Find the perimeter of the image

Mathematics

1 answer:

goblinko [34]3 years ago

3 0

Answer:

4x³ + 14x² + 8x + 6

Step-by-step explanation:

Perimeter = 2(L + W) = 2(7x² + 4x + 3 + 2x³) = 14x² + 8x + 6 + 4x³

4x³ + 14x² + 8x + 6

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If sin A = 3/8, find the value of cosec A - sec A.

alexira [117]

Answer:

$\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}$

Step by step explanation:

$\text{Given that,}\\\\~~~~~~\sin A = \dfrac 38 \\\\\implies \sin^2 A = \dfrac 9{64}\\\\\implies 1 - \cos^2 A = \dfrac{9}{64}\\\\\implies \cos ^2 A = 1 - \dfrac 9{64}\\\\\implies \cos^2 A = \dfrac{55}{64}\\\\\implies \cos A =\pm\sqrt{\dfrac{55}{64}}\\ \\\implies \cos A = \pm\dfrac{\sqrt{55}}8\\\\$

$\implies \dfrac 1{\cos A} = \pm\dfrac{8}{\sqrt{55}}$

$\text{Now,}\\\\\csc A - \sec A\\\\=\dfrac{1}{\sin A}- \dfrac{1}{\cos A}\\\\=\dfrac 83 -\left(\pm \dfrac 8{\sqrt {55}} \right)\\ \\\text{Hence,}\\\\\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}$