The length of a rectanglar field is
[tex] {8x}^{2} + 20x.[/tex]
If the width is twice the length, what is the area of the field?
[tex] {8x}^{2} + 20x.[/tex]
If the width is twice the length, what is the area of the field?
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Answer:
128x⁴+640x³+800x²
Step-by-step explanation:
Length =8x²+20x
Length =8x²+20xBreadth = 2(8x²+20x)
Length =8x²+20xBreadth = 2(8x²+20x) =16x²+40x
Length =8x²+20xBreadth = 2(8x²+20x) =16x²+40xArea of a rectangle = l×b
Length =8x²+20xBreadth = 2(8x²+20x) =16x²+40xArea of a rectangle = l×b =(8x²+20x)(16x²+40x)
Length =8x²+20xBreadth = 2(8x²+20x) =16x²+40xArea of a rectangle = l×b =(8x²+20x)(16x²+40x) =128x⁴+320x³+ 320x³+800x²
Length =8x²+20xBreadth = 2(8x²+20x) =16x²+40xArea of a rectangle = l×b =(8x²+20x)(16x²+40x) =128x⁴+320x³+ 320x³+800x² =128x⁴+640x³+800x²
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