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Are there meant to be any brackets in there?

The three streets  form a right triangle

The east-west   and  north-south streets  are legs  and  the  street  that bears  N 32° E  is  the hypotenuse

We  can find the frontage  along the hypotenuse, H,  as follows

H/sin 90  =  50/ sin 32

H = 50/sin 32  ≈ 94.35 m

The lfrontage along the north-south street is

sqrt ( 94.35^2 - 50^2 )  ≈ 80 m

The total fencing needed  =  50 + 80 + 94.35   ≈  224.35 m

How many 9-step paths are there from E to G?

Second soultion:

How many 9-step paths are there from E to G?

\(\begin{array}{|rcll|} \hline \text{Permutations of (hhhhhvvvv)} \\ \hline && \dfrac{(5+4)!}{5!4!} \\\\ &=& \dfrac{9!}{5!4!} \\\\ &=& \dfrac{5!6*7*8*9}{5!4!} \\\\ &=& \dfrac{6*7*8*9}{4!} \\\\ &=& \dfrac{6*7*8*9}{2*3*4} \\\\ &=& \dfrac{6*7*8*9}{2*3*4} \\\\ &=& \dfrac{3024}{24} \\\\ &=& 126 \\ \hline \end{array}\)

Find the number of integer solutions to \(a^2 + b^2 = 625\).

\(\begin{array}{|rclcrcl|} \hline a^2 + b^2 &=& 625& \text{or}& \mathbf{a^2 + b^2} &=& \mathbf{25^2} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{3^2+4^2} &=& \mathbf{5^2} \\\\ (3*5)^2+(4*5)^2 &=& (5*5)^2 \\ \mathbf{15^2+20^2} &=& \mathbf{25^2} \\ \hline \end{array}\)

\(a=15~ \text{and}~ b= 20 \)

There are other solutions for example: \(7^2+24^2 = 25^2\quad \text{ (See: Pythagorean triple )}\)

We will have a right triangle  with AB , AC being the legs  and BC  the  hypotenuse

To find  BC  we have

BC = sqrt [ 500^2  + 700^2]  ≈ 860.23 mi

To find  the  bearing  we  have

tan  BCA   = (500/700)  = 5/7

arctan (5,7)  =  BCA  ≈ 35.53°

I'm not sure  how  you want to denote the bearing, but  one  way is

N 35.53° W

111  117  135  153  159  171  177  195  315  333  339  351  357  375  393  399  513  519  531  537  555  573  579  591  597  711  717  735  753  759  771  777  795  915  933  939  951  957  975  993  999  Total =  41

11 , 12 , 15 , 24 , 36 , Total =  5

It  must take him  10 min to   travel one way on the  bus  since  it takes him 20 min in total when he goes by bus both ways

Therefore.....when he  walks and takes the bus  the time  he  walks 1/2 the total distance  must  be  40 - 10  = 30 min

Therefore....if he walked the  whole distance, it would take  him twice as long = 1  hour

The distance, M, from  the base of the mountain to the  nearest bank can be found as

tan 52.6  =   532/ M

Rearrange as

M =  532 / tan 52.6  ≈ 406.7 ft

The distance,N,  from  the base of the mountain to the far bank can be found as

tan 34.5  =  532/ N

N = 532 / tan 34.5  ≈ 774.1  ft

The width of the river  =   N - M  =    774.1 - 406.7   ≈  367.4  ft

√3sin x  =  cos x+ 1      square both sides

3sin^2 x   =  cos^2x  + 2cos x  + 1

3 ( 1 - cos^2x)  = cos^2x  + 2cosx  + 1

3 - 3cos^2 x = cos^2x + 2cos x + 1        rearrange as

4cos^2 x + 2cos x - 2  =  0         divide through by 2

2cos^2 x + cos x - 1  =  0     factor as

(2cos x - 1) ( cos x + 1)  =  0

So either

2cos x - 1  = 0                              

2cos x  = 1                                                            

cos x = 1/2                                                                

And this is true at  x = 60° and x =  360 - 60 = 300°

or

cos x + 1  =  0

cos x  = - 1

And this is true at x = 180°

Because we squred both sides  we need to check these solutions

√3 sin (60)  =  1 + cos (60) 

√3 (√3/2)  =  1 + 1/2

3/2  = 3/2      true

√3 sin (180)  = 1 + cos (180)

0 =  1  - 1

 0 = 0   true

√3 sin (300)  = 1 + cos (300)

√3 (-1/2)  = 1 + 1/2

-√3/2  =  3/2     not true

So...the  only solutions  are  x  = 60°   and  x   = 180°

Thank you :)

Nice, Cal  !!!!

50*0.4*0.3=6

The answer is 6 :D

I don't know an algebraic method of solving this.....(but there may be one)

Look  at the graph here  : https://www.desmos.com/calculator/ctrgxe2g8u

This will be true when  x  =  3 + 2/3   =  11/3

You do not need to examine previous answers.

My answer was there many hours before you started yours.

It would be impossible for you to overlook that an answer already existed.

By answering in full you reinforced what the students already know.

This is the best cheat site available.

They can see that even key answerers will not support each other in trying to change this reputation.

Do you notice that the original asker has been silent?

He/she has copied your answer and handed in the assignment.

Can you help me with this?

P and q are complex numbers such that |p| = 2/5x + 2, |q| = −3/4x − 1 , and |p+q| = 2x + 4 .

On what interval must x fall?

a) [-2, -4/3]

b) (-∞, -4/3]

c) [-2, ∞)

d) [-4/3, -60/47]

I got the first one, but I don't know if it's right. If someone could help, that would be greatly appreciated!

Connect A  to to top left  vertex  of the rotated square....call this point B

And call the top right  vertex of the right triangle formed = C

Then the area of this right triangle ABC = 1/2  area of the square  = 1/2

The other small green area will be a 45 - 45 - 90  right triangle  with  legs  of diagonal lengthof the square - the side length of the square =   (√2 - 1)

So  the  area of this triangle  = (1/2)(√2 - 1)^2  = (1/2) ( 2 + 1 - 2√2)  =  (1/2) (3 - 2√2)

So...the green area =   1/2  + (1/2) ( 3 - 2√2)  = (1/2) ( 1 + 3 - 2√2 ) = (1/2) (4 - 2√2)  =

(2 - √2)  units^2

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:)

Good luck!