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Hmmmmm.... Hey Chris.... what about:

"Their total age is two more than nine times the age of Vera"

L =  2V     

4L - 4 = E  ⇒  4L = E + 4

4L =  8V  = E + 4   ⇒  E  = 8V - 4

So

L  +  E  +  V   = 9V  + 2

2V  + (8V  - 4)  + V  =  9V + 2

11V - 4  = 9V + 2

11V  - 9V  =  2 + 4

2V  = 6

V = 3

L = 2V =  6

E = 8V - 4  =  20

Thanks to the guest for pointing out my earlier (stupid) error  !!!

"social distancing"  at greater than 100 ft?

L = 2v

e = 4L -4   = 8v-4

v

summed =    9v + 2

2v    + 8v-4     + v   = 9v+2

11v  - 4 = 9v+2

2v = 6

v = 3

L = 2v = 6

e = 8v-4 = 20

3x^2  - 6x  -  2              factor out the 3

3 ( x^2  - 2x  - 2/3)      

Take 1/2 of 2  = 1  square it  =1......add and subtract within the parentheses

3 ( x^2  - 2x + 1   -2/3  -1)           factor the first three terms.....simplify the  rest

3 [ ( x  -1)^2   - 5/3  ]         distribute the "3" across the terms in brackets

3( x - 1)^2  -  5

a =  3

h = 1

k = -5

And their sum  =    -1

Nice method, heureka    !!!!

A+ 2 = B -2         A + 2  =  C * 2           A + 2 =  D/2

A + 4  = B          (A+ 2) / 2  = C          2(A + 2)  = D

So  

A  +  B   +  C  +   D   =  36

A  +  (A + 4)   +  (A + 2) /2     + 2(A + 2)  =  36      multiply through  by 2

2A  +  2(A + 4)  + (A + 2)  + 4(A +2) =  72

2A  + 2A +  8  + A + 2  + 4A + 8  =  72

9A  +  18 =    72

9A  = 72 - 18

9A  = 54

A = 6          B =  10      C  =  4    D  = 16

A*B*C*D  =   6 (10) (4) (16)   = 3840

Note that minor arc AB  =2 (36°)   = 72°

And  arc  ABM  is 7/18 of the  circumference  =  (7/18) * 360   =140°

So minor arc BM =   140  -72    =  68°  = 2*angle BAM  =  angle BAC

So  angle   ABC  =  180  - 36 -  68  = 76°

(f-2)  = (10)(s-2)     f= 10s - 18

f+5 = 4(s+5)

f= 4s +15        equate the two f definitions     (10s-18) = 4s+15        solve for s = 5.5     so f = 37     Difference = 31.5 years

A quadrilateral has angles of q, 5q + 15, 3q + 25, and 3q + 70. Find the largest angle of the quadrilateral, in degrees

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

∠A = 20.83333334º              1q

∠B = 119.1666667º               5q + 15

∠C = 87.5º                             3q + 25

∠D = 132.5º                           3q + 70

x + 1/y  =  1               y +  1/z  =  1

(xy + 1) / y  = 1        (yz  + 1) /  z  = 1

xy +  1 =  y              yz +  1  = z

xy =  y -1                 1 = z - zy

x = (y-1)/y                1 = z ( 1 - y)

                                 z  = 1/(1-y)

So

xyz  =    (y-1) / y  *   y  *  1/(1-y)       =    (y-1)  /  (1-y)  =   -(1-y)  /( 1-y)  =    -1 

I need help with this system:
\(x(y + z) = 39\\ y(x + z) = 60\\ z(x + y) = 63\)

What is x?

\(\begin{array}{|lrcll|} \hline (1) & x(y + z) &=& 39 \\ (2) & y(x + z) &=& 60 ~\text{or}~ \mathbf{yx+yz = 60} \\ (3) & z(x + y) &=& 63 ~\text{or}~ \mathbf{zx+yz = 63} \\ \hline (1)+(2)+(3): &x(y + z)+ y(x + z)+z(x + y) &=& 39+60+63 \\ & x(y + z) +yx +yz +zx+zy &=& 162 \\ & x(y + z) +x(y+z) +2yz &=& 162 \\ & 2x(y + z) +2yz &=& 162 \quad | \quad :2 \\ & x(y + z) +yz &=& 81 \quad | \quad x(y + z) = 39 \\ & 39 +yz &=& 81 \\ & yz &=& 81-39 \\ & \mathbf{yz} &=& \mathbf{42} \\ \hline & \mathbf{yx+yz} &=& \mathbf{60} \quad | \quad \mathbf{yz = 42} \\ & yx+42 &=& 60 \\ & yx &=& 60-42 \\ & \mathbf{yx} &=& \mathbf{18}~\text{or}~ \mathbf{y=\dfrac{18}{x}} \\ \hline & \mathbf{zx+yz} &=& \mathbf{63} \quad | \quad \mathbf{yz = 42} \\ & zx+42 &=& 63 \\ & zx &=& 63-42 \\ & \mathbf{zx} &=& \mathbf{21}~\text{or}~ \mathbf{z=\dfrac{21}{x}} \\ \hline & \mathbf{yz} &=& \mathbf{42} \\ & \dfrac{18}{x}*\dfrac{21}{x}&=& \mathbf{42} \\ & 42x^2 &=& 18*21 \\ & x^2 &=&\dfrac{18*21}{42} \\ & x^2 &=& \dfrac{3*6*3*7}{6*7} \\ & x^2 &=& 3*3 \\ & \mathbf{x} &=& \mathbf{3} \\ \hline & \mathbf{y} &=& \mathbf{\dfrac{18}{x}} \\ & y &=& \dfrac{18}{3} \\ & \mathbf{y} &=& \mathbf{6} \\ \hline & \mathbf{z} &=& \mathbf{\dfrac{21}{x}} \\ & z &=& \dfrac{21}{3} \\ & \mathbf{z} &=& \mathbf{7} \\ \hline \end{array}\)

Multiply both sides of the first equation by \(y \) and both sides of the second equation by  to obtain \(z\)

\(\begin{align*} xy+1 &= y \\ yz+1 &= z. \end{align*}\)

Substituting \(xy+1\) for \(y\) in the second equation, we find

\((xy+1)z+1=z,\)

which simplifies to

\(xyz+z+1=z.\)

Subtracting \(z+1\) from both sides, we find that \(xyz=z-(z+1)=\boxed{-1}.\)

x+y = 25    y = 25-x

x/y = 4

x/(25-x) = 4

x = 100 - 4x

5x = 100

x = 20      y = 5   

z cannot equal zero......   1/z    in second equation

Is it 3?

8 ft³  * (12 in/ft)³/ (39.37 in/m)³  = .2265 m³

J

1/10*10

Since they are all different, we can count each one separately

Each ball has 4 spots possible to be placed in.

So 4^4=256 is the answer

2x / ( x - 2)  = -6/ ( x + 4)     cross-multiply

2x ( x + 4)   =  -6 ( x  - 2)         simplify

2x^2 + 8x  = -6x   +  12        rearrange as

2x^2 + 14x  - 12   =  0          divide through  by 2

x^2 + 7x   - 6  = 0            add 6 to both sides

x^2  +  7x   =   6

Complete the square on  x

Take 1/2  of 7 =  7/2      square this =  49/4and add to both sides

x^2  +  7x    + 49/4   =  6 +  49/4      factor the left side, simplify the  right

(x + 7/2)^2  =  73/4              take both roots

x + 7/2  =    ±√73/2        subtract 7/2 from  both sides

x =   ±√73/2  - 7/2

x + y  =  10     Subtract  y  from both sides of this equation to get     x  =  10 - y

Now since   x = 10 - y   we can substitute  10 - y  in for  x  in the second equation.

x²  -  y²   =   56

(10 - y)²  -  y²   =   56

(10 - y)(10 - y)  -  y²   =   56

100 - 20y + y²  -  y²   =   56

100 - 20y   =   56

-20y   =   -44

y   =   2.2

Using this value of  y  and the fact that   x  =  10 - y   we can find the value of  x

x   =   10 - y   =   10 - 2.2   =   7.8

So there is only one ordered pair of real numbers that satisfies the system of equations, and it is  (7.8, 2.2)

Let  f  be the age of the father and  s  be the age of the son.

The ratio of the age of the father to the age of the son 2 years ago was  10:1

The ratio of   f - 2   to   s - 2   =  10:1

f - 2 : s - 2   =   10:1

(f - 2) / (s - 2)  =  10/1

(f - 2) / (s - 2)  =  10

f - 2   =   10(s - 2)

f - 2   =   10s - 20

f   =   10s - 18

In 5 years, the age of the father will be  4  times the age of the son.

f + 5   =   4(s + 5)

f + 5   =   4s + 20

f   =   4s + 15

Now since   f  =  10s -18   and   f  =  4s + 15   we can say...

10s - 18  =  4s + 15

10s - 4s  =  15 + 18

6s  =  33

s  =  5.5

Now we can find the value of  f  using this value of  s

f   =   4s + 15   =   4(5.5) + 15   =   37

the difference in the age of the father and son   =   f - s   =   37 - 5.5   =   31.5

The rectangular coordinates of (r^2,2theta) are (20,6).

So we have the multiples of 5 from 4000-5995.

Dividing each multiple of 5 by 5 we get 800-1199.

1199-800+1=400 multiples