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3 years ago

The length of a rectangle is 14.5 in and the width is 3.2 in. What is its areA

Mathematics

1 answer:

nexus9112 [7]3 years ago

8 0

Answer:

46.4 in ^2

Step-by-step explanation:

We know the area of a rectangle is found by

A = l*w where l is the length and w is the width

A = 14.5 * 3.2

A = 46.4 in ^2

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<h3>Solution:</h3>

$\rightarrow\sf((p(x) {x}^{2} + 2x + 4)( {x}^{2} + 2x + 4)) \\ = \sf {ax}^{2} + bx + c \\ = \sf(p(x) {x}^{2} {x}^{2} + p(x) {x}^{2} (2x) + p(x) {x}^{2} \times 4 + 2x \times {x}^{2} + 2x(2x) + 2x \times 4 + {4x}^{2} + 4(2x) + 4 \times 4) \\ = \sf( {x}^{4} p(x) + {2x}^{3} p(x) + {4x}^{2} p(x) + {2x}^{3} + {8x}^{2} + 16x + 16) \\ \rightarrow \large\boxed{\sf\red{{{x}^{4} p(x) + {2x}^{3} p(x) + {4x}^{2} p(x) + {2x}^{3} + {8x}^{2} + 16x + 16}}}$

<h3>Answer:</h3>

$\large\boxed{\sf{\red{{x}^{4} p(x) + {2x}^{3} p(x) + {4x}^{2} p(x) + {2x}^{3} + {8x}^{2} + 16x + 16}}}$

$\color{red}{==========================}$

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