All Answers 358514 Answers

Hayley, (-1.8,0) is on the x axis!  you should have plotted it to see that!

Where is you special guest gardian angel.   LOL

Hayley, there seems to be a guest who is seriously looking out for you  :D

He/she seems to be a good teacher too :)

Maybe you will ace your maths exams next year :)

Thanks guest :)

Hayley, I think you messed this one up.  

Nevermind, you have post heaps of really good answers.   

Find the y-intercept and the x-intercept for the equation.

3x+4y=48

What is the y-intercept?  x=0

so

4y=48

y=12

What is the x-intercept? y=0

so

3x=48

x=16

Hi Solveit  :))

The function g is defined as g(x)=ax^2+bx-6a, where a and b are positive constants. The function has a line of symmetry at x=-3. Which of the following could NOT be a coordinate pair on the graph of y=g(x)?

A) (0,-3)

B) (0,3) 

C) (-3,0)

D) (3,3)

E) (-3,-3)

\(y=ax^2+bx-6a\)

     \(\frac{-b}{2a}=-3\\ -b=2a*-3\\ -b=-6a\\ b=6a\\ \mbox{so the equation becomes}\\ y=ax^2+6ax-6a\\ y=a(x^2+6x-6) \)

If x=0 then 

y=a(-6) <0     so      (0,3) CANNOT be a point but  (0,-3) could be

If x=-3

y=a(9-18-6) <0    so    (-3,0) CANNOT be a point but (-3,-3) could be

If x=+3

y=a(9+18-6)>0   so  (3,3) can be a point

SO

(0,3) and (-3,0)   cannot be points on the parabola.

What the h**l?

Why do you even ask that question it is so dumb. DO NOT DO THAT TODAY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

It was annoying enough yesterday and I am sick and tired of it all it does is make ya look and sound like idiots

No you're stupid it is 19! And don't do that again today.

Actually I sent at 8:20am in my time.🐰 LOL!!!!!!!!!!!!!!!!!!!!!

You seriously sent this at 2:20am? What s wrong with you? 🐷

1st one is Estelle Orb from 39 clues cahills vs vespers 

2nd one is a candle

        Let me know if I got it right!

Super fun way to test you knowledge!!!!!!!!!👏🏽👏🏽👏🏽👏🏽👏🏽👏🏽🙊bravo

Thanks.☺️I don't suppose you could give a summary of how to use a fricking calculator?lol

I do not remember ever hearing the word "radicand" before   ://

I just looked it up. :D

\(\sqrt{x}-4 \qquad \mbox{Radicand is x}\\~\\ \sqrt{x-4} \qquad \mbox{Radicand is } (x-4)\\~\\\)

GOOD LUCK

Hayley, you could have said a little more than that I think!     

No real number times by itself can give a negative answer, therefore   sqrt(-81)  has no real solution.

-3-6(-4k+5)=-12+3k

\(-3\;\;\;\:\qquad -6(-4k+5)=-12+3k\\ -3+12\;\;\;-6(-4k+5)=-12+3k+12\\ 9\;\;\;-6(-4k+5)=+3k\\ \mbox{divide both sides by 3}\\ 3\;\;\;-2(-4k+5)=+1k\\ 3\;\;\;+8k-10=+1k\\ 8k-7=k\\ 7k=7\\ k=1\)

Yes

26C5/67C5 =

\(\begin{array}{rcll} 26C5 &=& \frac{26!}{5!\cdot (26-5)! } \\ &=& \frac{26!}{5!\cdot 21! } \\ &=& \frac{21!\cdot 22\cdot 23\cdot 24\cdot 25\cdot 26}{5!\cdot 21! } \\ &=& \frac{22\cdot 23\cdot 24\cdot 25\cdot 26}{5! } \\\\ 67C5 &=& \frac{67!}{5!\cdot (67-5)! } \\ &=& \frac{67!}{5!\cdot 62! } \\ &=& \frac{62!\cdot 63\cdot 64\cdot 65\cdot 66\cdot 67}{5!\cdot 62! } \\ &=& \frac{63\cdot 64\cdot 65\cdot 66\cdot 67}{5! } \\\\ \frac{ 26C5 } {67C5} &=& \dfrac{ \frac{22\cdot 23\cdot 24\cdot 25\cdot 26}{5!} } { \frac{63\cdot 64\cdot 65\cdot 66\cdot 67}{5!} }\\ &=& \frac{22\cdot 23\cdot 24\cdot 25\cdot 26}{5!} \cdot \frac{5! }{63\cdot 64\cdot 65\cdot 66\cdot 67}\\ &=& \frac{22\cdot 23\cdot 24\cdot 25\cdot 26}{63\cdot 64\cdot 65\cdot 66\cdot 67}\\ &=& \frac{7893600 } { 1158917760 } \\ &=& \frac{ 115 } { 16884 } \\ &=& 0.00681118218 \end{array}\)

         f(2x)=[f(x)]^2

Given the equation above, if f(1)=3, what is f(2) ?

\(f(2x)=[f(x)]^2\\~\\ f(1)=3\\~\\ f(2*1)=[f(1)]^2\\~\\ f(2)=[f(1)]^2\\~\\ f(2)=[3]^2\\~\\ f(2)=9\)

Thanks kids, I did enjoy reading this thread :)

Yes, of course Hayley.  !!

I didn't understand that at all.