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Why do you seem so disappointed lol

OK....no prob....

Oh it was 3*a I overlooked that my bad I thought I was second guessing myself guess I was thanks CPhil

3a  = 18          divide both sides by  3

a =  6    

Expand   (x + 2) (x - 3) ( x + 5) =

( x^2  - x - 6) ( x + 5)  =

x^3 - x^2  -  6x + 5x^2 - 5x - 30 =

x^3 + 4x^2 - 11x - 30

So

a = 4      and     b = -30

Here is my attempt at this:

1/5 x 20 =4 pieces that have nuts in them
2/5 x 20 =8 pieces that are plain.

2/5 x 20 =8 pieces that have caramel in them.

3/5 x 20 =12 pieces that Andy will receive.
1/5 x 20 =4 pieces that Bert will receive.
1/5 x 20 =4 pieces that Charles will receive.

Since Andy cannot eat any nuts, his 12 pieces must be plain or caramel.
And since Charles also cannot eat nuts, his 4 pieces must be plain or caramel. Since he prefers caramel his 4 pieces must be caramel.
Since Bert is indifferent to plain or caramel, his 4 pieces must be nuts. 

The remaining 4 caramels + 8 plain ones must have gone to Andy.                                                                                

3.For what values of j does the equation (2x+7)(x-5)=-43+jx have exactly one real solution?

2x^2 - 3x - 35  =  -43 + jx      rearrange

2x^2 - (3 + j)x + 8 = 0

This will have one solution when the disriminant = 0  .......so....

(3 + j)^2 - 4(2)(8)  = 0

(3 + j)^2 - 64 = 0    

(3 + j)^2 = 64    take both roots

3 + j = ±8

3 + j = 8            or           3 + j = -8

j = 5                                   j = -11

2.Find all solutions to the equation x^2+29=10x.

x^2 - 10x + 29 = 0      complete the square on x

x^2 - 10x + 25  =  -29+ 25

(x -  5)^2 = -4         take both roots

x - 5 =  ±√-4

x - 5 = ±2i 

x = 5 ± 2i

Here's a few

1.Let r and s be the roots of x^2-6x+2=0 Find (r-s)^2

(r - s)^2 = r^2 - 2rs + s^2   =  r^2 + s^2 - 2rs      (1)

The sum of the roots  =   6/1 = 6

So (r+ s)^2 = r^2 + 2rs + s^2 = 36     (2)

The product of the roots = 2   = rs

So 2rs = 4

Therefore....using (2)

r^2 + 4 + s^2 = 36

r^2 + s^2 = 32

Therefore...plugging all of this into (1)....we have that

(r - s)^2  =  r^2 + s^2 - 2rs =    32 - 4   =  28

The magniude of the resultant is

√ [ 6^2 + 8^2 ]  =  

√100 =

10 km / hr

Thank you so much! I haven't had Calculus yet. My school does not teach a calculus class thankfully. I appreciate your help!

f(something) = -3, so, from the graph, something1 = 0 and something2 = -4

”something” is f(f(x)), so f(f(x)) = 0 and -4

This means f(somethingelse1) = 0 and f(somethingelse2) = -4

Fom the graph we see somethingelse1a = 2 and somethingelse1b = -6

Also somethingelse2 = -2 (only one value)

This means f(x) = 2, -6, -2

When f(x) = 2, x = 3 and -7

f(x) never reaches -6

When f(x) = -2, x = 1 and -5

So sum = 3 - 7 + 1 - 5 = -8

\(\text{You use the word velocity. Velocity is a vector.}\\ \text{Using a coordinate system where north is positive }y \text{ and east is positive }x \text{ we have}\\ \vec{v} = \dfrac{-12\hat{j} + 5\hat{j}}{0.25}= -28\hat{j}~km/hr\)

"What is the End Behavior and Average Rate of Change for each function?"

I wouldn't expect you to know the average rate of change unless you have had Calculus

Thus....unless.....you will just have to take my word for this part of the answer

1. f(x) = x²

End behavior = up at both ends.....average rate of change =  2x

2. f(x) = |x|

End behavior = up at both ends   ......average rate of change =    x / √x^2

3. f(x) = √x

End behavior....on the left end, x =0 and y = 0     at the right end, "up"  .....avg rate of change =  1 / [2√x ]

4. f(x) = log x

End behavior .....tends toward neg  infinitiy as x approaches 0 from the right.....up on the right

Avg rate of change = 1 / x

5. f(x) = 2^x

End behavior.......tends toward y = 0 on the left side .....positive infinity on the right

Avg rate of changle =   Ln (2)*2^x

6. f(x) = 1/x

End behavior........tends toward y = 0 on the left  and  tends toward neg inf as x approaches 0 from the left

Not defined at x = 0

Tends toward positive infinity as x approaches 0 from the right.....and then toward y = 0 on the right

Avg rate of change =  log x

\(\text{the numbers are all distinct so they can be ordered}\\ \text{the median will be the average of the 5th and 6th largest numbers}\\ \text{the 5th number must be at least 5 and at most 9}\\ \text{our minimal average sequence is thus}\\ 1,2,3,4,n,20-n,20-n+1,20-n+2, \dots 20-n+4 \\ \text{with mean}\\ \dfrac{10+20+80-4n+10}{10} = \dfrac{120-4n}{10} = 12 - \dfrac 2 5 n\\ \text{This is clearly minimized by choosing }n \text{ as large as possible}\\ \text{The largest possible }n \text{ is 9, and thus the minimum average is }\\ 12 - \dfrac{18}{5} = \dfrac{42}{5}\)

Im sorry but that answer is incorrect

*****

The original question asked to find asymptots for f(g(x)) . Thus after finding the asympoths for f(z),  one needs to compute for which values of x g(x)= -1 and g(x) = 3.

anyone?

de nada !

Thx. I almost messed that up really bad.

Not sure who told you that, but it is incorrect.....sorry

Hypothetical example:   IF you move 800 east  and then   400 west THEN your displacement would be 800 - 400 = 400       This ONLY works with displacements along a LINE.

For displacement, I was told to subtract the final position - the initial position. 

Doing that I got

-200km North

You're welcome....good luck with Physics !