CPhill

3) In triangle ABC, AB=17, AC=*, and BC=15. Let D be the foot of the altitude from C to AB. Find the area of triangle ACD.

If AC was supposed to be 8, we have a right triangle....

Its area  = (1/2) product of the legs  = (1/2) AC * BC  =  (1/2) 8 * 15  =  60  units^2

2) In a certain iscoceles right triangle, the altitude to the hypotenuse has length 4sqrt2. What is the area of the triangle?

Area  =  (1/2)hypotenuse * (altidude drawn to the hypotenuse)   (1)

Area  (1/2)  product of the leg lengths    (2)

Equate (1) and (2)  and let the leg lengths  = L  and we have

(1/2) hypotenuse * ( 4sqrt 2 )  = (1/2) L^2     multiply through by 1/2

hypotenuse * ( 4sqrt(2)   =  L^2

The hypotenuse  =  sqrt  ( L^2 + L^2)  =  sqrt (2 L^2 )  =   Lsqrt 2

L sqrt 2 * 4* sqrt 2  = L^2    divide through  by L

4 sqrt 2 * sqrt 2  = L

4*2  = L

8 = L

So...the area  =  (1/2) L^2  = (1/2) * 8^2  =  32 units^2

x^2 / 9  +  y^2 /25  = 1    can be transformed to  25x^2 + 9y^2  = 225    (1)

y  = 4x + k    (2)

The slope of a tangent line at any point on (1)  can be found as

50x + 18y y'  = 0

y'  =  -50x / [ 18y]   =  -25x / [ 9y]

And we are looking for  where the slope of a tangent line  = 4

So

-25x / 9y  = 4

-25x = 36y

y = (-25)/(36) x       sub this into  (1) for y

25x^2  + 9 (-25/36 x)^2 =  225

25x^2  + 9 (625/1296)x^2  = 225

4225/144 x^2  = 225 

x^2  =225 * 144 / 4225     take both roots

x = 15 * 12 / 65  = 180/65  =36/13

Or

x = -36/13

Subbing either value into  (1) to find  y we have

25 (36/13)^2  + 9y^2  = 225

32400 / 169  + 9y^2  = 225

32400 / 169 + 9y^2  = 38025/169

y^2  = [38025 - 32400 ] / [ 9 * 169]

y^2 = [5625] / [ 9 * 169]      take both roots

y = 75 / [ 3 * 13 ] =  75 / 39  = 25 / 13

OR

y = -25/13

So....the slope of the tangent line to the ellipse = 4  at   (-36/13 , 25/13)  and (36/13. -25/13)

Writing an equation of one tangent line using the first point we have

y  = 4 ( x + 36/13)  + 25/13

y = 4x + 144/13 + 25/13

y = 4x   + 169/13

y = 4x + 13

And writing the equation of the other tangent line we have that

y = 4 (x - 36/13) - 25/13

y = 4x - 144/13  - 25/13

y = 4x -169/13

y = 4x - 13

Note the graph here :  https://www.desmos.com/calculator/syelncmges

When k  = 0 .....the graph intersects the ellipse at two points

However when k < -13  ...the tangent line is shifted to the right of the ellipse

And when k > 13....the tangent line is shifted to the left of the ellipse

No prob  !!!

I'll give this a shot

We can transform the given equation  into

9x^2  + 4y^2   = 36

This is an ellipse centered at the origin

We have

9x^2  - 18x + 4y^2 + 16y  =  c    ..... [ complete the square on x , y]

9 ( x^2 - 2x + 1)  + 4 ( y^2 + 4y + 4)  = c + 9 + 16

9 ( x - 1)^2  +  4 (y + 2)^2  =   [ c + 9 + 16 ]     divide both sides by 36

(x - 1)^2 /  4   +  (y + 2)^2 / 9  =  [ c + 9 + 16] / 36

Since we want the right side to be  1,  c  must  = 11

So  we have the equation

(x - 1)^2 / 4  +  (y + 2)^2 / 9   =  1

So....the translation of the original graph  is  1 unit to the right and  2 units down  =   vector ( 1 , -2 )

So...a =  1 , b = -2   and  c  = 11

That isn't the equation of an ellipse...it's the equation of a hyperbola....

The system will have infinite solutions when both equations are  exactly the same

Note that we can add 4y to both sides of the second equation and get

6x + 8y   =  c

Notice that the left side is just a multiple  [ of 2 ] of the left side of the  first equation

So...multiplying the right side of the first equation by 2  gives us 14  = c  and both equations will be exactly the same.....so....the system will have infinite solutions!!!!

Notice that we have four corner rectangles in the darker strip, each with dimensions of  2 x 1  = 2 unit^2

So....the total area of these  = 8 units^2

The other parts of the dark gray strip would be rectangles with the same dimensions as the rectangles  comprising the area of the lighter gray strip...so...their total areas  =  100

So...the  area of the darker gray strip  = 100 + 8  =  108 sq units

 Corrected answer  !!!

.

100 integers from 500 - 599 contain at least one 5

And in each of the 499 integers

_5, _15, _25, _35, _ 45, _55 , _65, _ 75, _ 85 , _ 95

In each hundred  from 1 to 499 contain at least one 5

And there are 5 sets of these = 5 * 10  =  50

And in each hundred from 1 to 499 we have the  integers

_5_   where the last digit is not a 5

And we have 9 of these in every hundred

So...we have  5 * 9  = 45 of these

So....100 + 50 + 45  = 195

Just as  Dimitristhym  found  !!!