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There are 33 possible sequences of flips.

i'm kinda sick but one of my relative answered it here:

10 seconds *  1 m/s =   10 meters

10 meters / .25m/w = 40 waves

Distance covered by New train = distance covered by old train+20% of distance covered by old train

Or, Distance covered by New train =300 miles+20% of 300miles.

                    = 300 miles +( 20/100)×300

                    = 300 miles +20×3 miles

                   = 300 miles+60miles

                 = 360 miles. Ans.

Add 2x to both sides: 7x + 4 = 18

Subtract 4 on both sides: 7x = 14

Divide by 7:         x = 2

A Board foot is    1 sq ft   1 inch thick

1)       1/3 ft  *  1/2 ft    x 12   = 2 board feet

2)       1/2 ft * 1/6 ft  * 10   = ___________

Can you do the rest?

Add 2x to both sides of the equation

Subtract 4 from both sides of the equation

     then divide both sides of the equation by 7

Yeppers.....That's it !   

ABC is an equilateral triangle with the length of a side of 45 (45 is evenly divisible by 5 and 9)

Point G is the centroid, so the length of AG = 2/3(AN)

Use AD, AE, and angle EAG to calculate the angles.

BC = 2[sqrt(9² - 3²)] = 2√72

CE = (3 * BC) / 9 = 4√2

AE = sqrt(AC² - CE²) = 7

[ACD] / [ABE] = 5/9

(-8)² = +64

-8² = -(8²) = -64

2/9 of the pupils in Primary 5 scored an 'A' for a test. The ratio of the number of pupils who scored a 'B' to the number who scored an 'A' was 5 : 3. The rest of the pupils scored a 'C' and a 'D' in the ratio of 6 : 5. The pupils who scored a 'B' were 48 more than the pupils who scored a 'C'.

(a)   How many pupils were there in Primary 5?

(b)   How many pupils failed the test if the failure was a 'D' grade?

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A = 2/9                       A = 6/27

A : B = 3 : 5                                           A + B = 16/27

3/5 = (6/27) / B          B = 10/27

C + D = 11/27            C = 6/27

C : D = 6 : 5

                                  D = 5/27

B / C = 5 / 3

5 / 3 = (48 + C) / C            C = 72

B = 48 + C                         B = 120

C / D = 6 / 5                       D = 60

                                          A = 72

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                                  Total = 324 

Ratio:  A : B : C : D = 6 : 10 : 6 : 5

Consider the following diagram:

What are the lengths of AE and CE?

\(\begin{array}{|rcll|} \hline CD^2 + 3^2 &=& 9^2 \\ CD^2 &=& 81 - 9 \\ CD^2 &=& 72 \\ CD &=& \sqrt{72} \\ \hline \end{array} \begin{array}{|rcll|} \hline BD^2 + 3^2 &=& 9^2 \\ BD^2 &=& 81 - 9 \\ BD^2 &=& 72 \\ BD &=& \sqrt{72} \\ \hline \end{array} \begin{array}{|rcll|} \hline CB &=& CD+BD \\ CB &=& \sqrt{72}+\sqrt{72} \\ CB &=& 2\sqrt{72} \\ CB^2 &=& 4*72 \\ \mathbf{CB^2} &=& \mathbf{288} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline AE^2+CE^2 &=& 9^2 \\ \mathbf{CE^2} &=& \mathbf{81-AE^2} \\ \hline \end{array} \begin{array}{|rcll|} \hline (9+AE)^2+CE^2 &=& CB^2 \\ (9+AE)^2+81-AE^2&=& 288 \\ 81+18AE+AE^2+81-AE^2&=& 288 \\ 81+18AE+81&=& 288 \\ 18AE&=& 288-2*81 \\ 18AE&=& 126 \\ \mathbf{AE}&=& \mathbf{7} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{CE^2} &=& \mathbf{81-AE^2} \\ CE^2 &=& 81-7^2 \\ CE^2 &=& 81-49 \\ CE^2 &=& 32 \\ CE^2 &=& 16*2 \\ \mathbf{CE} &=& \mathbf{4*\sqrt{2}} \\ \hline \end{array}\)

Using the pythagorean theorem:

a^2 + b^2 = C^2

60^2 + 60^2 = (distance to 2nd base)^2

3600 +3600 = (distance to 2nd)^2          

7200 =  (distance to 2nd)^2        

Distance to 2nd = √7200 = 60√2 ≈ 84.85 ft.   distance between home plate and second base   

From the above question, it states the distance from home plate to second base is 46ft. Therefore, to calculate the distance from the pitcher's mound to second base we will subtract.

84.85 ft - 46 ft =   subtract distance from home plate

38.85 ft               distance from the pitcher's mound to second base

What is -8 squared?  

A minus times a minus is a plus, so it is +64.  You can verify this on your calculator.  

From 1st base to 2nd base is a 60 ft leg of a right triangle

   the hypotenuse would be from 2nd base to home plate

      use pythagorean theorem      d^2 = 60^2 + 60 ^2      d = from second to home = 84.85 ft     You can finish, Yah?

ABC has ∠BCA = 2∠BAC. Side BC is extended through C to a point D so that AC = CD. If AB = 3 and BC = 2, find AD.

```````````````````````````````````````````````````````````````````````

AD = 3.75

Sides: a = 2   b = 2.5   c = 3

Area: T = 2.48
Perimeter: p = 7.5
Semiperimeter: s = 3.75

Angle ∠ A = α = 41.41° = 41°24'35″ = 0.723 rad
Angle ∠ B = β = 55.771° = 55°46'16″ = 0.973 rad
Angle ∠ C = γ = 82.819° = 82°49'9″ = 1.445 rad

In the first bag, there are 13 yellow beans  (50% of 26).

In the second bag, there are 7 yellow beans (25% of 28)

In the third bag, there are 15 yellow beans (50% of 30).

There is a total of 13 + 7 + 15  =  35  yellow beans.

There is a total of  26 + 28 + 30  =  84  beans.

The percentage of yellow to the total is:  35 / 84  =  41.666666666%.

x^2  -6x + k - 3 = 0

the two numbers added = -6          call them a and 2a       a + 2a = -6

                                                                                             a = - 2

     and          a * 2a = k-3

                        2a^2  = k-3

                        8 = k-3

                           k = 11

14 degrees  plus  30 minutes  plus 52 seconds:

14<sup>o</sup>  +  30/60<sup>o</sup>  +  52/3600<sup>o</sup>  =  14.51444444<sup>o</sup>     =     (14 + 463/900)<sup>o</sup>

The largest degree of f(x) is 4.

The largest degree of g(x) is 6

Multiplying g(x) by a constant does not increase the degree of g(x).

Adding together these polynomials give a result whose highest degree is 6.

Thanks Alan,

Here is a matrix solver calculator.  Of course, you would  need to learn how to use it.,

First:  write  1 / [ 2sqrt(2) + 3sqrt(3) ] with a rational denominator by multiplying both

          the numerator and the denominator by the conjugate of the denominator: 

                   2sqrt(2) - 3sqrt(3):

     1 / [ 2sqrt(2) + 3sqrt(3) ]   ·   [ 2sqrt(2) - 3sqrt(3) ] /  [ 2sqrt(2) - 3sqrt(3) ]

=   [ 2sqrt(2) - 3sqrt(3) ] / [ 8 - 27 ]

=   [ 2sqrt(2) - 3sqrt(3) ] / - 19

=   [ 3sqrt(3) - 2sqrt(2) / 19

Second:  wrtie  sqrt(2) + sqrt(3)  with a denominator of 19:

=   19sqrt(2) / 19  +  19sqrt(3) / 19

Add together these two results:  [ 3sqrt(3) - 2sqrt(2) / 19  +  19sqrt(2) / 19  +  19sqrt(3) / 19

=  [ 22sqrt(3) + 17sqrt(2) ] / 19

a  =  17     b = 22     c  =  19