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Taya2010 [7]

3 years ago

xFormula1" title="show \: that \: |3 + 10| = |3| + |10| " alt="show \: that \: |3 + 10| = |3| + |10| " align="absmiddle" class="latex-formula">.

Mathematics

1 answer:

Tanya [424]3 years ago

3 0

13=13
i hope this help

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Allisa [31]

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$formulae \: to \: be \: used : \\ \cos 2 \theta = 2 { \cos}^{2} \theta - 1 \\ and \\ \cos 2 \theta = 1 - 2 { \cos}^{2} \theta \\ \\ now \: let \: us \: prove \: the \: given \: trigonometrical \: \\ identity. \\ \\ \frac{1 + \cos2A}{1 - \cos2A} = \cot \: A. \cot \: A \\ LHS = \frac{1 + \cos2A}{1 - \cos2A} \\ \\ = \frac{1 +2 { \cos}^{2} A- 1 }{1 - (1 - 2 { \sin}^{2} A)} \\ \\ = \frac{1 +2 { \cos}^{2} A- 1 }{1 - 1 + 2 { \sin}^{2} A} \\ \\ = \frac{2 { \cos}^{2} A }{2 { \sin}^{2} A} \\ \\ = \frac{{ \cos}^{2} A }{ { \sin}^{2} A} \\ \\ = { \cot}^{2} A \\ \\ = { \cot} A .{ \cot} A \\ \\ = RHS \\ \\ hence \: proved.$

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4. Using the formula for the volume of a cylinder, , find h when V = 141.3 and r = 3. Use pi = 3.14.

olya-2409 [2.1K]

$\text{The formula of a volume of a cylinder}\\\\V=\pi r^2h\\\\\text{We have}\ V=141.3\ \text{and}\ r=3.\ \text{Substitute}\\\\\pi(3^2)h=141.3\\\\9\pi h=141.3\qquad|\text{divide both sides by}\ 9\pi\\\\h=\dfrac{141.3}{9\pi}\qquad|\text{use}\ \pi\approx3.14\\\\h=\dfrac{141.3}{9\cdot3.14}\\\\h=\dfrac{141.3}{28.26}\\\\\boxed{h=5}$

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3 years ago