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23.   The degree of the polynomial will tell you the number of roots .....however....some may be repeated....(multiplicities )

24. Power functions have the form  y  =  ax^b

So   (1/4)x^0.5    is a power function  where a  = (1/4)  and b = 0.5

2000  = A ( 1 + .07/2)^(20*2)

2000 = A ( 1.035)^(40)  

2000 /  ( 1.035)^(40)  =  A  ≈  $505.14

at least one is incorrect. 

8000  =  A  (1.10)^20

Divide both sides by  (1.10)^20

8000/(1.10)^20  =  A  ≈  $1189.15

C  =  22pi ft

C  =  2 * pi  * r       ....so....

22 pi   =  2 * pi * r         divide out pi

22 =  2 * r         divide by 2

11 ft  =  r         22 ft   =    d

And   Area   =  pi  (r^2)   =  121 pi   ft^2

Thanks, Tiggsy...that's one I'm going to study closely.....!!!!

The area  of a rhombus  =   product of the diagonals / 2

So

120  =   (10) D  / 2

120  = 5 D   divide both sides by 5

24  = D    =  the other diagonal

And the diagonals bisect each other......and these bisected diagonals form legs of a right triangle  and the side of the rhombus  forms the hypotenuse

So....the legs are  10/2  and 24/2  =   5   and 12

And by the Pythagorean Theorem.....

5^2 + 12^2  =  side^2

√ [ 5^2 + 12^2 ]  = side

√  169  =  side

13  = side

And since there are 4 sides....the perimeter is 4 * 13  =   52 units

The four interior angles will sum to 360“

So  call the smallest angle x  and the largest angle 2x

So

90 + 90  +  x  +  2x  =  360

180  +  3x   =   360     subtract 180 from both sides

3x  =  180    divide both sides by 3

x  =  60°

And the largest angle  = 2 (60°)   =  120°

21.    I believe we can use synthetic division if we have a linear factor as a divisor.....but....it might not always work if we have a divisor polynomial that cannot be factored into separate linear factors

22.   The conjugate  2  - √ 3  will also be a root.....this is always true  with roots of the form a + √ b

23.   If we have a single variable polynomial, the highest power  found on that variable is the polynomial degree

Love these Posts. Already said that but will say it again. 😁😁

lol thx

HAHAHA!!!!....that's a lot of midpoints   !!!!

Midpoint of  PQ  =  (8  + 48) / 2  =  56/2  = 28  =  B

So  midpoint of   BQ   =  ( 28   + 48)/ 2  =  76/ 2  =  38  =  C

And midpoint  of PC  =  (8 + 38) / 2  =   46/ 2   =  23   =  D 

Thanks!

3x + 2y = 5

5x + 3y = 7

We can choose to eliminate y thusly  :

Multiply the first equation  by  3   and the second by -2

9x     +   6y   =   15

-10x  -   6y    =  -14      add these

-1x   =  1       divide both sides by -1

x  =  -1

And using   3x +  2y  =  5   to find y  we have

3(-1)  + 2y   = 5

-3 + 2y   =  5       add 3 to both sides

2y  =  8     divide both sides by 2

y  =  4

So  {x, y}   =  {-1, 4 }

4^x  - 2^x  =  56

(2^2)^x  - 2^x  = 56

2^(2x) -  2^x  -  56  = 0

(2^x)^2 - 2^x  - 56   = 0

Let  u    = 2^x         so      u^2  =  (2^x)^2

u^2  -  u  -  56   = 0       factor

(u - 8) ( u + 7)  = 0

So    either

u  -  8   =   0               or           u +  7    =  0

u =  8                                        u  =  - 7             

Substitute back

2^x  =  8                                  2^x   =  - 7       this isn't possible

x  =  3

Check

4^3  -  2^3   =  56  ??

64   -  8     =  56        correct  !!!

One thing about these identities is that there may be more than one way of proving them

So...let's try this....

 2cos^2 (x) = cos(2x) +1 

2( 1 - sin^2 (x))  =  cos (x + x)  + 1

2 - 2sin^2 (x)    =   cos(x)cos(x) - sin(x)sin(x)  + 1      subtract 1 from both sides

1 - 2sin^2(x) = cos^2(x) - sin^2(x)          add 2 sin^2x to both sides

1  =  cos^2(x)  + sin^2(x)        which is true

Here's an alternative, vector type solution.

Let $\displaystyle \hat{\imath}$ be a unit vector in the direction PQ and $\displaystyle \hat{\jmath}$ a unit vector perpendicular (upwards in the diagram), to this.

Then, $\displaystyle \underline{PB}=2\cos(60)\hat{\imath}+2\sin(60)\hat{\jmath}=\hat{\imath}+\sqrt{3}\hat{\jmath}$ , and $\displaystyle \underline{PC}=3\hat{\imath}$ . (All angles in degrees.)

In the triangle PBC, $\displaystyle \underline{PB}+\underline{BC}=\underline{PC}$, so $\displaystyle \underline{BC}=3\hat{\imath}-(\hat{\imath}+\sqrt{3}\hat{\jmath})=2\hat{\imath}-\sqrt{3}\hat{\jmath}$ .

Let the vector $\displaystyle \underline{CA} = \alpha\hat{\imath}+\beta\hat{\jmath}$ , then since CA is at right angles to BC, the scalar product $\displaystyle \underline{CA}.\underline{BC}=0$ ,

so,  $\displaystyle 2\alpha-\sqrt{3}\beta = 0$ ...................... (1).

The vector $\displaystyle \underline{AQ}$ will be some fraction of the vector $\displaystyle \underline{RQ}$ , $\displaystyle \underline{AQ} = k\underline{RQ}$, say,

so $\displaystyle \underline{AQ}=k(5\cos(60)\hat{\imath}-5\sin(60)\hat{\jmath})=k \left(\frac{5}{2}\hat{\imath}-\frac{5\sqrt{3}}{2}\hat{\jmath}\right)$.

Finally, (almost), in the triangle CAQ,  $\displaystyle \underline{CA} +\underline{AQ}= \underline{CQ}$ ,

so, $\displaystyle \alpha\hat{\imath}+\beta\hat{\jmath}+k\left(\frac{5}{2}\hat{\imath}-\frac{5\sqrt{3}}{2}\hat{\jmath}\right)=2\hat{\imath}$ .

Equating coefficients, $\displaystyle \alpha +\frac{5k}{2}=2\quad \text{ so, }\quad\alpha = 2-\frac{5k}{2}$,

and $\displaystyle \beta -\frac{5k\sqrt{3}}{2}=0 \quad \text{so}\quad \beta = \frac{5k\sqrt{3}}{2}$.

Substituting into (1),  $\displaystyle 2\left(2-\frac{5k}{2}\right)-\sqrt{3}\left(\frac{5k\sqrt{3}}{2}\right)=0$ , so $\displaystyle 4-\frac{25k}{2}=0,\quad k=\frac{8}{25},$

so $\displaystyle AQ = \frac{8}{25}\times5=\frac{8}{5}=1.6$ .

Tiggsy

5 + 3/5   =   28/5

So  we have

(28 / 5)  / (4 / 5)     invert the second fraction  and multiply

(28 / 5) * (5 / 4)   = 

( 28 / 4 ) *  ( 5 / 5)  =

7   *    1   =

7

The answer would be 7.

You can turn the mixed number 5 3/5 into a fraction (28/5) and then divide. 

:D

Since DC  can be considered as one base of parallelogram   ABCD  and  EG is the height drawn to this base.....the area  is  DC * EG =  27 * 16  = 432 units^2

But BC  can be considered as another base  and  FH  is a height drawn to this base....and since the area is constant......we have that

432  = BC * FH     but   AD  = BC  = 18  ....so....

432  = 18 * FH      divide both sides by 18

24  = FH

No problem :D

Yeah, Julius....I forgot about  the second  "≤" ....2pi is also a solution   

Thanks for noticing this  !!!

Oh also you said sin x =0 at 0 and π, but sin x hits zero and 0, π and 2π. Correct?

Good job, Ghosty  !!!