacent)}^{2} + {(opposite)}^{2} }} \\ { \tt{ {(7 + 3)}^{2} = \{( {x}^{2} + 9) \} {}^{2} + {y}^{2} }} \\ { \tt{100 = ( {x}^{4} + 18 {x}^{2} + 81) + {y}^{2} }} \\ { \tt{ {y}^{2} = 100 - {x}^{4} - {18x}^{2} - 81 - - - - (a) }}$

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${ \tt{ {y}^{2} = {7}^{2} + {x}^{2} }} \\ { \tt{ {y}^{2} - = 49 + {x}^{2} - - - - (b) }}$

Equate (a) and (b):

${ \tt{100 - {x}^{4} - 18 {x}^{2} - 81 = 49 + {x}^{2} }} \\ { \tt{100 - 49 - 81 = {x}^{4} + 19 {x}^{2} }} \\ { \tt{ - 30 = {x}^{2}( {x}^{2} + 19x)}} \\ { \boxed{ \tt{for \: \: {x}^{2} = - 30...imaginary \: value }}} \\ {\boxed{ \tt{for \: \: ( {x}^{2} + 19x ) \: = - 30}}} \\ { \tt{ {x}^{2} + 19x + 30 = 0 }} \\ { \tt{}}$

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1 year ago

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