CPhill

We have ..... Length * Width = Area .....so.....

(x +2) (x + 7) = 84   simplify

x^2 + 9x  + 14  = 84       subtract 84 from each side

x^2 + 9x - 70   = 0        factor

(x - 5) (x + 14)  = 0

And serring each factor to 0, we have that x = 5  or x = -14

Reject the second solution....so.....the length = (5 + 2) = 7   and the width = (5 + 7)  = 12

43,560 sq ft

x/8+x/10=9/8    multiply through by the common denominator of 8 and 10   = 40

40(x/8) + 40(x/10) = 40(9/8)

5x + 4x  = 45       sinplify

9x = 45       divide both sides by 9

x = 5

23 / 27  = about 85.2%

√(25/81)  = √25 / √81  =    5/9

cos2x+5cosx+3=0   ....note that cos2x = 2cos^x - 1  ....so we have.......

2cos^x - 1 + 5 cosx + 3  = 0     simplify

2cos^x + 5cosx + 2   = 0      factor

(2cosx + 1) (cosx + 2)  = 0     setting each factor to 0, we have......

2cosx + 1 = 0                             and                           cosx + 2 = 0

subtract 1 from each side                                            subtract 2 from each side

2cosx = -1                                                                cos x = -2

divide both sides by 2                                                this has no solution

cosx = -1/2

And the solution to this is  x  = (2/3)pi + 2pi * n    and     x = (4/3)pi + 2pi * n ....   where n is any integer

Area _trapezoid =  [height/2] * [ sum of the bases]  .....so we have.....

[5/2] * [7 + 6]  =

(5/2]*[ 13]  =

65/2  sq ft = 32.5 sq ft

(7 + 2/3) /  5 =

(23/3) / (5/1)  =  [ remember, that when we divide fractions, we "keep the first,"  "change to multiplication" and "flip" the second fraction ]......so, we have.......

23/3   * (1/5)   =

23/15 

log₃ 4    .....we can use the "change-of-base" rule here    ......we have.....

log 4 / log 3   =  about  1.26186

10m^2 - 12m - 12 = 0      divide every term by 2

This will not factor.......using the onsite solver and the quadratic formula, we have these two solutions :

$${\mathtt{5}}\left[{{m}}^{{\mathtt{2}}}\right]{\mathtt{\,-\,}}{\mathtt{6}}{m}{\mathtt{\,-\,}}{\mathtt{6}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{m}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{39}}}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{{\mathtt{5}}}}\\
{\mathtt{m}} = {\frac{\left({\sqrt{{\mathtt{39}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{{\mathtt{5}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{m}} = -{\mathtt{0.648\: \!999\: \!599\: \!679\: \!679\: \!6}}\\
{\mathtt{m}} = {\mathtt{1.848\: \!999\: \!599\: \!679\: \!679\: \!6}}\\
\end{array} \right\}$$