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The   slant height  of  the cone   will  =  the  height of  the  equilateral triangle

This  slant height  =  sqrt  [ 6^2 - 3^2  ]  =  sqrt  [ 36 - 9 ]  =  sqrt (27 ) 

The  circumference of the  cone  = side of the  triangle = 6

We can find the radius, r, of the  cone as

circumference =  2pi *r

6 = 2pi r

r =  6/ (2pi)  =   3/pi

Heigt of  cone =   sqrt  [  [sqrt (27)]^2  - (3/pi)^2  ]   =  sqrt [ 27 - 9/ pi^2 ]  =  sqrt  [ 27pi^2 - 9]  / pi  =

sqrt  [ 9 (3 pi^2 - 1 ) ] /pi  =     (3/pi) sqrt ( 3pi^2 - 1 )

Volume of  cone =  (1/3) radius^2  * height   =   

(1/3)  ( 3/pi)^2 *  (3/pi) sqrt ( 3pi^2  - 1) =

(9/ pi^3)sqrt (3pi^2 -1) ≈   1.55 units^3

Second blank

sin^2 x  - 1 

___________

(sin (x) + 1)^2

Note  that   sin^x  + cos^2x  = 1  →     sin^x - 1  =   -cos^2x

Third blank

-(cos^2 x)

__________

(sin (x) + 1)^2

--Find the other 2 sides of a triangle

--Find the area of a triangle

--Multiply the area by 2 and then divide it by 100

Law of Cosines

x^2  = b^2  + c^2  - 2(b)(c)cos( CAB)

x^2  = 15^2 + 17^2  - 2(15)(17)cos (108°)

x^2  =  225 + 289  - 2 (255)cos(108°)

x^2  = 514 - 510cos(108°)

x = sqrt  [  514 - 510 cos(108°) ]  ≈    25.9

tan (theta)  = -1/2     cos (theta)   >  0

This angle is in the 4th quadrant

x = 2      y = -1        r  = sqrt  (2^2  + (-1)^2 )  = sqrt  (5)

sin (theta)  =  -1/sqrt (5)        cos (theta)  =  2/sqrt (5)

sin  (2 theta)  =   2sin (theta)cos(theta)  = 2  (-1/sqrt (5)) ( 2/sqrt (5) )  =   -4/5

cos (2 theta)  =   cos^2(theta)  - sin^2(theta)  =  (2/sqrt (5))^2  - (-1/sqrt (5))^2  =  (4/5) - (1/5)  = 3/5

tan ( 2 theta)  =    sin (2 theta)  / cos ( 2 theta)  =    (-1/sqrt (5) ) / (2/sqrt (5))   =    -1/2

I assume the  violet paint is  composed of 1/3 red paint and 2/3  blue paint

6/2 =   3 quarts of  violet paint

3 (1/3)  = 1 quart red paint

2(2/3)  =  2 quarts blue paint

The area of a parallelogram can be found by multiplying side times the height to that side.

Since you will get the same answer using whatever side you want to use:

Area  =  FH x AD  =  EG x DC

--->        FH x 18  =  16 x 27

--->                FH  =  24

(Solved by Geno)

Vertex  = (0,0)

Focus = (-6,0)

The focus is to the left of the  vertex.....this parabola will open to the left

We have  the  form

-4px =  y^2

p = distance  from focus to vertex  = sqrt [ (-6-0)^2  +  (0 - 0)%2 ] =  sqrt (36)  = 6

The equation is

-4 (6)x = y^2

-24x  = y^2         which also can be writtten

x = (-1/24)y^2

See the graph here :   https://www.desmos.com/calculator/qgqpf58id8

To avoid confusion....I'm subbing  theta  for x

0° < theta < 90°    means that  x is in the  first quadrant......so theta/2 will also be in the  first  quadrant

tan theta = 2sqrt (6)   means that   y  =2sqrt (6) = sqrt (24)   and  x = 1

r =  sqrt  ( x^2  + y^2) =   sqrt  ( 1^2  + (sqrt (24))^2 ) =  sqrt (  1 + 24)  = sqrt (25)  = 5

cos theta =  x/ r=     1/5

sin  (theta/2)  =  sqrt [  ( 1 - cos (theta) )  /2  ] =    sqrt  [ (1 - 1/5) /2 ]  = sqrt  [ (4/5) /2 ] =  sqrt [ 4/10] = (2/10)sqrt (10)  =    sqrt(10)/5

cos (theta/2)  =  sqrt  [ (1 + cos (theta)) / 2  ] =    sqrt  [ (1 + 1/5) / 2 ]  = sqrt  [ (6/5)/2 ] = sqrt ( 6 ) /sqrt (10)  =

sqrt (6)sqrt (10) / 10  =  sqrt (60) / 10  =  2sqrt (15) / 10  =  sqrt (15)/5

tan (theta/2)  =   1 -cos (theta) / sin (theta)  =   (1 - sqrt (10)/5)  / (sqrt (15)/5)  =  

(5 -sqrt (10)) / sqrt (15)   =   [ 5sqrt (15) -sqrt (10)sqrt (15) ]  /15    =     [ 5sqrt (15)  - sqrt (150 ] / 15

Link::::https://web2.0calc.com/questions/in-the-figure-below-abcd-nbsp-is-a-parallelogram-ad-18-nbsp-and-dc-27-nbsp-segments-eg-nbsp-and-fh-nbsp-are-perpendicular-to-sides-of

x = 78.46304097º

x/2 = 39.23152048º

sin x/2 = 0.632455532               cos x/2 = 0.774596669                tan x/2 = 0.81649658

Help, Please??

EDIT: Ok, never mind! I figured it out mysef! 

Here's my solution:

We know that the points that he labels with the number 25 has a distance of sqrt(25) = 5 from the origin. Thus, using the distance formula and calling the points that satisfy the condition (x, y), we have:

sqrt(x^2+y^2) = 5. Sqaring both sides gives us x^2+y^2 = 25. See if you can solve it from there! 

Hint: We have to find the number of points (x, y) that satisfy the equation x^2+y^2 = 25. We also have to remember that x and y can be negative numbers too, but they all have to be real integers. Listing the possibilities out, we get

(x, y) = (0, 5), (5, 0), (0, -5), (-5, 0), (3, 4), (-3, 4), (3, -4), (-3, -4), (4, 3), (-4, 3), (4, -3), and (-4, -3). How many pairs is that? 

12 times (I did this problem yesterday lol)

i ihink it's 4

They literally said it wasn't 4 you idiot

Sure, I can try! First, we can call the side length of the square "x". The area of this orginal square would be x^2. Next, if the length of the side of the square was doubled, then one side would have a length of 2*x = 2x. The area of this new sqaure would be (2x)^2 = 4x^2. Therefore, the ratio of the areas of the original sqaure and the area of the new sqaure is x^2 / 4x^2 = 1/4, or 1:4. I hope that helped! 

3.1536 × 10^6

Can you do it in an algebraic way?

First, we cross multiply for all three equations to get 

a^2 = b, b^2 = 2c, and c^2 = 4a. Thus, we know that a = sqrt(b) from the first equation and b = sqrt(2c) from the second. Next, using the last equation, we find that 

c^2 = 4a ---> c^2 = 4sqrt(b). We already found that b = sqrt(2c), so we can substitute it in to get c^2 = 4*sqrt(sqrt(2c)) = 4*∜2c. Further simplifying it, we get c^8 = 512c ---> c^7 = 512, so c equals the seventh root of 512. 

Lastly, we plug this value of c into the third equation to obtain a = about 1.48599429.

Where is "as shown" ??

Original square =1^2 =1

2nd square         =2^2 =4

Ratio =1:4

Follow the instructions on this page:  https://www.mathsisfun.com/algebra/trig-cosine-law.html

S =Flying speed of the bird

W =Wind speed

112.5 / [S + W] = 1.5 hours................(1)

112.5 / [S  - W] = 2.25 hours..............(2)

Now, you have 2 simultaneous equations. Can you solve them?