All Answers 354599 Answers

15,000 x 1.1^8 - 15,000           =

15,000 x 2.14358881 - 15,000 =

32,153.83 - 15,000                   =17,153.83

17153.83215

The variable can be anything UP TO but NOT including 9

it cannot = 9   it can only be LESS than 9

m<9

The last one pictured...there isn't any letters assigned to the graphs in your post.

Anything MORE than 8 does not INCLUDE 8...the last graph show this. The hollow dot shows 8 is not included.

I believe your answer should have no more or less # os sig digs than the numbers given , which each have 3 sig digits.

SO I think it should be 38.3

Hectictar,

thank you very much!!..may I ask one more thing please, how would we approach finding Sin78?..Thank you very much for your time..

It's the first one ... y < 27  

because it's the only one that has y less than 27, which must happen if (7.06 * y) is to be less than or equal 27.

Can't be the 2nd one because if y is already > 27, then (7.06 * y) would be a lot greater than 27. 

Can't be the 3rd one because y can't be allowed to be 27, or (7.06 * y) would be a lot greater than 27.

Can't be the 4th one because y can't be 27 or more because then (7.06 * y) would be a lot greater than 27. 

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2.)

An other ex. is this: \(\sigma=(1,3,5)(2,7)(11,1,4)(5,6,2,8,9)(7,3)(4,5,2,1)\)
The answer is \((1,11,3,2,4,6,7,5,8,9)(10) \Rightarrow (1,11,3,2,4,6,7,5,8,9)\)

Start with \(x = 1 \text{ in } k(x)\)

\(\small{ \begin{array}{|ccccccccccc|c|cccccc|} \hline f &\circ& g &\circ& h &\circ& i &\circ& j &\circ& k & x & k(x) & j(k(x)) & i(j(k(x))) & \tiny{h(i(j(k(x))))} & \tiny{g(h(i(j(k(x)))))} & \tiny{f(g(h(i(j(k(x))))))} \\ \hline (1,3,5)&\circ&(2,7)&\circ&(11,1,4)&\circ&(5,6,2,8,9)&\circ&(7,3)&\circ&(4,5,2,1) &\color{red} 1 & 4 &4 & 4 & 11 &11 & 11 \\ & & & & & & & & & & & 11 & 11 &11 & 11 & 1 &1 &3 \\ & & & & & & & & & & & 3 & 3 &7 & 7 & 7 &2 &2 \\ & & & & & & & & & & & 2 & 1 &1 & 1 & 4 &4 &4 \\ & & & & & & & & & & & 4 & 5 &5 & 6 & 6 &6 &6 \\ & & & & & & & & & & & 6 & 6 &6 & 2 & 2 &7 &7 \\ & & & & & & & & & & & 7 & 7 &3 & 3 & 3 &3 &5 \\ & & & & & & & & & & & 5 & 2 &2 & 8 & 8 &8 &8 \\ & & & & & & & & & & & 8 & 8 &8 & 9 & 9 &9 &9 \\ & & & & & & & & & & & 9 & 9 &9 & 5 & 5 &5 &\color{red}1 \\ \hline & & & & & & & & & & & \color{red}10 & 10 &10 & 10 & 10 &10 &\color{red}10 \\ \hline \end{array} } \)

Permutation cycles:
(1, 11, 3, 2, 4, 6, 7, 5, 8, 9) (10)

(2)  Here is a sketch of a single face of the regular pentagonal pyramid:

Remember there are  5  faces total because a pentagon has  5  sides. So....

lateral area  =  5 * area of triangular face

lateral area  =  5 * (1/2)(18)( s )

We can find  s  with the Pythagorean theorem:

9² + s²  =  16²

s²  =  16² - 9²

s²  =  175

s  =  √[ 175 ]

s  =  5√7

lateral area  =  5 * (1/2)(18)( 5√7 )

lateral area  =  225√7  square units