x+y=9 y=2x
Subbing the second equation into the first for y, we have
x + (2x) = 9 simplify
3x = 9 divide by 3 on both sides
x = 3
And y = 2x = 2(3) = 6
So....our solution is
(3, 6)
y = x + 3
y - 3 = x
In the second equation we can rearrange it as
This is the same as the first equation........so we have infinite solurions on this line
The total outcomes possible = C(9,2) = 36
But, we're just interested in choosing 2 of the 3 dimes = C(3, 2) = 3
So....the probability is 3 / 36 = 1 / 12 = 8.33 %
Here you go.......https://www.desmos.com/calculator/qi3h0wixx7
Notice that the " < " means the line itself isn't included in the solution
15.6% = .156
So
.156N = 13 divide both sides by .156
N = 13/ .156 = 83 1/3
400 = e^(3t) take the ln of each side
ln 400 = ln e^(3t) and we can write
ln 400 = (3t) ln e and ln e = 1 so we can disregard this.....so we have
ln 400 = 3t divide both sides by 3
ln 400 / 3 = t = about 1.997
Let x be each side of the fountain....so we have
L x W = 700
(x + 7)(x + 7) = 700 simplify
x^2 + 14x + 49 = 700
x^2 + 14x - 651 = 0
Using the on-site solver, we have
$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{14}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{651}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\mathtt{10}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{7}}}}{\mathtt{\,-\,}}{\mathtt{7}}\\ {\mathtt{x}} = {\mathtt{10}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{7}}}}{\mathtt{\,-\,}}{\mathtt{7}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{33.457\: \!513\: \!110\: \!645\: \!905\: \!9}}\\ {\mathtt{x}} = {\mathtt{19.457\: \!513\: \!110\: \!645\: \!905\: \!9}}\\ \end{array} \right\}$$
So...each side of the fountain is about 19.46 ft
And the area is (19.46)^2 = 378.7 sq ft
See my answer here....http://web2.0calc.com/questions/how-do-i-solve-this_10
10^100
In other words......1 followed by 100 zeroes.....
N/6 + 8 - 3 = 7 simplify
N/6 + 5 = 7 subtract 5 from both sides
N/6 = 2 multiply both sides by 6
N = 12 .....and that's the number