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Rosala, we may sometimes forget your birthday, but none of us will forget you!

This may be the post to which Melody is referring.

Note that  

√[ 4 +  √ [16 + 16a] ]     can be written as

√[ 4 +  4√ [1 + a] ]   =

2 √[ 1+  √ [1 + a] ]

Let this  = 2x

So we have

2x  + x  =  6

3x  = 6

x = 2

So...

√[ 1+  √ [1 + a] ]    = 2     square both sides

1 + √ [1 + a]  =  4          subtract 1 from both sides

√ [1 + a]   = 3          square both sides again

1 + a  =  9    →    a  =  8

1. 3y+4x=16                        

    2x-y=18

Here's the graph of this one :https://www.desmos.com/calculator/ca7j4iemr4

2. y = -5x+6

    10x+2y=12

Rearranging the first equation  as 5x + y = 6 

Multiplying by 2   on both sides

10x + 2y  = 12  produces the same equation as the second one.....thus, these are the same line....so we will have infinite solutions

Let b₁  = a

So....b₃  =   ar*2   and b₅  = ar*4

So

ar^2 -   a  =   16  →    a [r^2 - 1] = 16   →    a  = 16 / [ r^2 - 1 ]   (1)

ar^4 - ar^2  =  144→  a [ r^4 - r^2]  =  144 → a = 144 / [ r^4 - r^2]     (2)

Equating (1)  and (2)  we have that

16 / [ r^2 - 1 ]   =  144 / [ r^4 - r^2]    cross-multiply

16  [ r^4 - r^2]  =  144 [ r^2 - 1 ]    simplify

16r*2 [ r^2 - 1] = 144 [r^2- 1]       divide through  by  16

r^2 [ r^2 - 1]  = 9 [r^2 - 1]       divide through by [ r^2 - 1]

r^2  =  9   →   r  =  3  or  -3

If r  = ±3,  then b₁ =  a =  16 / [ 3^2 - 1]  =  16 / 8  =  2 

The range is the set of all possible outputs that a function can produce. 

\(f(x)=4|x+5|\)

The absolute value of any number will can result in a positive number or 0, and 4 multiplied by a positive number or 0 does not change this property at all. Therefore, the range is the following:

\(\text{Range}:\{\mathbb{R}|\hspace{1mm}y\geq0\}\)

This means that the range can be all nonnegative numbers. In interval notation, it would look like the following:

\(\text{Range}:[0,+\infty)\)

How do we find the intercepts? To do it without a graph, you can figure out by setting x=0 and y=0 and solving in each case. Let's do that.

Let's do the exact same process. This time, however, we set y=0 to find the x-intecept.

x = 0.8333r

10x = 8.333r

100x = 83.333r

100x - 10x = 83.333r - 8.333r = 75

90x = 75

18x = 15

6x = 5

x = 5/6

Express the following in the form a + bi, where a and b are real numbers:      sqrt(i)

\(\sqrt i=a+bi\\ i=(a+bi)^2\\ i=a^2+2abi-b^2\\ i=(a^2-b^2)+2abi\\ so\\ a^2-b^2=0\;\; and\;\ 2ab=1\\ (a-b)(a+b)=0\\ a=\pm b\\ but \;\;since\;\; 2ab=1, \;\;a=b\\ 2a^2=1\\ a=\pm\frac{1}{\sqrt{2}}\\ a=\pm\frac{\sqrt 2}{2}\\ so\\ \sqrt{i}=\pm(\frac{\sqrt 2}{2}+\frac{\sqrt 2i}{2})\)

11(5/6)

thatnk you so much

(n - 2) 180 / n  = 176

Multiply both sides by n

(n - 2)180  = 176n

Simplify

180n - 360  = 176n

Add 360 to both sides, subtract176n from both sides

4n  = 360     

Divide both sides by 4

n  = 90    [sides]

Let the time taken by your friend =t minutes, then you have:

3t =4(t -5), solve for t

t = 20 minutes taken by your friend

20 - 5 = 15 minutes taken by you.

And this makes sense since you can cover 1 miles in 15 minutes, while it takes her 20 minutes to cover 1 mile.

P.S. The reason that you don't need to divide by 60 is that the ratio of 3:4 remains the same, whether you divide or not.

At 3 mph....your friend covers  3 / 60   miles per minute

At 4 mph......you cover  4/ 60    miles per minute

So....every minute....you close    [  4 - 3 ]/ 60   =  1/ 60   miles  on your friend

And when she has been walking for 5 minutes......she will be  5 * [3  / 60 ]  = 15/60  = 1/4 of a mile ahead of you when you start 

So......the number of minutes, m, it will take to catch your friend can be found by :

( 1/60) m  = 1/4      multiply both sides by 60

m =  60  / 4   =    15 minutes

A constant term is the term that is unchanging. 6, for example, is a constant. 12.5 is a constant; they don't change. 

To figure out \(f(g(x))\), let's break this down bit by bit. Since we already know that \(g(x)=x^3+5x^2+9x-2\), this means that \(f(g(x))=f(x^3+5x^2+9x-2)\).

Let all sides have length 1

Distance BD = sqrt(2)

Half of BD = sqrt(2)/2

cos(ABD) = (sqrt(2)/2)/1 

ABD = cos^-1(sqrt(2)/2) → 45 degrees

thank you!!! 

Rewrite your recursive equation as U_n = U_n+1 + 2

U₆ = 3

U₅ = 3 + 2 → 5

U₄ = 5 + 2 → 7

... etc.

Area - switch the a with an o

No problemo!

yes!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Okay thanks

I figured it out

Edit: 63 was correct

Okay thanks

Bye Bye