CPhill

2. Find the length of a side of an equilateral triangle of area 10.2 m^2.

Area  =  (1/2)* side^2 * [√3 / 2 ]

10.2 = [√3/4 ] * side^2     multiply both sides by   4/ √3

[10.2 * 4 ]  /  √3    =   side^2

40.8 / √3  = side^2   take the square root  of both sides

√ [  40.8 / √3 ]  = side  ≈ 4.853 m

1. A kite of area 40 m^2 has one diagonal 2 m longer than the other. Find the lengths of the diagonals.

Area_kite  = D1 * D2  / 2    let  D2  = D1 + 2

40  =  D1 * [ D1 + 2]  / 2         multiply both sides by 2

80 =  D1 *  [D1 + 2] 

80 =  [D1]^2  + 2D1         rearrange

[D1]^2  + 2D1  −  80   = 0       factor

( D1   +  10)  (D1 −  8 )  =  0        set the factors to 0  →  D1 = −10m  [reject ]     or   D1  = 8 m

And D2 =  D1  + 2   =  8 + 2    = 10 m

Thanks, Tiggsy......!!!!!

w = 4(2t - 9)     divide by 4 on both sides

w /4  = 2t - 9    add  9 to both sides

w/4  +  9  =  2t         divide through by 2

w/8   +  9/2   =   t

Five times seven and seven times three
Add to my age and it will be
As far above six nines and four
As twice my years exceeds a score!  

Let x be your age.....and we have .....[5*7  +  7*3  +   x ]   =  [56  + x]        ....so.....

[56 + x]  −  6*9 +  4   =  2x   −  20      [a score = 20]

56 + x  −  58    =   2x   −  20

x  − 2  = 2x  −  20        rearrange

18 =  x  = your age

3 1/2 divided by 2 1/3

3 1/2 =   7/2

2 1/3 =  7/3

So....

[7/2]  / [ 7/3]  =

[7/2]  * [3/7]   =

3/2    =    1 + 1/2

Very nice, Majid....!!!!

y = 1 / [x + 2]       multiply   both sides by   x + 2

y [x + 2]  =  1       divide both sides by  y

x + 2   = 1 / y       subtract 2 from both sides

x = 1 / y  − 2       swap x and y

y   =  1/x − 2          for y, write f⁻¹(x)

f⁻¹(x)  = 1/x  − 2

Here's the graphs  :    https://www.desmos.com/calculator/thzey4cwwc

y'+6y/(x+2)=1/(x+2)²   multiply both sides by (x+ 2)^6

y' (x + 2)^6  +  6y(x + 2)^5  =  (x + 2)^4

Let u   = y (x + 2)^6    →  du =   y' (x + 2)^6  +  6y(x + 2)^5  

So we have

du   = (x + 2)^4    integrate both sides

∫ du   = ∫  (x + 2)^4 dx

u =  (x + 2)^5 / 5    +  C₁

y (x + 2)^6  =  (x + 2)^5 / 5    +  C₁        divide through by   ( x + 2)^6

y  =   1 / [5( x+ 2)]  +  C₁ / ( x + 2)^6

y  =   1 / [5x + 10]  +  C₁ / ( x + 2)^6

xy'+2y=x²-3        multiply both sides by x

x^2y'+2xy=x^3-3x

Let u  =  x^2*y   →  du   =  x^2y'  + 2xy

So we have

du  = x^3-3x      integrate both sides

∫ du   = ∫ x^3 -3x dx

u  = x^4/4  - (3/2)x^2  + C₁

x^2*y = x^4/4  - (3/2)x^2  + C₁     divide through by   x^2

y  =  x^2/4  - (3/2)  +  C₁ / x^2