All Answers 358098 Answers

Jerk...:)

Ok, nice wrap. Yawwwwnn

Lucky! My mom isn't even letting me have the summer off... She says I'll get extra credit for doing school through the summer...

Today was my last exam of the fall semester 

Now I've 3 weeks holiday 

Here are some of my favorite posts:

You could use the graphing calculator at: https://www.desmos.com/calculator 

Or, you could choose a number of values of x, then calcvulate the corresponding values of f(x) and plot one against the other by hand or in a piece of graphing software.

You should get something that looks like:

You're still on:)?

Lol Instagram, homeys means like, friends/family/bros...:)

theyre amazing arnt they

hi r u bored

hello bro

Dear Bree,

       You may have forgotten, but they have NFAS, need for attention syndrome. You cannot blame them to badly...:)

Hello! Are u having a good day/night?

Y= 9

y^3 = 4x^2

If x is doubled, by what factor does y increase?

\(\begin{array}{lcll} \boxed{~ \begin{array}{lcll} y^3 &=& 4\cdot x^2 \end{array} ~} \end{array}\\ \begin{array}{rcll} \\ (y\cdot f)^3 &=& 4\cdot (2x)^2 \\ y^3\cdot f^3 &=& 4\cdot 4 \cdot x^2 \qquad & | \qquad 4 \cdot x^2 = y^3\\ y^3\cdot f^3 &=& 4\cdot y^3\\ f^3 &=& 4\\ \mathbf{f} &\mathbf{=}& \mathbf{\sqrt[3]{4}}\\ \end{array}\\\)

sqrt(sqrt(x-5) + x) = 5 

Find all real solutions 

The only real solution is 21

Solve for x:
sqrt(sqrt(x-5)+x) = 5

Raise both sides to the power of two:
sqrt(x-5)+x = 25

Subtract x from both sides:
sqrt(x-5) = 25-x

Raise both sides to the power of two:
x-5 = (25-x)^2

Expand out terms of the right hand side:
x-5 = x^2-50 x+625

Subtract x^2-50 x+625 from both sides:
-x^2+51 x-630 = 0

The left hand side factors into a product with three terms:
-((x-30) (x-21)) = 0

Multiply both sides by -1:
(x-30) (x-21) = 0

Split into two equations:
x-30 = 0 or x-21 = 0

Add 30 to both sides:
x = 30 or x-21 = 0

Add 21 to both sides:
x = 30 or x = 21

sqrt(sqrt(x-5)+x) => sqrt(sqrt(21-5)+21)  =  5:
So this solution is correct

sqrt(sqrt(x-5)+x) => sqrt(sqrt(30-5)+30)  =  sqrt(35) ~~ 5.91608:
So this solution is incorrect

The solution is:
Answer: | x = 21

sqrt(sqrt(x-5) + x) = 5 

Find all real solutions 

\(\begin{array}{lcll} \sqrt{ \sqrt{x-5} + x } &=& 5 \qquad &|\qquad \text{(square both sides)} \\ \sqrt{x-5} + x &=& 5^2 \\ \sqrt{x-5} + x &=& 25 \qquad &|\qquad - x \\ \sqrt{x-5} &=& 25-x \\ \sqrt{x-5} &=& 25-x \qquad &|\qquad \text{(square both sides)} \\ x-5 &=& (25-x)^2 \\ x-5 &=& 25^2-2\cdot 25x+x^2 \\ x-5 &=&625-50x+x^2 \\ 625-50x+x^2 &=& x-5 \qquad &|\qquad - x+5 \\ 625-50x+x^2 - x+5 &=&0 \\ x^2 -51x +630 &=&0 \\ x_{1,2} &=& \frac12 \cdot ( 51\pm \sqrt{51^2-4\cdot 630} ) \\ x_{1,2} &=& \frac12 \cdot( 51\pm \sqrt{2601-2530} )\\ x_{1,2} &=& \frac12 \cdot( 51\pm \sqrt{81} )\\ x_{1,2} &=& \frac12 \cdot ( 51\pm 9 )\\\\ x_{1} &=& \frac12 \cdot ( 51 + 9 )\\ \mathbf{ x_{1} } &\mathbf{=}& \mathbf{30}\\\\ x_{2} &=& \frac12 \cdot ( 51 - 9 )\\ \mathbf{ x_{2} } &\mathbf{=}& \mathbf{21} \end{array}\)

Find the first term, the common difference, and tn for a series having Sn= 4n−3n^2.

\(\begin{array}{lcll} \boxed{~ \begin{array}{lcll} S_n &=& 4n-3n^2 \end{array} ~}\\\\ \end{array}\\ \begin{array}{lcll} t_n &=& S_n - S_{n-1} \\ t_n &=&4n-3n^2 - [ 4(n-1)-3(n-1)^2 ] \\ t_n &=&4n-3n^2 - 4(n-1) + 3(n-1)^2 \\ t_n &=&4n-3n^2 - 4n + 4 + 3(n^2-2n+1) \\ t_n &=&-3n^2 + 4 + 3(n^2-2n+1) \\ t_n &=&-3n^2 + 4 + 3n^2-6n+3 \\ t_n &=& 4 -6n+3 \\ \mathbf{t_n} & \mathbf{=} & \mathbf{7 -6n} \\\\ t_1 &=& 7 - 6\cdot 1 \\ t_1 &=& 7-6\\ \mathbf{t_1} &\mathbf{=}& \mathbf{1}\\\\ d &=& t_n - t_{n-1} \\ d &=& 7-6n -[ 7-6(n-1)]\\ d &=& 7-6n - 7 + 6(n-1)\\ d &=&-6n + 6(n-1)\\ d &=&-6n + 6n -6\\ \mathbf{d} &\mathbf{=}&\mathbf{-6}\\ \end{array}\)

Sorry:

\(\sin{( 2x )} = 2\cdot \sin{( x )}\cdot \cos{( x )}\\ \frac12 \cdot \sin{( 2x )}= \frac12 \cdot2\cdot \sin{( x )}\cdot \cos{( x )}\\ \frac12 \cdot \sin{( 2x )}=\sin{( x )}\cdot \cos{( x )}\\\)

(cos x + 1)(√2 cos x - 1) = 0      set each factor to 0

cos x + 1 = 0       add 1 to  both sides

cos x  = -1

And this happens at  pi  on  [0, 2pi]  

And

√2 cos x - 1  = 0      add 1 to both sides

√2 cos x  = 1          divide both sides by √2

cos x =    1 / √2

And this happens at   pi/4    and 7pi/4    on [0, 2pi]

.................................................

A shopper at a clearance sale is pleased to discover some rolls of wrapping paper on sale for $4.41 each. The original price tags read $6.62. What is the discount, as a percentage?         ASAP PLEAASEEEEEE

1 - (4.41/6.62) X 100=33.38%

idk

I think it's -5 not sure