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As follows

"Suppose f(x) is a polynomial of degree 4 or greater such that f(1)=2, f(2)=3, and f(3)=-1. Find the remainder when f(x) is divided by (x-1)(x-2)(x-3)."

f(x) = p(x)*(x-1)(x-2)(x-3) + r(x)    where r(x) is the remainder (which must be a quadratic at most).  say r(x) = a*x^2 +b*x + c

2 = a(1²) + b(1) + c          or   a + b + c = 2             (1)

3 = a(2²) + 2b + c            or   4a + 2b + c = 3          (2)

-1 = a(3²) + 3b + c           or  9a + 3b + c = -1          (3)

(2) - (1):    3a + b = 1           (4)

(3 - (2):     5a + b = -4          (5)

(5) - (4)    2a = - 5     or    a = -5/2 

Use this in (4) to find       b = 1 - 3(-5/2)  or   b = 17/2

Use these two in (1) to find c.  I'll leave the rest to you.

Use this to help

\(R_1=Pink,\quad R_2=Blue,\quad R_3=Green, \quad R_4=Purple\)

Area of circle is    100pi

Area of rectangle in the middle is 12*10=120

B+B+G+G+Pu-Pu+120 = 100pi

simply to get

B+G=50pi-60

Pink=G+B-Purple +120

Pink+ purple = G+B+120

substitute

Pink+ purple = 50pi-60+120 = 50pi+60

(Pink+Purple) - (Blue+green) = (50pi+60) - ( 50pi-60) = 120 units squared.

I really did not need to do all this.  

Just looking at the pic I can see that         pink - blue - green +Purple = 120

Draw it up and let the side length of the square be 2. That is an easy number to work with.

Find lengths AC and AP and AQ using Pythagoras's theorem and basic logic.

What is angle ACB ?

Now it is easier to find angle CAP and then double it than it is to find angle QAP.

You can find angle CAP using either sine rule or cosine rule.   (the answer will be approx.)

Edit (added):  you want sine of it, not the angle itself, so that is easier and it will be exact.

Now Age26 you should have the info you need, if you want to clarify something then ask.

Please no one answer over me.

This polygon has 24 sides.

Let r and s be the roots of 3x^2+4x+12=0. Find r^2+s^2.

Hello Age2!

\(3x^2+4x+12=0\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {-2 \pm \sqrt{16-4\cdot 3\cdot 12} \over 2\cdot 3}\)

The parabola has no real solution.

\(x=-\frac{2}{6}\pm \frac{i}{6}\cdot \sqrt{128}\\ x=-\frac{2}{6}\pm \frac{i}{6}\cdot \sqrt{2\cdot 2^6}\\ x=-\frac{2}{6}\pm \frac{8i}{6}\cdot \sqrt{2}\\ \color{blue}x=-\frac{1}{3}\pm \frac{4}{3}\cdot \sqrt{2}\cdot i\\ \)

\(r=-\frac{1}{3}+ \frac{4}{3}\cdot \sqrt{2}\cdot i\\ s=-\frac{1}{3}- \frac{4}{3}\cdot \sqrt{2}\cdot i \)

\(r=-\frac{1}{3}+ \frac{4}{3}\cdot \sqrt{2}\cdot i\\ s=-\frac{1}{3}- \frac{4}{3}\cdot \sqrt{2}\cdot i \)

\(r^2=\frac{1}{9}-\frac{8}{9}\cdot \sqrt{2}\cdot i+\frac{16}{9}\cdot 2\cdot 1\\ s^2=\frac{1}{9}+\frac{8}{9}\cdot \sqrt{2}\cdot i+\frac{16}{9}\cdot 2\cdot 1\)

to be continued

  !

Find all positive integer values of c such that the equation x^2 - 5x + c only has roots that are real and rational. Express them in decreasing order, separated by commas.

Hello Guest!

\(x^2 - 5x + c =0\)

         p        q

\(x=-\frac{p}{2}\pm \sqrt{(\frac{p}{2})^2-q}\)

\(x=2.5\pm \sqrt{6.25-c}\)

\(c\in \mathbb{N_0}\ |0\le c\le6\)

Let the area of the square XSWZ be 529 u².          side = √529

Area of the square QEZA = 529 - 273 = 256 u²      side = √256

Yellow area     A = [23(23-16)] /2 + [16(23-16)] /2  

help please somone find the value of x and y

Hello Guest!

\(\alpha=180°-150°\\ \alpha=30° \)

\(sin\ \alpha=sin\ 30° \\sin\ 30°=\frac{5}{y}\\ 0.5=\frac{5}{y}\\ \color{blue}y=10\)

\(x=\sqrt{10^2-5^2}\\ x=\sqrt{75}=\sqrt{3\cdot 5^2}\)

\(x=5\cdot \sqrt{3}\approx 8.660254\)

  !

Your answer is correct IF the vertical lines are parallel.

1 - 68^2  -  66^2 =4,624  -  4,356 = 268

(1001, 1771, 2002, 2772, 3003, 3773, 4004, 4774, 5005, 5775, 6006, 6776, 7007, 7777, 8008, 8778, 9009, 9779) >>Total = 18 such numbers

My original answer (V = 1/3 * Ah ≈ 40.033 u³) is NOT correct!!!                                                                                          (link:: https://web2.0calc.com/questions/volume-question-please-explain)

The correct answer is:    V_(CFAH) = 40 u³  

Let   m∠DAQ  =  a

Let   m∠PAQ  =  b

Let   m∠PAB  =  c

Notice that    △DAQ  ≅  △BAP,    and so    m∠DAQ  =  m∠PAB    which means    a  =  c

Then since the measure of an interior angle of a square is 90°...

a + b + c  =  90°

                               Since  a = c  we can substitute  a  in for  c

a + b + a  =  90°

                               Combine like terms

b + 2a  =  90°

                               Subtract  2a  from both sides to solve for  b

b  =  90° - 2a

And so....

sin(b)   =   sin(90° - 2a)

                                             Now we can use the rule that says:  sin(90° - θ)  =  cos(θ)

sin(b)    =   cos(2a)

                                             Now we can use the double-angle formula for cos

sin(b)   =   1 - 2sin²(a)

                                             And  sin(a)  =  opposite / hypotenuse  =  1 / sqrt(5)  **

sin(b)   =   1 - 2(1/5)

sin(b)   =  3/5

**If you would like more explanation on this part please feel free to ask!

A = 6 * 13 - (3 * 6) * 1.5 = 51 in²

sin(PAQ) = 0.6 

The sine of angle PAQ is 0.6

ST * SU = SV * SW           solve for SV using quadratic formula

You are right.

My answer   4C2*3  *  2    was correct but my simplification of it had a careless error

4C2=6 (not 4)

6*3*2= 36  

Instead of perpendicular, read vertical. 

A number is doubled     2x

and then the product is increased by 6       2x+6

This sum is then multiplied by -3           (-3)(2x+6)

and the result is 54 greater than three times the opposite of the number.      (- 3)(2x+6) - 54  = (3)(-x) 

(if the opposite of a number is the number * -1 )         x = -24

Srry. It's wrong because the answer key says it's 3 times 6 times 2 which is 36.

C(3 1) is three and C(4 2) is 6 so 3 times 6 times 2 (two ways to pair up the men and women left)

It's fine at least I got it now :-)

        (322 * 5280 *  5280 ft²) / 7 200 000 000 persons = ..........    ft² / person

Something's missing here!!!

The sum of those two (given) angles is 180º, but that does not necessarily mean that the perpendicular lines are parallel!!!