All Answers 356776 Answers

idk how with polynomials...

1. centimeter

2. kilogram

3. meter

4. liter

5. gram

6. milliliter

Just expand it our and then collect like terms.  I thought you cold do that yourself :/

I'm sorry, but I need a full polynomial form, not factored form.

\((2x-3i)(2x+3i)(x+8)(x+2)\\ =(4x^2-9i^2)(x+8)(x+2)\\ =(4x^2+9)(x+8)(x+2)\\ =(4x^2+9)(x^2+10x+16)\\ =4x^2(x^2+10x+16)\quad +\quad 9(x^2+10x+16)\\ etc\)

you to dark

have a good day man

I did try Gino's method as well but it didn't work.  IPerhaps I have problems with primary school arithmetic  

Heureka has it working below. I will check what he has done. Thanks Heureka. 

Thanks also to Hectictar and Gino, yes that was what I was one of the posts I was referring too.   

It was good to see Dragonlance's comment too. Another of our old favourites who has moved on.   

Thanks Chris but I still do not know the relevance of the overbar...

Oh my god this is so helpful! Thanks so much guest! You should join Web2.0, the people here are very nice and very helpful!

Yay. Good for you! I remember when I joined the Top Users List. (Only a few months ago in fact...)

And at that time, you only needed to have over 95 points to be in the Top Users List. Now you have to be over 140...

What is the degree of the function that generates the data shown?

\(\displaystyle \begin{array}{|r|rrrrr|} \hline x & y\\ \hline -3 & 159 \\ & & -130 \\ -2 & 29 & & 100 \\ & & -30 & & -72 \\ -1 & -1 & & 28 & & 48 \\ & & -2 & & -24 & \\ 0 & -3 & & 4 & & 48 \\ & & 2 & & 24 & \\ 1 & \color{red}d_0=-1 & & 28 & & 48 \\ & & \color{red}d_1=30 & & 72 & \\ 2 & 29 & & \color{red}d_2=100 & & 48 \\ & & 130 & & \color{red}d_3=120 & \\ 3 & 159 & & 220 & & \color{red}d_4=48 \\ & & 350 & & 168 & \\ 4 & 509 & & 388 & & \\ & & 738 & & \\ 5 & 1247& & & & \\ \hline \end{array}\)

\(\displaystyle \begin{array}{|rcll|} \hline y(x) &=& \binom{x-1}{0}\cdot {\color{red}d_0 } + \binom{x-1}{1}\cdot {\color{red}d_1 } + \binom{x-1}{2}\cdot {\color{red}d_2 } + \binom{x-1}{3}\cdot {\color{red}d_3 } + \binom{x-1}{4}\cdot {\color{red}d_4 } \\\\ &=& \binom{x-1}{0}\cdot ( {\color{red}-1 } ) + \binom{x-1}{1}\cdot {\color{red}30 } + \binom{x-1}{2}\cdot {\color{red}100 } + \binom{x-1}{3}\cdot {\color{red}120 } + \binom{x-1}{4}\cdot {\color{red}48 } \\\\ &=& -1 \\ &&+~ \underbrace{30 \cdot(x-1)}_{\text{max degree } 1} \\ &&+~ \underbrace{100 \cdot\left(\frac{x-1}{2}\right)\cdot \left(\frac{x-2}{1}\right)}_{\text{max degree } 2} \\ &&+~ \underbrace{120 \cdot \left(\frac{x-1}{3}\right)\cdot \left(\frac{x-2}{2}\right)\cdot \left(\frac{x-3}{1}\right)}_{\text{max degree } 3} \\ &&+~ \underbrace{48 \cdot\left(\frac{x-1}{4}\right)\cdot \left(\frac{x-2}{3}\right)\cdot \left(\frac{x-3}{2}\right)\cdot \left(\frac{x-4}{1}\right)}_{\text{max degree } \color{red}4} \\ &=& \mathbf{2x^4-3} \\ \hline \end{array}\)

expand see WolframAlpha:

If the difference \(d_i\) is constant, so the degree of the polynomial to calculate the coefficients is \(i\).

\({\color{red}d_4}=48\) is constant, so the degree is \({\color{red}4}\)

If the difference \(d_i \) is constant, so the degree of the polynomial to calculate the sum of the coefficients is \(i+1\).

\(\displaystyle \begin{array}{rrrrrr} x & y\\ -3 & 159\\ & & -130 \\ -2 & 29 & & 100\\ & & -30 & & -72\\ -1 & -1 & & 28 & & 48\\ & & -2 && -24 & \\ 0 & -3 & & 4 & & 48\\ & & 2 && 24 \\ 1 & -1 & & 28 & & 48 \\ & & 30 & &72 \\ 2 & 29 & & 100\\ & & 130\\ 3 & 159\\ \end{array} \)

If the function is a polynomial tabulated exactly (no decimal approximations), its degree can be found using a difference table, (see above).

If the function is linear, the entries in the first difference column (the third column in the table), will be constant

If it's a quadratic the entries in the second difference column will be constant, and so on.

In the table above, entries in the fourth difference column are constant, so it's a quartic.

The table with its differences can then be used to calculate the coefficients of the polynomial.

The change in momentum is 2(30-0) kg.m/s → 60 kg.m/s

The rate of change of momentum (= force) = 60/0.01 N → 6000 N

.

These symbols often confuse me, too.....

The second one says that AC is congruent to PR....however....the triangles are similar, not congruent, so this is false....for the same reason, the fourth one is also false....the triangles would have to be congruent for BC = QR

The "formula" to figure the new points is as follows :

New point  =  ( Vertex of triangle - Dilation Center)*Scale Factor + Dilation Center

So we have

[ ( -3 , 7) - ( 5, -5)] *14  + (5, -5)  =

[ (-8, 12) ] *14  + ( 5, -5)  =

(-112, 168) + (5, -5) =

(-107, 163)

[ ( -7 , -5) - ( 5, -5)] *14  + (5, -5)  =

[ (-12, 0) ] *14 + (5, -5) =

(-168, 0) + (5, -5) =

(-163, -5)

[ ( 9 , 3 ) - ( 5, -5)] *14  + (5, -5)  =

[ (4, 8) ] *14 + (5, -5)  =

(56, 112) + (5, - 5)  =

(61, 107 )

Here's the pre and post images :

139 / 65   ≈  2.14 hrs

261 / 70  ≈  3.73 hrs

167 / 55  ≈  3.04 hrs

Geno's method is known as the sum of differences method......this would have probably worked here, but it's a bit of work.....I like your approach better......

4 / 5    =   x  /  18      where x is the ounces of orange juice

Muliply both sides by 18

(18) * 4 /5  =   72/5    =  14.4 oz  of orange juice

Let angle A  = x

So B  = 3x

And C = 6x - 1

And these sum to 180°.....so

x + 3x + 6x - 1  = 180

10x  - 1  =  180      add 1 to both sides

10x  =  181       divide both sides by 10

x =  18.1° = A 

B =   3 (18.1)  =  54.3°

C =  6(18.1) - 1   =  107.6°

the side length of the cube   =   \(\sqrt[3]{216}\)  cm   =   6 cm

So we know...

volume of water + volume of cube   =   10 * 12 * 6  cm³   =   720 cm³

volume of water +  216 cm³   =   720 cm³

volume of water   =   504  cm³

10 cm * 12 cm * original height   =   504  cm³

original height   =   504 / 120  cm   =   4.2 cm           

These angles are supplementary, so...

(x + 1)°  +  (9x - 21)°   =   180°

x + 1 + 9x - 21   =   180

                                         Combine  x  and  9x  to get  10x , and combine  1  and  -21  to get  -20 .

10x - 20  =  180

                                         Add  20  to both sides of the equation.

10x   =   200

                                         Divide both sides of the equation by  10 .

x   =   20

So....

x + 1   =   20 + 1   =   21

9x - 21   =  9(20) - 21   =   159

area of triangle  =  (1/2) * base * height

The problem tells us...

base  =  27 mm

height  =  48 mm

So to find the area, just plug in the values we know for base and height.

area of triangle  =  (1/2) * 27 * 48      mm²        ....Can you finish it?  

area of trapezpoid   =   (first base + second base) / 2  *  height

area of trapezpoid   =   (6 + 9) / 2  *  4

area of trapezpoid   =   15 / 2  *  4

area of trapezpoid   =   60 / 2

area of trapezpoid   =   30       cm²

For two angles to be supplementary, they need to add up to 180 degrees.

180-156=24

The measure of the supplement of a 156° angle is 24°.