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Hi Lightning  

\(P(x)=ax^2+bx+c\\ P(4-\sqrt{11})=a(4-\sqrt{11})^2+b(4-\sqrt{11})+c=0\\\qquad \text{Where a,b and c are integers}\\ a(4-\sqrt{11})^2+b(4-\sqrt{11})+c=0\\ a(16+11-8\sqrt{11})+b(4-\sqrt{11})+c=0\\ a(27-8\sqrt{11})+b(4-\sqrt{11})+c=0\\ 27a+4b=-c\quad(1)\\ -8a\sqrt{11}-b\sqrt{11}=0\\ -8a-b=0\\ b=-8a \quad (2)\\ \text{sub two into one}\\ 27a+4*-8a=-c\\ -5a=-c\\ c=5a\)

So the equation becomes

\(P(x)=ax^2-8ax+5a\\ P(x)=a(x^2-8x+5)\\ \)

Find P(3) and P(4) and do the division.

You can finish it   

(oh, you do need to check my working, I have not done so. )

Maybe he/she means that they need help on their prep for their final exam?

THAT is funny.......the only thing I have posted is here.    Nothing has been sent anywhere or anyway else.

TheFA has complained that you are harassing him/her.

Just thought I'd let you know.   

Thanks EP,

Guest, perhaps you had something like this in mind?

...and TheFA is cheating.....

Cause we are suckers ?    

cos c ~ .279

I might as well post my solution.

We can use a very handy formula, which is very important to remember, \(A=rs\), where \(A\) is the area of the figure, \(r\) is the inradius when it is a complete circle, and \(s\), the semi-perimeter of the figure, or you can call it as, "half the perimeter."

Now, this is where the tricky part comes in. we can reflect the second part of the figure, like this:

After drawing this figure, we can find the area of one triangle, the multiply it by two, to find the area of the rhombus.  Drawing the altitude of sixteen to the base, we find that the area is \(\frac{1}{2}*16*15=120\). Multiply it by two since there are two triangles, to get \(2*120=240.\)And, using the Pythagorean Theorem, that gives us \(\sqrt{8^2+15^2}=\sqrt{17^2}=17.\) Multiply it by four to find the whole(full) perimeter, and divide it by two to find the semi-perimeter, to get \(17*2=34.\) Now, we found \(A\) and \(s\), and we divide it to get \(r\), to get \(\frac{240}{34}=\boxed{\frac{120}{17}}.\)

A spinner has four equally sized sectors that are numbered 1, 2, 6, and 6. The spinner is spun twice and the product of the outcomes is found.

A fair decision is to be made about which one of four flowers will be planted.

The flower options are peonies, roses, asters or petunias.

Which description accurately explains how a fair decision can be made in this situation?

Let's look at a table of possible outcomes

     1    2    6     6

1   1    2    6     6

2   2    4   12   12

6   6   12  36   36

6   6   12  36   36

Now  ....let's consider each  situation

If the product of the spins is 6, peonies will be planted. If the product is 12, roses will be planted. If the product is 36, asters will be planted. If the product is 1, 2 or 4, petunias will be planted.

P(prod = 6)  =1/4     P(prod = 12)  = 1/4     P( prod = 36)  =  1/4  P(prod = 1, 2 or 4) = 1/4

We can stop here.....this provides a fair division....each flower has an equal chance of being planted

1. A hat contains 5 balls. The balls are numbered 1, 3, 4, 5, and 7. One ball is randomly selected and not replaced, and then a second ball is selected. The numbers on the 2 balls are added together.

A fair decision is to be made about which one of two candy bars to purchase, using the sum of the numbers on the balls.

The candy bar options are Choco Delight or Go Nuts.

Which description accurately explains how a fair decision can be made in this situation?

If the sum of the balls is less than 9, purchase Choco Delight. If the sum is 9 or greater, purchase Go Nuts.

If the sum of the balls is even, purchase Choco Delight. If the sum is odd, purchase Go Nuts.

If the sum of the balls is a factor of 40, purchase Choco Delight. If the sum is not a factor of 40, purchase Go Nuts.

If the sum is a multiple of 4, purchase Choco Delight. If the sum is not a multiple of 4, purchase Go Nuts.

1 3 4 5 7

Possible outcomes

(1, 3)  (1, 5)  (3,4) (3,7)   (4,7)

(3,1)   (5, 1)  (4,3)  (7,3)  (7, 4)

(1, 4)  (1, 7)  (3,5) (4,5)  (5, 7)

(4, 1)  (7,1)  (5, 3)  (5,4) (7, 5)

Let's look at each situation 

P(sum< 9)  = 12/20     P (sum 9 or greater)  = 8/20   not fair

P(even sum)  = 12/20    P(odd sum)  = 8/20    not fair

P(sum is a factor of 40)  = 10/20   P (sum is not not a factor of 40) = 10/20

P(sum is a multiple of 4)  = 8/20    P (sum is not a multiple of 4) = 12/20

The third  answer provides a fair decision

Why are we answering your EXAM questions?????

This is an ARITHMETIC series with common difference, d = 14     a1 = -215

a47 = a1 + (47-1)(14)

       = -215 + 46(14) = 429

S₄₇ = n(a1+a47)/2 = 47(-215 +429)/2 = 5029

Since we have rational coefficients....

If 2 - √5  is a root then so is 2 + √5

If 4+ √10  is a root then so is  4 - √10

If 14 - 2√7  is root then so is 14 + 2√7

Lastly....if - √2  is a root then so is √2

So.....the smallest possible degree is  8

= 5^(-8+-10) = 5^-18

deleted...didnt really have  the info

i got the wrong answer. Do you know what the correct one is?

Since you don't specify base b, that means the answer will be the same in any base from 2 to 10:

11011_2 * (2 - 1) + 1001 = 100100 = 36_10

11011_3 * (3 - 1) + 1001 = 100100 = 252_10

11011_4 * (4 - 1) + 1001 = 100100 = 1040_10

11011_5 * (5 - 1) + 1001 = 100100 = 3150_10

11011_6 * (6 - 1) + 1001 = 100100 = 7812_10

11011_7 * (7 - 1) + 1001 = 100100 = 16856_10

11011_8 * (8 - 1) + 1001 = 100100 = 32832_10

11011_9 * (9 - 1) + 1001 = 100100 = 59130_10

11011_10*(10-1) + 1001 = 100100 = 100100_10 

4.

y  is positive when  x  is in the interval  (-∞, -2)

y  is positive when  x  is in the interval  (0, 4)

3.

The yearly attendance at a ballpark is shown in the table.

Which answer describes the average rate of change from Year 2 to Year 5?

average rate of change from year  2  to year  5   =   \( \frac{\text{333,700 people - 298,300 people}}{\text{5 years - 2 years}}\)

average rate of change from year  2  to year  5   =   \( \frac{\text{35,400 people}}{\text{3 years}}\)

average rate of change from year  2  to year  5   =   \( \frac{\text{11,800 people}}{\text{1 years}}\)

average rate of change from year  2  to year  5   =   11,800 people per year

\(\text{$3$ combos of the all one color}\\ \text{$3\cdot 2=6$ combos of 2 colors}\\ \text{$1$ combo of all 3 colors}\\ \text{$10$ in total}\)

EP: The question says "Less than a 1000". So, Cphill's answer is correct.

Thank you, I understand fully now. 

Ahemm, mods, pls lock now.

5 + 6 means there are 11 balls in total.

Of these 11, 5 are white, so the probability of the first ball that's drawn being white is 5/11.

I'll assume the ball isn't put back into the box. So now you have 10 balls left, 4 of which are white.

('cause you already removed one white ball, the total number of balls and the number of white balls goes down)

Which means that the second pull has a 4/10 chance of being a white ball.

Both of these events need to happen, so you multiply the chances:

(5/11) * (4/10) = 2/11