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helpful, thank you, you helped me learn a bit too :D

Task 6

Let  N  = 37

We can write this as  (x + y)  =  (49-  12)

So....note that

(49 -  12) ^2  =    49^2  - 2(49*12) + 12^2  =   2401  - 1176  + 144    = 1369  =  37² 

Task 5

Let  the polynomial  be    x^2  - 4x  + 4

Let the  linear binomial be    ( x - 2)

Part 1

               x  -  2

x - 2   [   x^2  -  4x  + 4  ]

              x^2   -2x

             _____________

                       -2x   + 4

                       -2x   + 4

                       _______

                                 0

Part 2

f(a)   = f(2)  =   2^2  - 4(2)  + 4   =    4 - 8  + 4   =  0

Part 3 

Since  we obtained a "0"  with polynomial long division......then x  = 2   is  a  root. The Remainder Theorem says that if f(a)  = 0, then  "a" is a root.  This is borne out by the fact that f(2) = 0  where  "a"   = 2.

Thank you so much!! I am not good with these! Thanks for your help!

tangent P  =   opp/adj =  24/18

tangent Q  = adj/opp = 18/24

tan 26/10

tan B  = 10/26

Note...on this silly  question.......one leg of the right triangle is LONGER than the hypotenuse...IMPOSSIBLE!!!!

Last one   .....we can use the tangent here 

tan 19  =   height  / 4100     multiply both sides by  4100

4100 * tan 19   = height  ≈   1412 ft

Thank you so much!!!

First one  ....Note that

sin L  = 21/75  ⇒    sin^-1 (21/75)   = L

cos L  = 72/75 ⇒  cos-1 (72/75)  = L

tan L = 21/72  ⇒  tan-1 (21/72) = L

Second one

sin A  = 35/37

cos A  = 12/37

thanks again

Task 3

y  =   (x + 3)

        ______

         (x - 2) 

The horizontal asymptote  occurs at  y  = 1

The vertcal asymptote occurs at  x  = 2

The x intercept is  at  x  = -3

The y intercept is  at   (-3/2)

Here's a  graph : 

https://www.desmos.com/calculator/4l3egciy58

ty very much

Here's a few

Task 2

y  = x^2       (quadratic)

y = x            (linear)

Sub  the   second equation into the first for y

x  = x^2       subtract x from  both sides

0  = x^2  - x      

x^2  - x   =  0   factor

x ( x - 1)  =0

Set each factor to 0  and solve for  x

x  = 0        x  - 1  =  0   

                add 1 to  both sides

x = 0          x  = 1

So....when x  = 0, y  = 0

And when x  = 1, y  = 1

So....the solutions are  (0,0)  and (1,1)

Here's a graph ;

https://www.desmos.com/calculator/ldpimov7kd

Parm how do u know my "tone" is? You are not sitting next to me right now.

Okay! Thank you for the help. I was not sure!

First one  ...this is a  30-60-90 right triangle

The side opposite  the  60° angle will be  √3 * times the side opposite the 30° angle  = √3 * 8  = 8√3  =  √192

So....you were correct....!!!!!!!

Second one

We can use the tangent, here ;

tan (20°) = 4 / x         we can rearrange this as

x  =  4  / tan(20°)  = 10.99

Let  O  = (0,0)

Let A  = (2,3)

Let B = (1,1)

Using the "tip to tail" method....let   C  = (2 + 1, 3 + 1)  = (3,4)

So......the length of diagonal  OC  =   √ [ (3 - 0)^2  + (4-0)^2  ] =   √[9 + 16 ] = √25  =  5   units

And the length of diagonal AB  = √ [ (2-1)^2 +  ( 3 - 1)^2 ]  = √ [ 1^2  + 2^2]  =  √5 units

The perimeter  is   2*OA  = 2 * √ [ 2^2 + 3^2 ] = 2 *√13    +    2* AC  = 2* √ [ (2 - 3)^2 + (3 -4)^2]  = 2 √2

So....the total perimeter is     2  [√13  + √2 ]  ≈  10.04  units

Here's a pic :

6)When a square with area 4 is dilated by a scale factor of k we obtain a square with area 9. Find the sum of all possible values of k.

7)In triangle ABC, AB = BC = 5 and CA = 6.

 What is the perimeter of triangle A3B3C3?

8)D, E and F are the midpoints of BC, CA, and AB respectively. A dilation maps triangle ABC to triangle DEF. What is the scale factor of this dilation?

9)In triangle XYZ, XY = 20, YZ = 15 and XZ = 25. Y is reflected over XZ to Yprime. Find YYprime.

I dont want to be mean but i dont think this has anything to do with math so please stop posting non math problems on here

Remember that    a(x1, y1)   - b(x2, y2 ) =   ( ax1 - bx2, ay1  - by2 )

a(3,4)  - b(2,4)   =  (8,10)

So we have that

3a  - 2b  = 8      and

4a  - 4b  = 10

Multiply the first equation by - 2  all the way through

-6a  + 4b  =  -16     

4a  -   4b   =  10        add these

-2a   =  -6     divide both sides by  -2

a  = 3

And using   3a  - 2b  = 8  to  find b, we have that

3(3)  - 2b  = 8

9  - 2b  = 8 

-2b  = -1

b  = 1/2

OK....no prob   !!!!

Here's one possibility  [ but maybe not the only one ]

The sum of the interior angles  = 540°

Let the smallest angle  = 60°

Let the next two angles   =  60.5°    and 61.5°

Note that the median  =  [ 60.5  + 61.5)  / 2  = 61°

Thw sum of the angles, so far, is  182° 

So...the sum of the other two angles will be     540  - 182  =  358°

Let the next two angles  = 179°   

The sum of the angles is 

[60 + 60.5 + 61.5 + 179 + 179]  =  540°

The median  =  61°    and   there is only one mode,  179°

Thank you! Thank you! I've been stuck on them. I appreciate the help!

First one :

We know an opposite side and an adjacent side....the tangent seems most appropriate, here

So we have

tan x  = 11/9       use the tan  inverse  (atan) to find x

atan (11/9)  =  x ≈  50.71° = 51°

Second one....

tan Y  = 6/7          tan Z  = 7/6

Third  one :

We know an angle, the hypotenuse and we are looking for an adjacent side to the angle....the cosine is what we need

cos (26)  = x / 10   multiply both sides by 10

10 cos (26)  = x   ≈  9

Fourth one :

We know an angle, the adjacent side....and we are looking for the hypotenuse....the cosine, again....

cos36  = 12 / x       divide both sides by cos 36......multiply both sides by  x

x   =  12  / cos 36  ≈  14.8

Last one

We know an angle, an adjacent side....and we are looking for the opposite side.....the tangent seems good

tan  40  = x / 5       multiply both sides by  5

5 tan (40)  = x  ≈  4.2