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FinnZ [79.3K]
3 years ago
6

Find the volume of the triangular prism

Mathematics
1 answer:
gulaghasi [49]3 years ago
3 0

Answer:

V = 5280 in^3

Step-by-step explanation:

The volume is found by

V = Bh  where B is the area of the base and h is the height

first we need to find the area of the base

The base is the triangle

We need to find the base dimension for the triangle

34^2 = 16^2 + b^2

34^2 -16^2 = b^2

900 = b^2

30 =b

B = 1/2 bh

  = 1/2 (30) *16

B =240

V = Bh

V = 240 * 22

V = 5280 in^3

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Answer:  \frac{2x(5x + 1)}{4x - 1}

<u>Step-by-step explanation:</u>

\frac{4x^{3}-12x^{2}}{4x^{2}+7x - 2} ÷ \frac{2x^{2}-6x}{5x^{2}+11x + 2}

= \frac{4x^{2}(x - 3)}{(4x - 1)(x + 2)} ÷ \frac{2x(x - 3)}{(5x + 1)(x + 2)}

= \frac{4x^{2}(x - 3)}{(4x - 1)(x + 2)} x \frac{(5x + 1)(x + 2)}{2x(x - 3)}

= \frac{2x(5x + 1)}{4x - 1}

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Find the 10th term of the sequence defined by the rule, f (n) = 4n - 3.
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Considering only the values of β for which sinβtanβsecβcotβ is defined, which of the following expressions is equivalent to sinβ
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Answer:

\tan(\beta)

Step-by-step explanation:

For many of these identities, it is helpful to convert everything to sine and cosine, see what cancels, and then work to build out to something.  If you have options that you're building toward, aim toward one of them.

{\tan(\theta)}={\dfrac{\sin(\theta)}{\cos(\theta)}    and   {\sec(\theta)}={\dfrac{1}{\cos(\theta)}

Recall the following reciprocal identity:

\cot(\theta)=\dfrac{1}{\tan(\theta)}=\dfrac{1}{ \left ( \dfrac{\sin(\theta)}{\cos(\theta)} \right )} =\dfrac{\cos(\theta)}{\sin(\theta)}

So, the original expression can be written in terms of only sines and cosines:

\sin(\beta)\tan(\beta)\sec(\beta)\cot(\beta)

\sin(\beta) * \dfrac{\sin(\beta) }{\cos(\beta) } * \dfrac{1 }{\cos(\beta) } * \dfrac{\cos(\beta) } {\sin(\beta) }

\sin(\beta) * \dfrac{\sin(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}}{\cos(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}} * \dfrac{1 }{\cos(\beta) } * \dfrac{\cos(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}} {\sin(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}}

\sin(\beta) *\dfrac{1 }{\cos(\beta) }

\dfrac{\sin(\beta)}{\cos(\beta) }

Working toward one of the answers provided, this is the tangent function.


The one caveat is that the original expression also was undefined for values of beta that caused the sine function to be zero, whereas this simplified function is only undefined for values of beta where the cosine is equal to zero.  However, the questions states that we are only considering values for which the original expression is defined, so, excluding those values of beta, the original expression is equivalent to \tan(\beta).

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