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4 years ago

Need some help on this question please help!

Mathematics

2 answers:

galina1969 [7]4 years ago

7 0

oof- thats really hard

Masteriza [31]4 years ago

3 0

The answer should be D. The problem says the praying mantis can only rotate from 8-25°, and 6^2 is 36. All of the other answers are in the acceptable interval stated in the problem.

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Find the height of an equilateral triangle with sides of length 12

MAVERICK [17]

Answer:

Answer

sqrt(108)

6 sqrt(3)

10.3923

I don't know which answer you want.

Step-by-step explanation:

Drop a perpendicular from the top angle to the base.
The base is cut into 2 equal parts.
Each part is 12/2 = 6
The perpendicular, as its name implies, meets the base at 90o.
You can use the Pythagorean Theorem to find the height.

h^2 = side^2 - (1/2 b)^2

h^2 = 12^2 - 6^2

h^2 = 144 - 36

h^2 = 108

Take the square root of both sides

h = sqrt(108)

h = sqrt(2*2 * 3 * 3 * 3)

h = 2 * 3 sqrt(3)

h = 6sqrt(3)

h = 6 * 1.7321

h = 10.3923

4 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=prove%20that%5C%20%20%5Ctextless%20%5C%20br%20%2F%5C%20%20%5Ctextgreater%20%5C%20%5Cfrac%20%7B

inysia [295]

$\large \bigstar \frak{ } \large\underline{\sf{Solution-}}$

Consider, LHS

$\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered} \\ \\ \text{So, using this identity, we get} \\ \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered} \\$

So, using this identity, we get

$\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

can be rewritten as

$\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}$

<h2>Hence,</h2>

$\begin{gathered} \\ \rm\implies \:\boxed{\sf{ \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}$