All Answers 358087 Answers

Yeah....we couldn't both be wrong......LOL!!!!

Hi Chris and Sabi,

I want to try this one too :)

The parabola at the top is concave up so it is the one with the positive leading coefficient.

so you want

$$\\4<(\frac{1}{2}x^2+7)-(\frac{-1}{4}x^2+3x)<7\\\\
4<(\frac{2x^2}{4}+7)-(\frac{-x^2}{4}+3x)<7\\\\
4<\frac{2x^2}{4}+7+\frac{x^2}{4}-3x<7\\\\
4<\frac{3x^2}{4}+7-3x<7\\\\
-3<\frac{3x^2}{4}-3x<0\\\\
-12<3x^2-12x<0\\\\
-4 -4 x^2-4x>-4\qquad and \quad x(x-4)<0\\\\
x^2-4x+4>0\qquad and \quad x(x-4)<0\\\\
(x-2)^2>0\qquad and \quad x(x-4)<0\\\\$$

$$y=(x-2)^2$$    is a concave up parabola with a double root at x=2 so it is going to be positive for all valures of y except  x=2

$$y=x(x-4)$$  is a concave up parabola with roots at x=0 and x=4 . It will be BELOW the xaxis (negative) between the 2 roots.      Negative for   $$0

These will both be true when    $$0      EXCEPT  when x=2

This is identical to CPhill's answer so by the law of averages we are probably both correct.   LOL

Remembering that a^-1  = 1/a     .... we have

4^-1   = 1/4

10^-1  = 1/10

So we have

1/4  - 1/10      and getting a common denominator of 20, we have

5/20  - 2/20  =  3/20

x - y - 3 = 0      This is a line...to graph it, pick some value for y, say 0.....then plugging this into the equation makes x = 3

So...we have one point on our graph, i.e.,  (3, 0)  ....now pick another value for y, say 1......then, putting this into the equation forces x to be 4....and we have a second point → (4, 1)

$${\mathtt{540}}{\mathtt{\,\times\,}}{\mathtt{0.3}} = {\mathtt{162}}$$

30% of 540 is 162

youre wrong geno3141

actually, the Muslims discovered algebra

God invented algebra

log₂y  =  (1/3)log₂27

Since a multiplier goes into the log expression as an exponent:

log₂y  =  log₂27^1/3

An exponent of 1/3 represents a cube root, so 27^1/3 = 3  (the cube root fo 27 is 3).

log₂y  =  log₂3

So:  y = 3.

Since there are  6  different letters, you have  6  choices for the first letter.

Now, since there are  5  letters remaining, you have  5  choices for the second letter.

Then, you have  4  choices for the third letter;  3  choices for the fourth letter;  2  choices for the fifth leetter; and only  1  choice for the sixth, and last letter.

Multiplying these choices together:  6 x 5 x 4 x 3 x 2 x 1  =  6!  =  720 choices.

$${\mathtt{1}}{\mathtt{\,\times\,}}{{\mathtt{1}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}-{\mathtt{6}} = {\mathtt{145}}$$

$${\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}-{\mathtt{6}} = {\mathtt{145}}$$

$${\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\times\,}}-{\mathtt{6}} = {\mathtt{145}}$$

$${\mathtt{x}}{\mathtt{\,\times\,}}-{\mathtt{6}} = {\mathtt{145}}$$

$${\mathtt{x}} = -{\mathtt{24.166\: \!666\: \!666\: \!666\: \!666\: \!7}}$$

$${\mathtt{58}}{\mathtt{\,\times\,}}{\mathtt{45\,678\,963}} = {\mathtt{2\,649\,379\,854}}$$

k 6 characters in the name robert

6*6=36

overall, there should be 36 possible unique arrangements for the letters ROBERT

haha, you main egos have done lots of mess and damage to the world, and now we alt egos have been sent by our spiritual leaders to clean up after you

now the Age Of Great Empires has ended, the Sole Ruling Age has begun, and youre right, we regret having no voice and only one leader

i as the alt ego, am crowned the Main Ego now!

alt ego, i told you only i can be the one to participate in surveys

when an alt. ego takes the place of the main ego, the main ego goes to the CIA

unless,

youre turning me into the CIA?

i am the greatest villain everyone has ever seen, yet i lose to my alt ego?

you know what youve caused, Alt Egos? youve caused your main egos reputations to fall!

*Royality Towers fall and burn, one by one, starting in Australia, then to Japan, China, Korea........and the last one falls at Rome, Italy

*as Royality falls, other empires fall, the main egos start to fall, including Heathers main ego, soviet vikings, and after that, all empires have fallen, and now the evil principal has all the power and control

now we have to work for the principal!!! this is total defeat for everyone!

you Alt Egos are to blame for this!!!

We just want to write  f(x) - g(x) > 0    ....in other words, we want all the values such that the height of the graph of f(x) at some value of x is greater than the height of the the graph of g(x) at the same x value

I possibly don't understand this, but I think you want to find the values where the differences in the height of the curves is more than 4 but less than 7....so we have....

[(1/2)x^2 + 7] - [(-1/4)x^2 + 3x] > 4   simplify

(3/4)x^2 - 3x + 7 > 4

(3/4)x^2 - 3x  + 3  > 0      and this is > 0 everywhere except  at x = 2 → ( -∞, 2) U (2, ∞)

Let b be the second term....so we have.....

3,  b,  3 + b, 3 + 2b, 6 + 3b

And the last term is equal to 3 ...so.....

6 + 3b = 3    

3b = -3     →   b = -1

And ths series is :

3, -1, 2, 1, 3

Let n be the first integer...and we have

n + (n + 1) + (n +2) +  ..... + (n + 38) + (n +39)

40n + [(39)(40)/2]  = 100

40n + 780  = 100    subtract 780 from both sides

40n = -680    divide both sides by 40

n = -17

We have the following system

9 = a₁ + d(5 -1)

-84 = a₁ + d(32 -1)       simplifying, we have

9 = a₁ + 4d

-84 = a₁ + 31d    subtract the second equation from the first

93 = -27d     divide both sides by -27

d = -31/9

Using the first equation to find a₁, we have

9  = a₁ + 4(131/9) 

a₁ = 9 - 4(-31/9) = 205/9    and this is the first term

So....the 23rd term is given by

205/9 + (-31/9)(23-1) = -53

There are [157 - 13] / 16 + 1  = 9 + 1 = 10 terms in the sequence

The mean will be the average of the middle two terms...so we have...

[(13 + 4*16) + (13 + 5*16)] / 2  = 85

[Notice that we could have found the same result by adding the first and last terms and dividing by 2 ]

SuperCell-Creators of games Clash Of Clans and Boom Beach

The sequence is given by

a_{n =}  a₁ -3(n - 1)    where a₁ is the first term and a_n is the nth term.....so we have......

-17 = 88 - 3(n - 1)      simplify

-17 = 88 - 3n + 3

-17  = 91 - 3n      subtract 91 from both sides

-108 = -3n        divide both sides by -3

36 = n   ....so...  35 terms appear before -17 

9 is the 1st term and 4 is the third

And the nth term is given by....    a₁(r)^(n-1 )    where a₁ is the first term....and r is the geometric multiplier.....so, we have

9(r)^(3- 1)  = 9(r)^(2)  =  4      divide both sides by  9

(r)^2 =  4/9    take the square root of both sides

r  = 2/3

So, b =  9(2/3)^(4 -1)  = 9(2/3)^3 =  8/3

Let a₁ be the first term....then the successive terms are

3a₁ + 4,  9a₁ + 16,  27a₁ + 52,  81a₁ + 160     

So .... 81a₁ + 160 = 403     subtract 160 from each side

81a₁ = 243    divide both sides by 81

a₁ = 3

The sum of the first n even integers is given by  (n)(n + 1)

The sum of the first n odd integers is given by n^2

So we have

(400)(401) - (400^2)   =

(400)(400 + 1)  - (400^2)  =

400^2 + 400 - 400^2   = 400