CPhill

Quadratic Trinomial  =  3 terms with highest power exponent  =  2

Cubic Trinomial  =  3 terms  with highest power exponent = 3

( c - d)^10      7th term

The 7th term is as folows :

C (10,6) * c^4 * d^6  =

210c⁴d⁶

BTW.... here's the full expansion  as a check ;

c^10 - 10 c^9 d + 45 c^8 d^2 - 120 c^7 d^3 + 210 c^6 d^4 - 252 c^5 d^5 + 210 c^4 d^6 - 120 c^3 d^7 + 45 c^2 d^8 - 10 c d^9 + d^10

 6x^2 + 5x

 x^2 + 5x^(1/5)......no  fractional powers allowed

 -x^2 + 5x

 -7x^2 + (5/3)x

 y

 4x^3 + y

2 + s

- x^2 - sqrt of x + 2    .......  sqrt ( x)  = x^(1/2).....no fractional exponents

A plolynomial    has constant coefficients associated with each variable and  non-negative integer powers on these variables

So....the polynomials are in  red

Point A is located at (1,4). Point P at (3,5) is 1/3 the distance from A to point B.What are the coordinates of point b

Let m be the x coordinate of B

So

x coordinate of A   +  3 times ( x coordinate of P - x coordinate of A)  =  m

1   +  3 ( 3 - 1)  =  m

1  + 3(2)  = m

1 + 6  = m

7  = m

Similarly

Let n be the y coordinate of B

So

y coordinate of A   +  3 times ( y coordinate of P - y coordinate of A)  =  n

4  +  3 ( 5 - 4)  =  n

4  + 3(1)  = m

4 + 3  = m

7  = m

So B =  (m, n)  =  (7, 7)

Draw a circle with a radius of 15  and a center at (-3, 2)

The equation of this circle is

(x + 3)^2  + (y - 2)^2   = 225

Let point  L  be located at   ( m, 0)    where m is positive

So...we can determine m  by letting  y  = 0  in in the above equation...so we have...

(m + 3)^2  + (0 - 2)^2  =  225

(m + 3)^2  + (-2)^2  =  225

(m + 3)^2  + 4   =  225     subtract 4 from each side

(m + 3)^2  =  221            take the positive root of both sides

m  +  3   =  √ 221          subtract 3 from  both sides

m = √ 221  - 3

So  L =  ( √ 221  - 3 , 0 )

And we can find P thusly

[ -3 +  (1/3) ( √ 221  - 3 - - 3) ,  2  + (1/3)(0 - 2) ]  =

[ - 3 +  (1/3) (  √ 221 )  ,  2 + (1/3)(-2)  ]  =

[ -3 + √ 221/3   ,  2 - 2/3 ]  =

[ -3 + √ 221/3  , 4/3 ]

See the graph here : https://www.desmos.com/calculator/ggoirs81vs

We can use the Law of Cosines here

x^2  =  10^2 + 10^2  - 2(10)(10)cos (48°)

x^2  =   200 - 200cos(48°)

x = √ [  200 - 200cos(48°) ]   ≈ 8.135

Point C is located at the origin. Point Q at (-1,-2) partitions CD in a ratio of 1:6. What are the coordinates of point D?

Think about this

QD  must be 6 times the distance CQ

Let  m be the x coordinate of D

So

x coordinate of Q +  6 times ( the difference in the x coordinate of Q - the x coordinate of the origin)  = m

So

-1 +  6 (-1 - 0)  =  m

-1 + 6(-1)  = m

-1 - 6  =  m

-7  = m

Similarly

Let  n be the y coordinate of D

So

y coordinate of Q +  6 times ( the difference in the y coordinate of Q - the y coordinate of the origin)  =  n

So

-2  + 6(-2 - 0)  =  n

-2 + 6(-2) = n

-2 - 12  = n

-14  = n

So....D  = (m ,n)  =  (-7 , -14 )

IMHO, It's easier to figure point Q if we realize that it also must be equal to 1/5 of the distance from H to J

So....we can find point Q as follows :

Q  =   

[ -5  +  (1/5) ( 5 - - 5) ,  -6 + (1/5) ( -1 - -6)  ]  =

[ -5 + (1/5) (10)  , - 6 + (1/5) (5) ]  =

[ - 5 + 2, - 6 + 1 ]

{ -3, - 5)  =  Q

1)  If this were a right triangle...it would have to be that

8^2 + 9^2  = 12^2      but

64 + 81  >  144

145 > 144

Thus....angle ABC isn't quite large enough to be 90°

And the other two angles are definitely < 90° each....so this is an acute triangle

2) In isosceles triangle APS, we have  angle P is equal to 54 degrees and AP = AS. Which is larger, AP or PS?

                                                       A

                                           P                   S

Angle  APS  = Angle  ASP......So....angle  PAS =  [ 180  - 2*54 ] =    180 - 108  = 72°

And in a triangle.....the greater side is opposite the greater angle......and angle PAS > angle ASP.....so ...... PS > AP

3)In actute triangle ABC, we know  AB = 7, BC = 8 and that CA is the shortest side. What is the smallest possible integer value of CA?

Since this is an acute triangle, we must have that

7^2  + CA^2 >  8^2

49 + CA^2  > 64

CA^2 > 15

CA > √15  ⇒    CA > ≈ 3.8

So....the least integer value for CA  is  4

4)In obtuse triangle ABC, we know  AB = 7, BC = 8 and that CA is the longest side. What is the smallest possible integer value of CA?

Since the triangle is obtuse and CA is the longlest side, we must have that

AB^2 + BC^2  <  CA^2

7^2  + 8^2  < CA^2

49 + 64 < CA^2

113 < CA^2

√113  < CA

CA > √113 ≈ 10.6

So ......   the smallest possible integer value for CA  is 11

5)Two diagonals of a parallelogram have lengths 6 and 8. What is the largest possible length of the shortest side of the parallelogram?

The diagonals will bisect each other......

So we have 4 triangles that have two sides of 3 and 4

And....by the triangular inequality, we have that

Longest side + Intermediate side > Shortest side

4 + 3  >  Shortest side

7 > Shortest side

Shortest side < 7

So......the largest possible value for the shortest side must be < 3

6)Two sides of an acute triangle are 8 and 15. How many possible lengths are there for the third side if it is an integer?

Longest possible side length  

8 + 15  >  Third side length

23 > Third side length

Third side length < 23

Shortest possible side length

8 + Shortest side length > 15

Shortest side > 7

So.....the possible integer side lengths are

8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 19,, 20 21, 22

So.....15 integer lengths are possible

Don't know how to do 7....maybe someone else has a solution......!!!

x = yz + y

Make y the subject.....note that the right side can be factored as

x  = y (z + 1)       now....divide both sides by (z + 1)   and we have that

x / ( z + 1)   =  y