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We have  that

m + m^2  =  15/4

And

m * m^2  =  a^3

m^3  = a^3

This  implies  that  m  =  a

So

a + a^2  =  15/4

a^2  +  a  =  15/4        complete the square on a

a^2 + a + 1/4  =  15/4  + 1/4

(a + 1/2)^2  =  4     take both roots

a + 1/2  =  ±√4

a + 1/2  =  2             or             a + 1/2   =    - 2

     a  = 3/2             or              a  = - 5 / 2

A circle is revolved about a diameter and the sphere obtained has numerically the same volume as the circle had area.  Find the circumference of the circle.

The formulas for area and for volume both use the radius, so let's find that first, and then figure the circumference. 

V  =  (4/3) (π) r³  

A  =  (π) r²  

Set them equal                                (4/3) (π) r³  =  (π) r²  

Divide both sides by (π) r²               (4/3) r  =  1 

Divide both sides by 4/3                    r  =  3/4   

Circumference is 2 π r                       C  =  2 π  r   =  2 x 3.1416 x 3/4  =  3.1416 x 3/2  =  about 4.7  

_.

( -1 - 8) i   +  (5 - - 6) j  =

-9 i  +  11  j

We have the  form

y  =  Asin (Bx )  + C

A   = the amplitude  =  2

C = the vertical shift  =  - 3

B  is  the  hardest to calculate

B  =   2pi  / period

B =   2pi  / 4

B =  pi/2

So....our function is  

y = 2sin[ (pi/2) x ]   -  3

Here's the graph  :  https://www.desmos.com/calculator/h2aglcdrbw

thank you so much for the clear and concise answer, I understand how to solve it now!!! I'm really thankful that there are people like you on the forum :)))

a)  Volume  =  (40 * 20 *  24) cm^3       =  19200 cm^3

b)  1L   =  1000cm^3

So  2L  = 2000 cm^3

So.....we can find how much the water wil rise  by  solving  this

(20 * 40  * h)  =   2000

800 *  h  =  2000     divide both sides by 800

h  =  2000/800  =  2.5 cm  =    height  that the water will rise

The sum   of  the roots  =  (2 - m)

So

a + b =  2m     square  both sides of this

a^2 + 2ab + b^2  = 4m^2      (1)

The product  of  the roots  = -( m + 3)

So

ab  = -(m + 3)

2ab =  -2(m + 3)      (2)

Sub (2)  into (1)

a^2  - 2 (m + 3)  + b^2   =  4m^2

a^2  + b^2  =  4m^2  + 2(m + 3) 

(2n + 1) + (2n + 3) + (2n + 5) + ... (2n + 47)  =  5280

There are 24 terms in the list:   (2n + 1) + (2n + 3) + (2n + 5) + ... (2n + 47)  

So, the left-hand side can be rewritten as  (2n)(24) + (1 + 3 + 5 + ... + 47)  =  5280

We can use the formula:  Sum  =  n(t₁ + t_n) / 2 to find the sum of  1 + 3 + ... + 47:

   Sum  =  24(1 + 47)/2  =  24(48)/2  =  576

Equation:  (2n)(24) + (576)  =  5280

                           48n + 576  =  5280

                                      48n  =  4704

                                          n  =  98

Now, to find the sum:  1 + 2 + 3 + ... + 98  using the formula:  Sum  =  n(t₁ + t_n) / 2

   Sum  =  98(1 + 98)/2  =  4851.

Note  that  the function will be maxiimized  when  (7/3)x + 1/6   =  (3pi/2)

So....the  max  temp   =  -6sin 3(pi/2) - 5  =   -6(-1)  - 5   =    1° F

And  the function will be  minimized whex  (7/3)x + 1/6 =  pi/2

So....the min temp  is   -6sin(pi2) - 5   =  -6(1)  - 5  =   -11° F

So....the temp falls  from 1°F to -11°F

Note  that we  can write this as

(101^2 - 97^2)   +  (93^2  - 89^2)  +   .....+  (5^2  - 1^2)   =

(101 - 97) (101 + 97)  +  ( 93 - 89) (93 + 89)  +  ( 85 - 81)(85 + 81) + (77 - 73) (77 + 73) + ..+ (5 - 1) (5 + 1)  =

4   [ 198   +  182  +  166   + 150 +   ....+  6 ]  

The  number  of terms  inside  the btackets  =   [198 - 6] / 16  + 1  =  12 + 1  =  13

So.....using  the sum of an arithmetic series.....the sum is

4 *  [ 198 + 6]  / (13/2)=

26  [ 204]  =

5304

And those other kids say their homework is hard and time-consuming. 

You have mastered the art of time and skill allocation. 

Clever you.

P (first card a 6  and  the secon a queen)   = (4/52) (4/51)  =  (1/13) (4/51)  =   4 / 663

hyp^2 =  (hyp-4)^2   + (hyp-8)^2

               h^2 -8h +16   + h^2 - 16h + 64

h^2 +24h -80 = 0      quadratic formula h = 4, 20      throw out 4      hyp = 20    leg1 = 12    leg2 = 16      20 + 12 + 16 = perimeter

\(\left(6x+\dfrac{1}{3x}\right)^6\\~\\ T_n=\begin{pmatrix} 6\\ n \end{pmatrix} (\frac{1}{3x})^n(6x)^{n-6}\\ \text{If you pull out the x terms you get } (\frac{1}{x})^n*x^{n-6} \\ \text{The constant term will have }x^0\qquad\\ \text{So solve for n}\\ etc\)

Please no other answerer finish this. 

In triangle(ABC),  sin(A) : sin(B) : sin(C)  =  2 : 3 : 4                     Find  cos(A + C)

In triangle(ABC), let    side a = 2x    side b = 3x    side c = 4x

Using the Law of Cosines:  cos(C)  =  ( a² + b² - c² ]) / ( 2·a·b )

cos(A)  =  [ (3x)² + (4x)² - (2x)² ] / [ 2·(3x)·(4x) ]  =  [ 9x² + 16x² - 4x² ] / [ 24x² ]  = 7/8

Using the Pythagorean Theorem:  sin(A)  =  sqrt(15) / 8

cos(C)  =  [ (2x)² + (3x)² - (4x)² ] / [ 2·(2x)·(3x) ]  =  [  -3x² ] / [ 12x² ]  =  -1/4

Using the Pythagorean Theorem:  sinC)  =  sqrt(15) / 4

Using the formula:  cos(A + C)  =  cos(A) ·cos(C) - sin(A)·sin(C)

                                                  =  (7/8) · (-1/4) - ( sqrt(15) / 8 ) · ( sqrt(15) / 4 )

                                                  =  -11/16

Find the hypotenuse length     sqrt ((-1)^2 + 3^2) =   sqrt 10  

          sin of the angle is  3/sqrt10       csc is 1/sin   =   sqrt10/3    or  1/3  sqrt10

Hey you must be in geometry for k12, im doing the same project.

For this, you have to use inverse tangent to find theta. The terms were opposite, adjacent. You take inverse tangent because you are looking for the angle.

Tan^-1(8/425) = 1.078 BUT!!!!

It is not 8 ft BECAUSE you probably just took the fence height and thats it. You can't do that because then the fence height itself is taking the 2 ft from the ground to the bat which doesn't count towards the angle. You take 8-2 to get the fence height from the BAT to the FENCE TOP.So, that would be 6ft. Also, the sentence did say "hit 2 ft above the ground" which means you wouldn't use those 2 ft for the height

Tan^-1(6/425) = 0.8088

As you can see, that angle elevation would count for the baseball field that has an angle elevation of less than 0.86 degrees, this is where this question came into play, basically answering that.

Ask me more questions if u need help cuz we could be in the same class

If you don't understand what I'm saying, email your teacher, they are happy when students communicate with them

The  #  of  subsets are

12C0  + 12C1  + 12C2  + 12C3 + 12C4   =

 1    +    12     +   66  +  220  +  495  =

794

A   Each cup of blue paint requires two cups of yellow paint to make the green paint.  

B   Each cup of yellow pant requires 1/2 cup of blue paint because the ratio is   2:1   which is tha same as   1 : 1/2 .

C   $54 / 6 cup =   $ 9 per cup .

Assuming that this  is a right triangle, C  is  the right angle

So

Tan C  =  Tan (90°)   =    0

2.A right cylinder with a base radius of 3 units is inscribed in a sphere of radius 5 units. Find the total volume, in cubic units, of the space inside the sphere and outside the cylinder.

The  only tricky thing  is  finding (1/2)  the  height of the cylinder.....this  equals

sqrt  ( radius of sphere^2  -  radius of cylinder^2 )  =  sqrt (5^2 - 3^2)  = sqrt (16)  =  4

So

Volume of sphere  =  (4/3)pi (5)^3

Volume of cylinder  =  pi (3)^2 * 8

So  the volume we are looking for  is

(4/3) pi ( 5)^3  - pi (3^2) * 8   =

pi  [ ( 4/3)(125)  - (9)(8) ]   =

pi  [ 500/3  - 72 ]  =

pi  [ 500 -216] / 3   =

(284 / 3)  pi    units^3   ≈  297.4  units^3