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If you plot the graph, you will find that the any horizontal line intersects the graph at most once.  So by the horizontal line test, the function has an inverse.

The sum of all possible values of a is 12.

Pretty easy, I'll help you get started

$i$ = $sqrt(-1)$

$i^2$ = $-1$

$i^3$ = $-sqrt(-1)$

$i^4$ = $1$

And it repeats like that.

This problem is very elementry and very engrossing.

x^2 + bx + c         when x = 0    f(x)=4    so c = 4

x^2 + bx + 4        when x = 1 the value is -10

1^2 + b + 4 = -10

b = -15

f(x) =    x^2 -15x+4

Arc TU is   72/360 ths of the entire circle

pi * 2r = entire circle circumf

72/360 * pi * 2r = 6pi

r = 15 units

Vertical asymtopes when denominator = 0

(x-1)(x-2) = x^2 -3x+2       a = -3   b = 2

oh yes, this helps Melody. Thanks!

8 = 2³

(2³)^x+1  = 2^3x+3

             = 2^3x * 2³

             = 7 * 8 = 56

thanks ;)

$5 * 5^b = 5^(b+1)$

$5^(b+1)$ = $5^(4(b - 1))$

$5^(b+1)$ = $5^[4b - 4]$

$b + 1$ = $4b - 4$

$3b$ = $5$

$b$ = $5/3$

a)

1 and 4 are the square numbers. 

So the chance that neither dice show them is 2/3*2/3 = 4/9, meaning the chance that at least one was shown is 5/9. 

b)

9 * 5

9 savory options, and 5 sweet ones. 

=^._.^=

Note that everything cancels out so that the final equation is 2/(n+1). 

2/(n+1) = 1/500

Can you take it from here?

=^._.^=

Oopsies, my bad. 

AC = AE + EC, not AD + EC. 

I'm sorry if I got you confused. 

=^._.^=

I will assume these are only positive numbers

A = 16     B = 96      Difference = 80

If the numbers can be negative

A = -96     B = 96     Difference = 192      **** edited ****

Need a little more info here !

   19 = 1/2 d     given       Multiply both sides by 2

2*`19 = 2 * 1/2 d

38 = d                   d book = 38 oz

3 years is  how many standard deviations away from 3.5 ??    .5 / .35 = 1.43 standard deviations below the mean====> z-score table = - 1.43

   from the negative  z-score table this represents .0764    or   7.64 %

The money amounts form an arithmetic sequence with d = -50 dollars

the SUM of an arithmetic sequence is given by    Sn=n(a1+an)/2    where in this case , n = 5   and a1 is the amount given to 1st place and an would be 5thplace (which is $200 less)

   so     SUM_{of prize money} = 5(1^{stplaceamount} + 1^{stplaceamount} - 200) /2      you can simplify this to

                Sum = 5 (2*1st -200)/2

                         = 5 *1st - 500            dollars

48 g out of 60 g ?%

Hello Guest!

\(60g:100\%=48g:x\\x\cdot 60g=48g\cdot 100\%\\ x= \dfrac{48g\cdot 100\%}{60g}\\ \color{blue}x=80\% \)

\(48 g\ of\ 60 g\ is\ 80\% .\)

or so:   \(\dfrac{48g}{60g}=\dfrac{4}{5}=0.8=\color{blue}80\%\)

  !

I think you got confused catmg ???

YES

AB = AC = AD     all radii.

I assume AE=7.5cm

radius = 7.5+5=12.5cm

Use pythagoras's theorum to find ED

then area of a triangle formula

I have answered on the original question post.    

Inititally 100

after one minute 50 left

after 2 minutes 25 bubbles left

after 3 minutes 2 bubles left

after 4 minutes 0 bubles left

the question is unclear as to whether the bubbles are blown at the end or at the beginning of each minute.

Let's say they are blown at the end of each minute.

So at (immediately after) the end of ten minutes there are 

100 bubbles that have just been blown

+ 50 that were blown a minute ago

+ 25 that were blown 2 minutes ago

+ 2 that were blown 3 minutes ago

and all the ones that were blown before that are gone.

You can add them up  :)

There are values A and B such that
\(\dfrac{Bx-11}{x^2-7x+10}=\dfrac{A}{x-2}+\dfrac{3}{x-5}\)
Find A+B.

\(\small{ \begin{array}{|lrcll|} \hline & \dfrac{Bx-11}{x^2-7x+10} &=&\dfrac{A}{x-2}+\dfrac{3}{x-5} \quad | \quad x^2-7x+10 =(x-2)(x-5) \\\\ & \dfrac{Bx-11}{(x-2)(x-5)} &=&\dfrac{A}{x-2}+\dfrac{3}{x-5} \quad | \quad * (x-2)(x-5)\\\\ & Bx-11 &=& A(x-5)+ 3(x-2) \\ & \mathbf{ Bx-A(x-5) } &=& \mathbf{ 3(x-2) + 11 } \\ \hline 1.~ x=0:& 0 +5A &=& -6+11 \\ & 5A &=& 5 \\ & \mathbf{A} &=& \mathbf{1} \\ \hline 2.~ x =5:& 5B - 0 &=& 9+11 \\ & 5B &=& 20 \\ & \mathbf{B} &=& \mathbf{4} \\ \hline \end{array} }\)

\(\mathbf{A+B = 5}\)