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x^2 - 8x + 5 = 0

x^2 - 8x + (8/2)^2 = (8/2)^2 -5

(x - 4)^2 = 16 - 5

(x - 4)^2 = 11

(x - a)^2 - b = (x - 4)^2 -11

This should be pretty simple... but anyways:

16 x 15 x 40 = 240 x 40 = 9600

(X - 5) + (X/2) + 6 = 13

(X - 5) + (X/2) = 7

X + (X/2) = 12

X = 8

3 : 4 : 6 ratio

 Wowwww I spotted a typo and Melody agreed!!!

Will it be fixed?

I get 8/2^7 !!!

Thank you that does make sense. I will solve more problems in this category

Hmm what does double trouble mean?

I'm not sure what "point slope" form is.

I'll give the form slope intercept.

y = ax+b

a is the slope, b is the y-intercept.

As we can see, the slope is (rise over run), and since the coordinates, (2,-1), and if you go up the next point on a grid is (1,1). Rise, 2, run, -1. 

Slope is thus -2.

y = -2x + b.

At a increase of slope 2, so for every 1 it goes left, it goes up 2.

We need to go left twice, so 2x2 = 4.

-1 + 4 = 3/

Thus the equation is

y = -2x + 3

Any grade between 65% and 94% on the 6th test will give you an average ranging from 75% to just under 80%.

CE = AC = 10x+18.

DE = 7x - 1

CD = 10x + 18 - (7x -1) = 3x +19

We don't actually need that, though.

Or we do:

3x + 19 = 7x - 1 since D is the midpoint of CE, so CD = DE

20 = 4x

x = 5

Now

BC = 9x-2

10x+18 - (9x-2) = x + 20

AB = x+20

x = 5

AB = 25

BC = 43

CD = 34

DE = 34

25 + 43 = 34+34 = 68 = 68!

Is the question: (x^2) * (y^3) =46656? OR: x^(2*y^3) =46656?

The average of his grades so far is: 77%

C grade: 75%-80%

Im not sure what "what grade would he need..." means, so I am guessing you mean percentages. Just test each of the integer degrees from there and you will get the span of percentages that will work to get him a C.

P(3) = 3^3 + 2(3^2) - 5(3) + 4

P(3) = 9 + 18 - 15 + 4

P(3) = 16

Wow thank you so much! I posted this question before and some random guest answered incorrectly, without any work either. So this helped me so much thanks!

answered.

I played with this on a spreadsheet for quite a while.

But this is that I ended up with

f(2)=1         n=2 works and for integers from 1 to 100 this is the only n that will give an output of 1

If f(n)=2    then this will work too.   So n can be 4          f(4)=2    f(2)=1   so    f(f(4))=1

by the same logic  f(8)   f(16)  f(32)  and f(64)  will all work.

so n can be  1,2,4,8,16,32 or 64

So I get 7.

I have broken my own guideline and given you a final answer, I hope you learn from it.

\(x^2-5x+4=(x-4)(x-1)\)

So this is what you need to solve:

\(\frac{A}{x-4}+\frac{B}{x-1}=\frac{2x+3}{x^2-5x+4}\)

Please allow asker guest to finish this for him/herself

Looks like a typo to me Cal  

A fair coin is flipped 7 times. What is the probability that at least 5 consecutive flips come up heads?

All heads   + 6 heads with 5 in a row + 5 heads in a row

All heads, just 1 way

6 heads

THHHHHH

HTHHHHH

HHHHHHT

HHHHHTH

4 ways

5 heads

TTHHHHH

HHHHHTT

THHHHHT

3 ways

Total number of possible tosses = 2^7

So what answer do you think I might finish with?

Please let HelpBot be the next to answer. 

Find the equation of the parabola that goes through the three points (0, 9), (1, 7) and (2, 7).

Hello Guest!

\(y=ax^2+bx+c\)

Put the coordinates of the three points in
the equation \(y=ax^2+bx+c\), and solve for a, b, and c.

I.    \(9=c\)

II.   \(7=a+b+9\)

III.  \(7=4a+2b+9\)

4 * II  .     \(28=4a+4b+36\)

  - III.         \(\underline{7=4a+2b+9}\)

                21  =           2b  + 27 

\(2b=21-27=-6\)

\(b=-3\)

\(7=a+b+9\)

\(7=a-3+9\)

\(a=1\)

The general equation  \(y=ax^2+bx+c\)

becomes the real equation \(y=x^2-3x+9\).

  !

Solve for p: 2(p + 1) + 8(p - 1) = 5p

Hello Guest!

\(2(p + 1) + 8(p - 1) = 5p \\ 2p+2+8p-8-5p=0\\ 5p=6\)

\(p=1.2\)

  !

The expressions are 1/(x - 1) and -3/(x - 4).