CPhill

y = (25 - x^2)^1/2    taking the derivative, we have

y' = (1/2)(25 -x^2)^(-1/2) (-2x) =  -x / √[25 - x^2]

And the slope of the line when x = 3 = -3/ √[25 - 3^2]  = -3/ √16 = -3/4

And the slope of the normal line will be the negative reciprocal of this = 4/3

So we have

y - 4 = (4/3)(x -3)

y = (4/3)x - 4 + 4

y = (4/3)x       and that's the equation of the normal line to the function at (3, 4)

See the pic here of the function, the tangent line to the function at (3, 4) and the normal line to the function going through the same point

My original amswer wasn't entirely correct.....you actually helped me to see MY dumb error...!!

Thanks, Melody......when I checked again, that second solution IS possible, too!!!

Here is the other triangle that Melody found....

 GOOD JOB, MELODY !!!!   A POX ON ME !!!!

343^(-4/3)  =

[343^(1/3)]^(-4) =

7^(-4)  = 

1/(7^4) =

1 / 2401

All of them.......

0 1 2 3 4 5 6 7 8 9

 -6x+y=4      rearrange  by adding 6x to both sides

y = 6x + 4   and that's in S-I form

Just hit the "2nd button" on the onsite calculator and then the"acos"  button......

If we do this correctly....we will arrive at a familiar formula....let's see...

ax^2+bx+c=0   subtract c from both sides

ax^2 + bx  = -c       divide through by a

x^2 +(b/a)x = -c/a     complete the square on x

x^2 + (b/a)x + [b^2] / [4a^2]  = -c/a + [b^2] / [4a^2]     factor the right side....get a common denominator on the left

[x + (b/2a)]^2  = [b^2 - 4ac] / [4a^2]   take the square roots of both sides

x + (b/2a)  = ±√[b^2 - 4ac]/[2a]    subtract (b/2a) from both sides

x = (-b±√[b^2 - 4ac]) / [2a]

Voila!!!....we have the Quadratic Formula..... !!!!

We'll have to use Heron's Formula to solve this.

Let  a = 26, b= 40 and x be the unknown side

s = the semi-perimeter  =  [x + 26 + 40]/2 = [x + 66]/2

So we have

200 = √[s(s -a)(s-b)(s-x)]

200^2  = [x + 66]/2 * [(x + 66)/2 - 26] * [(x + 66)/2 - 40]* [(x + 66)/2 - x] 

200^2 = [x + 66]/2 * [ x + 14]/2 * [x - 14]/2 * [66 - x]/2

200^2 = [x + 66] * [x + 14] * [x - 14] * [66- x] * (1/16)

40000 = [x + 66] * [x + 14] * [x - 14] * [66- x] * (1/16)

640000= [x + 66] * [x + 14] * [x - 14] * [66- x] 

We could brute force this.....but....I'm going to let WolframAlpha do the heavy lifting...

It returns two solutions

x = 2√89 ≈ 18.87   and  2√1049 ≈ 64.78

Both satisfy the Triangle Inequality.....so.....looking at the Law of Cosines

18.87^2  = 40^2 + 26^2 - 2(40)(26)cosΘ   

And the angle between the sides of 26 and 40 could be 22.6°

And using the Law of Sines, we have

sinΘ/40 = sin 22.6/18.87

sinΘ = 40sin22.6/18.87 =  54.55°

And the remaining angle is 180 - 22.6 - 54.55 = 102.85

But this is impossible because it would mean that the greatest angle is opposite the intermediate side

And using the Law of Cosines again, we have... 

64.78^2 = 40^2 + 26^2 - 2(40)(26)cosΘ 

So, the angle between the sides of 26 and 40 could be 157.38°

And using the Law of Sines again, we have

sinΘ/40 = sin 157.38/64.78 = 13.73°

And the remaining angle is 180 - 157.38 - 13.73 = 8.89°

So, the solution is

Side 64.78,  opposite angle = 157.38°

Side 40, oppsite angle 13.73°

Side 26, opposite angle 8.89°

So....the remaining unknown side is 64.78

Here's the approximate triangle

SEE MELODY'S ANSWER BELOW...THE SECOND SOLUTION IS POSSIBLE  !!!

Find three consecutive even integers such that the sum of the smallest integer and twice the second is 12 more than the third.

Let x be the first integer, x +2 be the second and x + 4  the third  ....so ....

x + 2(x + 2)  =  (x + 4) + 12  simplify

x + 2x + 4  = x + 16

3x + 4  = x + 16     subtract x, 4  from both sides

2x = 12     divide by 2 on both sides

x = 6  

x + 2  = 8

x + 4 = 10