All Answers 356067 Answers

Rewrite as follows: 

\({x^2}^2 = -4\)

Take the square root of both sides: 

\(x^2 = 2i\)

Take the square root again: 

\(\color{brown}\boxed{x = 1 + i}\)

SImplify as follows:

 \(\sqrt{{3^2 \over 5^2}^{-3}} - {\sqrt{x} \over 3^3} = {11^2 \over 3^3}\)

\({\sqrt {1 \over{3^2 \over 5^2}^3}} - {\sqrt{x} \over 3^3} = {11^2 \over 3^3}\)

\({\sqrt {1 \over{3^6 \over 5^6}}} - {\sqrt{x} \over 3^3} = {11^2 \over 3^3}\)

\({\sqrt {5^6 \over 3^6}} - {\sqrt{x} \over 3^3} = {11^2 \over 3^3}\)

\({5^3 \over 3^3} - {\sqrt{x} \over 3^3} = {11^2 \over 3^3}\)

\(5^3 - \sqrt{x} = 11^2\)

\(125 - \sqrt{x} = 121\)

\(\sqrt{x} = 4\)

\(\color{brown}\boxed{x = 16}\)

That is great.  

I like that you have used a link to Chris's answer.

Just one little suggestion.

Please put the address in the link so that people know where they are linking to.

I, for one, am wary about hyperlinking to unknown destinations.

No, you have misunderstood.  I did not explain myself well.

For most posts, what you did was perfect!

I have hidden the original post because it was a great string of questions and I do not want people to think it is ok to do that.

It is not ok, we are not here to do the entirity of people's homework for them.

That is the only reason I posted the pic, i did not want the original to be visible.

If it had been any normal post then what you did was perfect.

By giving a link you are automatically giving the writer the credit. 

Your welcome?!?!? Next time I'll try to credit the user- and put a picture if I'm not lazy-

Not trying to be rude or anything, but its decently easy, I copy and paste the whole thing, and look at results.

@below(Ahhhh the pictures to big)

Its ok,

Referring back to old answers is perfectly acceptable. ... usually encouraged.

I was a bit shocked to see how Omi had effectively done someones assignment for them but it was a long time ago.

If I saw a post like that today i would delete it but the individual answers are fine.

I think I will just add the answer, giving Omi credit for it of course. Just in case the assignment has been repeated in its entirity.

Thanks you for finding, and posting it.

\(Suppose\\ \dfrac{a}{b}=\dfrac{3}{5},\ \dfrac{b}{c}=\dfrac{15}{2},\ \dfrac{c}{d}=8\)

\(What\ is\ the\ value\ of \)    \(\dfrac{a}{d}\ ?\)

Hello Guest!

\(Suppose\\ \dfrac{a}{b}=\dfrac{3}{5},\ \dfrac{b}{c}=\dfrac{15}{2},\ \dfrac{c}{d}=8\)

Then

\(b=\dfrac{5a}{3},\ c=\dfrac{2b}{15},\ d=\dfrac{c}{8}\\ \dfrac{5a}{b}=3,\ \dfrac{2b}{c}=15,\ \dfrac{c}{d}=8\\ 5a=3b,\ 2b=15c,\ c=8d\)

\(b=\dfrac{5a}{3},\ c=\dfrac{2a}{9},\ d=\dfrac{a}{36},\ a\neq 0\)

\({\color{blue}\dfrac{a}{d}}=\dfrac{a}{\frac{a}{36}}=a\cdot\dfrac{36}{a}\color{blue} =36\)

Sorry, it lured me.

  !

(Should I post this? If not I will gladly delete this.)

This question has been asked here -------(address deleted)

Omi's answer from May 2018

Yep, Guest,  you  are correct.....thanks for pointing that out  !!!!

I forgot that the polygon is  convex.....n =16 would produce a concave polygon !!!

My bad  !!!!

Only one answer......n  = 9     !!!!

N = 16 takes the interior angle past 180 degrees doesn't it ?

Simplifying  by the binomial theorem, we get

x^3 + 12x^2 + 48x + 64 + x^3 + 15x^2 + 75x +125 = x^3 + 21x^2 + 147x + 343 + x^3 - 6x^2 + 12x - 8 

Subtract out the x^3 terms....simplify the rest

27x^2  + 123x + 189  =  15x^2 + 159x + 335

12x^2 -36x -  146 = 0    divide through by 2

6x^2  -18x - 73  = 0

Using the Quadratic Formula the two solutions  are  x ≈ -2.2969    or  x ≈ 5.2969

2x - 5  <= x + 12       rearrange as

2x - x <= 12 + 5       simplify

x <= 17

x =  ( -inf , 17)

THX, InhumanCalculator  !!!

Here's another approach :

https://web2.0calc.com/questions/halp_105#r1

We know that the tens digit cannot be 0, so we can count the total number of integers between 10 and 99 with 0 as a unit's digit:

10

20

30

40

50

60

70

80

90

Hope this helped. Let me know if I did anything wrong!

If the two lines are parallel, then they have the same slope. We can calculate the slope of the first equation to be -1/3 (remember, the slope in standard form is -A/B for Ax+By=C). Now that we know the slope, we can calculate the x and y intercepts using the point and extending it:

x-intercept = (-1,0)

y-intercept = (0,-1/3)

You can now plug this back into the equation and solve for A and B.

I hope this helped. Let me know if I did anything wrong!

Sum  of this arithmetic series  =    n   ( a1  + an)   / 2

n = number of sides

a1= 120  = first term

an  = 120 + 5 (n -1) = 115 + 5n   =  last term     

So  we  have    n [ 235 + 5n ] / 2      (1)

And the sum of the interior angles of an n-sided polygon =    (n - 2) (180)     (2)

Set (1)  = (2)

n [ 235 + 5n ] /2  = (n -2) (180)    simplify

235n + 5n^2  = 2 ( n -2)(180)

235n + 5n^2  = (n -2) (360)

235n + 5n^2  = 360n - 720

5n^2 - 125n + 720  =  0            divide through by 5

n^2 - 25n + 144   = 0          factor

(n - 16) ( n - 9)   = 0

Setting  each factor  to  0 and solve for n and we get two answers

n = 9     or  n =  16 

y int  =   3(0) - 6y = 18   ⇒     y = -3

x int   =   3x - 6(0) = 18  ⇒     x = 6

We can form a right triangle with legs of 3 and  6

The distance between  the two intercepts  is the hypotenuse = sqrt [6^2 + 3^2]  = sqrt (45)  =3 sqrt (5)

2(x - 8)^2  + 4   =

2 (x^2 - 16x + 64) + 4  =

2x^2 - 32x + 128 + 4  =

2x^2 -32x + 132

a = 2

b = -32

c = 132

So

3ax^2  + 2bx + c  =

6x^2- 64x + 132         

6  [ x^2 - (32/3)x + 22]       complete the square on the terms in the brackets

                                           Note : ([ 32/3] [1/2]) =  (16/3)^2   =256/9

6 [ x^2 - (32/3)x  + 256/9  + 22 - 266/9 ]  =

6 [ (x - (16/3)]^2  +  6 [ -58/9]  =

6 [ x - (16/3) ]^2  -116/3

h  = 16/3

THX for confirming my answer, BuilderBoi  !!!!

Since O  is the circumcenter  of    ABC,  then   AO = BO = CO =  the circumradius

Then angle ABC is an inscribed  angle in the  circumcircle  =  30°

But angle AOC  is  a central angle  intercepting  the same arc  as angle ABC

So  angle AOC  =  2 * angle ABC  =  2 * 30°  =  60°  

No, C Phill's answer is correct. 

By graphing the system, we find that the solution for the system is (5.2, -0.4), and \(5.2 \times -0.4 = \color{brown}\boxed{-2.08}\)