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First one.....just add  3 + 2.58 + 1.2  + 1.3 + 0.68 + 0.5 + 2.15.....the total is your answer

Second one

9.68   = 5%  of 9.68  =

9.68  + 9.68 * .05  ≈  10.16     >  10

Not enough   !!!

4)  

f(x)  = 3x + 3

g(x)  = 4x + 3

find  f (g(f(2)))

Work from the "inside out"

f(2) =   3(2) + 3  =  9

g ( f(2))   =  g (9)   =  4(9) + 3  =  39

Finally

f ( g ( f(2)))  = 

f ( 39)  =    3(39) + 3   =  40(3)   = 120

2) What is the greatest integer value of  b such that -4 is not in the range of  y = x^2 + bx + 12?

This parabola turns upward.....therefore...we want to  have the  vertex lie above  ( a, -4)  where a  is the x coordinate of the  vertex....so

a = -b / [ 2 (1) ]

2a  =-b

-2a  =b

So.....we want to solve this to find the y coordinate of the vertex

(a)^2  + (-2a)(a)  + 12  = -4

a^2 - 2a^2   = -16

-a^2   = -16

a^2 =16

a = ± 4

b achieves its greatest value when a  = -4   →    -2(-4)  = b  = 8

Note that  the graph here : https://www.desmos.com/calculator/6vgqqov8mg

shows that   when b  = 8, the  vertex is at  (-4, -4) and when b = 9 that part of the  graph lies below y = -4

But  when b = 7, the graph lies wholly above  y = -4

So....b =7  is the greatst integer value that guarantees that -4  is not in the range of  x^2 + bx + 12

(1)

5(2)^2  - 1/2  + 3   -  ( 2^2 - k)   =    2

20  - 1/2 + 3   - 4  + k  = 2

18 + 1/2 + k  = 2

(37/2)  + k  = 4/2

k =  4/2  - 37/2

k = - 33/2

11 1/12 - 2 2/3

like this

fg (x)   =    5

√[2x - 5]  / 10 / x     = 5      multiply  both sides  by x

√[2x - 5] * 10  = 5x              divide both sides by 5

2√[2x -5]  = x                    square both sides

4 (2x - 5)  = x^2               rearrange as

x^2  - 8x  + 20   = 0       complete the  square on x

x^2  - 8x + 16  =  -20 + 16

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

x - 4  = ±√[-4]

x - 4= ±2i

x = 4 ± 2i

Is this supposed to be :   11^(1/12)  - 2^(2/3)   ????

9/ 11 +  1/3

The common denominator  = 33   ...so we have

9 (3)            1 (11)

_____  +   _______  =

11 (3)          3 (11)

27          11

__  +    ___  =

33         33

38

__

33

12  =  3 * 2^2

18  = 3^2 * 2

36 =  3^2 * 2^2

Taking the greatest   powers on 2 and 3 that appear, we  have

3^2 * 2^2  =  36   = the common denominator.....!!!

OK....no prob....!!!

Well....if we  have    -17 / 136      the answer  MUST be  a negative  fraction......

However if the problem is  17 / 136, then   A is  correct....

Thank you for checking my first answer and helping me with the second! I was overthinking the second one

So, it's A ?

-17 / 136   =

-17/ ( 8 *17)   =

-1 /  8

question 2

What does the remainder theorem conclude given that  f(x)/x+3 has a remainder of 11?

Enter your answer by filling in the underlines

f(__) = ___

This says  that  f(-3)  = 11

First one....we are testing whether x + 1  is a divisor of the  numerator.....this is the same as  testing whether  -1  is a root....so.....

Sub  -1  into the the  numerato and we  get that

−12(-1)^17+3(-1)^5−9(-1)^2−1   =

12  - 3  - 9   -  1  =

-1

Aditya evaluates the numerator of the expression when x=−1 . He finds the remainder of the division to be −1 .

You are correct, Jenny!!!!

Yes, that's the problem

No problem :D

Hi maccabresubwoofer, 

First, \(\frac{1^1}{3} = \frac{1}{3} \)

next, \(\frac{2^3}{4} = \frac{8}{4} = 2\)

Finally, \(2+\frac{1}{3} = 7/3\)

but thats not a choice. And second, A and B are the same and D and E are the same...

I think you might mean \(1^{\frac{1}{3}} + 2^{\frac{3}{4}}\)

 But whatever.

-Evan

thanks so much!

Oops CPhill was more detailed Sorry!

Hi Guest, 

First, 

\(\frac{\sqrt{338}}{\sqrt{288}}+\frac{\sqrt{150}}{\sqrt{96}} = \frac{13}{2^2\cdot \:3}+\frac{\sqrt{150}}{\sqrt{96}}\)

Next, \(\frac{13}{2^2\cdot \:3}+\frac{\sqrt{150}}{\sqrt{96}} = \frac{13}{2^2\cdot \:3}+\frac{5}{2^2}\)

Simplifying again, \(\frac{13}{2^2\cdot \:3}+\frac{5}{2^2} = \frac{13}{12}+\frac{5}{2^2}\)

Lastly, \(\frac{13}{12}+\frac{5}{2^2} = \frac{13}{12} + \frac{5}{4} = \frac{13}{12}+\frac{15}{12} = \frac{28}{12} = \frac{7}{3}\)

Your welcome :P, Evan

√338      √ 150

____  +  _____    =

√288         √96

13√2         5√6

____  +    ____   =

12√2         4√6

13             5

__    +      __   =

12             4

13         15

__  +     ___     =

12         12

28

__    =

12

7

__

3

Hi Guest , 

You can assume that the bracelet only has 5 spot because the problem didn't say anything about the length.

Then, you think hm. How many ways can I put the first bead into the bracelet? 

Well because we have 5 beads there are 5 choices.

Next, you think after you put the first bead how many choices are there to put the second bead into the bracelet?

Because you already put one bead into the bracelet, there are only 4 choices.

You find that the patatern is 5, then 4, then 3, then 2, then finally 1 final choice.

Multiply them all together you get 120. Notice that \(120 = 5!\)

But since rotations and flips are not the same, you have to divide by 5, \(\frac{120}{5}=24\)

then you divide by 2 because flips aren't the same so the answer is \(\boxed{12}\)

-From Evan

Hi guest, think of the question as 4 parts.

First part is heads. The probability of that occurance is 1/2 because it is either heads or tails.

Second part is tails. The probability of that occurance is also 1/2 because it is either heads or tails.

Third part is heads. The probability is 1/2 same reasoning of the First and Second part

Fourth part is tails. The probability is 1/2 same reasoning.

Multiply them together, 

Your welcome, Evan