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See here, EP  :

https://en.wikipedia.org/wiki/Parallelogram_law

Sum =   (12)(13)  / 2  =   78

Divisible  by   1, 2, 3, 6 

(I assume  that 1 and 10  are inclusive)

f(c)  =  y

y =  -1

The graph only interesects  this value once

THAT is a really obscure law I did  NOT know about !!  

1+2+3+4 = 10 balls are accounted for already ...this leaves 2 balls to distribute

there are 4 ways to put the two in ONE box

there are 6 ways to distribute One and One ball amongst the four boxes

10 ways      (I think)

Not exactly the same problem, but very similar.....

See if this helps  :

https://web2.0calc.com/questions/pls-help_94109

xy = b    sub in the point given

2*8 = b = 16

y = a x^2 + b      sub in the point given....and b

8 = a (2^2) + 16

8 = 4a + 16

-8 = 4a

a = -2

-2,16

Parallelogram Law :  The  sum of  the squares  of the four sides of a parallelogram  = the  sum of the squares of the diagonals

So

2*7^2   + 2*9^2  =  8^2  +  x^2

98  + 162  =  64 + x^2

98  + 162  - 64   = x^2

196  = x^2

sqrt (196)  = x   = 14  =  the longer diagonal

1/7  =  0.142857 (repeating)

n=7   I think

\(m+{1\over m}=8\)

⇒ \((m+ {1\over m})^2=64\)

\(m^2+{1\over m^2}+2=64\)

\(m^2+{1\over m^2}=62\)

⇒\(m^2+{1\over m^2}+4 = 66\)

~Hope you got it. 

So it's given that r = 1/8 and S_∞ = 80

∵ We know that, 

\({S}_{∞} = {a\over 1-r}\)

\(80={a\over 1-{1\over 8}}\)

\(80={8a\over 7}\)

⇒ \(a=70\)

Similar question have been asked before twice check both of these links for reference - 

I think you missed a term, the GP should be : 1/2, 1/4, 1/8, 1/16,...

⇒ a = 1/2

    r = 1/2

    n = 6

\({S}_{n}={a(1-r^n)\over 1-r}\)

      \(= {{1\over 2}(1- {1\over 2^6})\over {1- {1\over 2}}}\)

      \(=1- {1\over 2^6}\)

      \(={63 \over 64}\)

\({S}_{6}= {63\over 64}\)

It may be more than 1 0 .....I think it is more like 1 0 1 1 0 1 1 1 0 0 1    people.

The domain is the scope of the 'x' values    [0,23]      which represents years.

   So the domain is   23 from 0 inclusive         represents years

https://web2.0calc.com/questions/our-club-has-10-members

Let $s_0, s_1, s_2, \ldots$ be the sequence we are seeking.  Write $s_0=a$ and $s_1=b$.  By the arithmetic mean assumption, we have that $2s_{n-1} = s_n + s_{n-2}$  if $n \ge 2$.  Thus, we may define the sequence as
\[
s_n =
\left\{
  \begin{array}{ll}
    a & \mbox{if $n = 0$}  \\
    b & \mbox{if $n = 1$}  \\
    2s_{n-1} - s_{n-2}  & \mbox{if $n \ge 2$.}
  \end{array}
\right.
\]

We claim that $s_n$ has the closed-form formula $s_n = nb - (n-1)a$.  We prove this by induction.  When $n=0$ the conjectured formula gives $s_0 = 0\cdot b - (-1)a = a$ which is correct.  When $n=1$ the conjectured formula gives $s_1 = 1\cdot b - (0)a = b$ which is also correct.  Now let $n\ge2$ and assume by induction that the conjecture formula is correct for all $k$ such that $0\le k < n$.  We have by its recursive definition and the inductive hyphothesis that 
\begin{eqnarray*}
s_n &=& 2s_{n-1} - s_{n-2} \\
    &=& 2((n-1)b-(n-2)a) - ((n-2)b - (n-3)a) \\
    &=& (2(n-1)-(n-2))b - (2(n-2)-(n-3))a \\
    &=& nb - (n-1)a
\end{eqnarray*}
and it agrees with the formula again.  Thus the conjectured formula is correct by induction.

By assumption $s_{27} = 94$ and $s_{94}= 39$.  This gives 2 equations $27b-26a=94$ and $94b-93a=39$.  Solving for $a$ we get $a=\frac{7783}{67}$.  By the way, $b=\frac{7728}{67}$ but the question didn't ask for it.
 

z^3-ci = -88 so that means that the imaginary part of z^3 is c and than the real part of z^3 is -88. So, x^3-3xy^2 = -88 so x=2,y=4 which means z=2+4i

y = yo + v0 t - 1/2 a t²                    vo = initial velocity     a = acceleration of gravity   t = time

    =  0   + 17.2 (0.5) - 12 (32)(0.5²)  = __________ meters

We have two equations

x - 2y  =  10  ⇒     x  =  2y + 10       (1)

(x +2)^2   +  ( y -7)^2   =  169         (2)

Put (1)  into (2)

(2y + 10 + 2)^2  +  ( y - 7)^2   =   169

(2y + 12)^2  + (y - 7)^2  =   169

4y^2  + 48y  + 144 + y^2 - 14y  + 49   = 169

5y^2 + 34y  + 193  =  169  

5y^2  + 34y + 24  =  0                   factor as

(5y  + 4)(y + 6)  =  0

Set  each factor to 0  and solve for y  and we have

y  = -4/5                                         and                       y =   -6

x = 2(-4/5)+ 10                                                           x = 2(-6) + 10

x = 42/5                                                                      x  = -2

The points are        (-2, -6)    and   (42/5 -4/5)

if x

if x>=a, the graph y=f(x) is the same graph as y=ax^2

line in the graph has positive slope, since the parobola only has nonegative values

so, a>0, and thus, the line in the graph passes through every single horizantal line <= a^2+2a

the parabola region passes through every single horizantal line >=a^3

a^2+2a>=a^3

a+2>=a^2 (since a>0 and can divide by a)

0>a^2-a-2

0>(a-2)(a+1)

-1<=a<=2

greatest value of a is 2

JP

MathProblemSolver101

Please do not answer multiple questions like this. This kid is clearly cheating. 

These question still make me thinkg

I do them a bit at a time.

5 people take 27 minutes

so one person must take 27*5=135 minutes

so 18 people will take 135 divided by 18 = 7.5minutes

= 7 minutes 30 sec   just as answer guest said.

Use the formula