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See the following image :

Let AC   = 8     BC  = 15      AB   =17

The area of this triangle   = (1/2) (8)(15)  =   60

The  altitude of  this triangle  can be found as

60    = (1/2)  17 * height

120 /17   = height  = CD  

This height will form the radius of two  cones

The height of the smaller  cone can be found as   sqrt  [ 8^2  - (120/17)^2  ] = 64/17  = 

The height of the larger cone can be found as   17 - 64/17  =  225/17

So   when   triangle  ACD  is rotated about hypotenuse  AB   its  volume is

(1/3) pi (120/17)^2 * (64/17)   =   [307200 / 4913 ] pi

And when  triangle BCD is rotated  about hypotenuse AB  its  volume id

(1/3)pi  ( 120/17)^2 (225/17)  = [1080000 / 4913 ]  pi

So...the total  volume  is  ( [  307200 +  1080000]  / 4913  )  pi  =   

[1387200 / 4913 ]  pi   cm^3 

I've added an amended graph that might help you a bit.

 If A is an acute angle such that tan A + sec A = 3, then find cos A.

tanA ==> x

x + √(x² + r²) = 3     ==>   x = 1.333333333     or     4/3

angle A = tan^-1(4/3) = 53.13010235º

cos A = 3/5 

No prob  !!!

Thank you!

Multiplying  through by e  we  get

e^(Ln2 ^(4/3) )  = 2^(4/3)           and we  can write

e ^ [ (4/3) Ln 2 ]=  2^(4/3)       take the Ln of both sides

Ln e ^[ (4/3) Ln 2 ]  =  Ln 2^(4/3)         and we  can write

[ (4/3) Ln 2 ] Ln e   =  (4/3) Ln 2          ..... Ln e  =1....so  we  can ignore this

(4/3) Ln 2  =  (4/3) Ln 2       !!!!

If no triplets  are on the team =  C(11,5)  =  462  possible teams

If one triplet is on the team  C (3,1) *  C( 11,4)  =  990  possible teams

If two triplets are on the team =  C(3,2) * C ( 11, 3)  = 495 possible teams

So    462 + 990  +  495 =  

1947   different teams

0 ≤  # apples ≤ 9

Number of diagonals in a polygon is given by :

n ( n-3)  / 2    where n  = the number of  sides

So

n ( n- 3)  / 2  =   2n     simplify

n (n -3)  =  4n

n^2  - 3n  = 4n

n^2 - 7n  =  0

n ( n - 7)  =  0

Setting each  factor to 0  and solving for  n  produces   

n  = 0      reject

n =  7

Guest. Hi I nned help on this. If you could help that would be great.Its a hw question and i need it by tmrw.! I really apprecaite it.

In the diagram, 4ABC ∼ 4ACD. If AD = 12 and DB = 15, what is AC?

Sorry i dont have diagram. Closest i can get is this.AD and BC are on bottom of triangle btw.

(Sorry I do not have a picture or key for that operation thingy you have but I'll try my best)

so if |><| means (x+2)(y-3) and we're trying to find ((t)|><|(t+2))-((t+1)|><|(t+1)), first we must expand it

((t)|><|(t+2)) would have t represent x and (t+2) represent y so (t+2)((t+2)-3). We can expand that to t^2+t-2

Next we have ((t+1)|><|(t+1)) We do the same thing (t+1+2)(t+1-3) which expands to t^2+t-6.

(t^2+t-2)-(t^2+t-6)=-8

Pretty sure that's correct

tan D   =  4sinD

sinD               4sin D

______  =                            multiply both sides  by  cos D

cos D

sin D  = 4sinDcosD     rearrange as

4sinDcosD  - sin D   =  0   factor as

sinD  ( 4cos D - 1)    =  0

So...set each factor to  0  and solve

sin D  = 0  →  D = 0°  or 180°   (reject)

4cos D - 1   =  0

cos D =  1/4

But... cos D  =  sin F   = 1/4

Is there a picture of some sort or any information other than this? Because this is not enough information to answer the question unless the question is legit, then the answer would be 9 but I doubt it.

When y is 0 is when the ball hits the ground. We can treat this as a quadratic equation which will be

-16t^2+22t+45=0

factors into

(8t+9)(2t-5)=0

t=-9/8, 5/2

Since the answer can result in negative seconds, the answers is 5/2 seconds.

A

12                    20

B          16                       C

AB =  sqrt (20^2  - 16^2)   = sqrt ( 400 - 256)  = sqrt (144)  =  12

ΔABC ≈  ΔDEF

Since the triangles are similar, then sin F = sin C  =   12/20  =  3/5

Thanks!

sin x  = -0.31

x  ≈  -18  

First angle =  180 + 18 ≈  198°

Second angle  = 360 - 18 = 342°

Nunber of  full  rotations in 2500° =  2500/360  ≈  6.94

(.94)  * 360  ≈   339°

So  there  will  be   6(2)  + 1  = 13  values of   x  that satisfy sin x = -.031

tan A  +  sec A  = 3

sin A            1

_____ +   _____   = 3

cos A        cos A

sin A  +  1

_________ =    3               

   cos  A

sin A + 1  = 3 cos A          square both sides

sin^2 A + 2sin A  +  1  =  9cos^2A

sin^2 A  + 2sinA  + 1  = 9 ( 1 -sin^2A)

sin^2 A  + 2sin A +  1  =  9 - 9sin^2 A      simplify as

10sin^2A  + 2sin A  - 8    = 0    divide through by 2

5sin^2 A  + sina  -  4 =    0      factor as

(5sinA  - 4) ( sin A + 1)   = 0

Setting each factor to 0 and solving for A  gives us that

sin A = 4/5       or   sin A   =  - 1

If sinA  = -1....A =  270°...but the tangent  and sec are undefine at the angle

So

sin A =4/5

And cos A   =  3/5

Added  corrected answer  ......

.

This has to satisfy the triangle inequality  that says that the  sum of any two side lengths of a triangle are greater than the remaining side

So...let x be the possible unlnown side lengths.....and we have that

21 + 27  >  x               and           21 + x  >  27

48 > x                                                x > 6

So....the possible side lengths  are  [ 7 , 47 ]

The number of positive integers =  47 - 7  + 1 =   41

When it hits the gtound the height will  = 0

So

-4.9t^2  + 42t +  18.9  = 0

Using the quadratic formula we  have  that

t =   -42 ±  sqrt ( 42^2 - 4(-4.9)(18.9) )           -42 ±  46.2          -42 - 46.2

     ____________________________  =    ___________  =  ________   =   9 sec

                     2 (-4.9)                                          -9.8                   - 9.8 

-4.5659785e+41

In right triangle DEF, we have angle E = 90 degrees and tan D = 4*sin D. What is sin F?

Angle F = 90º - [cos^-1(1/4)]  ==>  ∠F = 14.47751219º

Let the shorter leg  = x

The longer leg  x + 8

The hypotenuse   x + 8 + 10   =  x + 18

By rhe Pythagorean Theorem, we have

x^2  +  ( x + 8)^2  =  (x + 18)^2     simplify

x^2  + x^2 + 16x  + 64  =  x^2  + 36x + 324

2x^2 + 16x  +  64  =  x^2  + 36x + 324

x^2 - 20x  - 260  =  0      complete the square on x

x^2 - 10x + 100  =  260 + 100

(x - 10)^2 =  360       take the positive roor

x - 10    = sqrt (360)

x - 10  = 6sqrt (10)

x  =  6sqrt (10)  + 10   = shorter leg

x + 8  =  6sqrt (10) + 18  = longer leg

x + 18 = 6sqrt (10) + 28    = hypotenuse

idk but good luck with this