Find the volume if the trapezoidal prism

Ответ:

kingofmortals2119

21.04.2020

Step-by-step explanation:

Formula for Volume of a Trapezoidal Prism. If the prism length is L,trapezoid base width B, trapezoid top width A, and trapezoid height H, then the volume of the prism is given by the four-variable formula: V(L, B, A, H) = LH(A + B)/2. In other words, multiply together the length, height, and average of A and B.

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Ответ:

curiouscadencee

16.11.2021

Mathematics

$\dfrac{\sin y}{1 + \cos y} + \dfrac{1 + \cos y}{\sin y} = 2\csc y$ for the left-hand side: $\text{lhs} = \dfrac{\sin y}{1 + \cos y} + \dfrac{1 + \cos y}{\sin y}$ we can try for a common denominator of $\sin y(1 + \cos y)$ multiply the numerator and denominator of $\frac{\sin y}{1 + \cos y}$ by $\sin y$ $\text{lhs} = \dfrac{\sin^2 y}{\sin y(1 + \cos y)} + \dfrac{1 + \cos y}{\sin y}$ multiply the numerator and denominator of $\tfrac{1 + \cos y}{\sin y}$ by $(1 + \cos y)$ $\text{lhs} = \dfrac{\sin^2 y}{\sin y(1 + \cos y)} + \dfrac{(1 + \cos y)^2}{\sin y(1 + \cos y)}$ combining $\text{lhs} = \dfrac{\sin^2 y + (1 + \cos y)^2}{\sin y(1 + \cos y)}$ expanding: $\begin{aligned} (1 + \cos y)^2 & = (1 + \cos y)(1+\cos y) \\ & = 1(1+\cos y) + \cos y (1+\cos y) \\ & = 1 + \cos y + \cos y + \cos^2 y \\ & = 1 + 2\cos y + \cos^2 y \end{aligned}$ inserting: $\begin{aligned} \text{lhs} & = \dfrac{\sin^2 y + (1 + \cos y)^2}{\sin y(1 + \cos y)} \\ & = \dfrac{\sin^2 y + 1 + 2\cos y + \cos^2 y}{\sin y(1 + \cos y)} \end{aligned}$ addition order does not matter; i move sin²y and cos²y together so that i can use the pythagorean identity $\begin{aligned} \text{lhs} & = \dfrac{\sin^2 y + \cos^2 y + 1 + 2\cos y }{\sin y(1 + \cos y)} \end{aligned}$ pythagorean identity says that $\sin^2 y + \cos^2 y = 1$ so $\begin{aligned} \text{lhs} & = \dfrac{\boxed{\sin^2 y + \cos^2 y} + 1 + 2\cos y }{\sin y(1 + \cos y)} \\ & =\dfrac{\boxed{1} + 1 + 2\cos y }{\sin y(1 + \cos y)}\\ & = \dfrac{2 + 2\cos y }{\sin y(1 + \cos y)} \end{aligned}$ factor out a 2 from the numerator $\begin{aligned} \text{lhs} & = \dfrac{2( 1+ \cos y) }{\sin y(1 + \cos y)} \end{aligned}$ the $(1 + \cos y)$ cancel and we are left with $\begin{aligned} \text{lhs} & = \dfrac{2 }{\sin y} \\ & = 2 \cdot \dfrac{1}{\sin y} \end{aligned}$ by definition, $\frac{1}{\sin y} = \csc y$ $\begin{aligned} \text{lhs} & = 2\csc y \\ & = \text{rhs} \end{aligned}$

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