CPhill

OK.....glad to help.....!!!

Let's see hoe tertre arrived at his solution

7x^(1/4) - ( x^1/4)^2 = 12     multiply through by -1

(x^1/4)^2 - 7(x^1/4) = -12

(x^1/4)^2 - 7(x^1/4) + 12 = 0         let  x^1/4 = a     and we have

a^2 - 7a + 12 = 0     factor

(a - 4) (a - 3) = 0      

Setting both factors to 0 and solving for "a" gives   a =4   or a = 3

So

x^1/4 = 4         or       x^1/4 = 3      take each side to the 4th power

x = 256          or       x =  81

And the sum of these is 337.....just as tertre found !!!!!

Let's take this by parts

Conjugate of13 + 2i  =  13 - 2i

So  (1 + i) (13 - 2i)  =  13 + 13i - 2i - 2i^2  =   15 + 11i

And the connjugate of 15 + 11i   =  15 - 11i

So

i (15 - 11i) =  15i - 11i^2  =   11 + 15i

And the conjugate of this  is   11 - 15i

(1 + i) z  - 2z(conjugate)  =  -11 + 25i

(1 + i) (a + bi) - 2(a - bi) = -11 + 25i

a + ai + bi + bi^2 - 2a + 2bi  = -11 + 25i

-a +3bi + ai - b  = -11 + 25i 

-(a + b) + (a + 3b)i =  -11 + 25i

Equate real and complex parts and we have this system

- (a + b) = - 11   ⇒  a + b = 11    ⇒  b = 11 - a     (1)

a + 3b = 25      (2)

Sub (1) into (2)  and we have that

a + 3(11 - a) = 25

a + 33 - 3a = 25

-2a = - 8

a = 4

And using (1)    b = 11 - a  = 11 - 4  = 7

So.....the complex number z is

4 + 7i

Note that angle C   = 180  - (38 + 44) = 180 - 82  = 98°

Using the Law of Sines, we  can  find DC as

DC / sin 38 = BD / sin  98

DC = 10 *sin 38 * sin 98 ≈ 6.097

And the area of BCD can be found as

(1/2) DB * DC sin BDC  =   (1/2) *10 * 6.097 sin 44   ≈  21.18 units^2      (1)

And angle ABD = 90 -  55  =  35°

Using the Law of Sines again

BD/sin A  =  AD /   sin ABD

10/ sin 90 = AD /sin 35

AD = 10 sin 35  ≈ 5.74

And the area of ABD can be found as

(1/2)BD * AD *sin ADB =

(1/2) 10 * 5.74 sin 55 ≈   23.5 units^2   (2)

So....the total area = (1) + (2)  = [ 21.18 + 23.5] units^2 ≈ 44.68 units^2 = 44.7 units^2

2. 

Triangle DEF is the medial triangle of Triangle ABC and Triangle XYZ is the medial triangle of Triangle DEF. If the perimeter of Triangle XYZ is 5, what is the perimeter of Triangle ABC?

A medial triangle has 1/2 the perimeter of its enclosing triangle....so

Perimeter of DEF = 2 * Perimeter of XYZ  = 10

Perimeter of ABC = 2 * Perimeter of DEF = 20

Ah....thanks, Rom....I didn't think about that   !!!

Call the lengths of the two legs x and y

We have this system of equations

x^2 + [(1/2)y]^2 = 25   ⇒  x^2 + (1/4)y^2 = 25    (1)

[(1/2)x]^2 + y^2 = 40    ⇒  (1/4)x^2 + y^2 = 40   (2)

Multiply equation (1)  through by -4 and we have this system

-4x^2   - y^2 =  -100

(1/4)x^2 + y^2 = 40       add these

-(15/4)x^2 = - 60

x^2 = 16

x = 4

And using (2) to find y, we have

(1/4)(4)^2 + y^2 = 40

4 + y^2 = 40

y^2 = 36

y = 6

So....the hypotenuse length is   sqrt (4^2 + 6^2 )  =  sqrt (52) = 2sqrt  (13)

And in a right triangle, the median drawn from the right angle is 1/2 the hypotenuse length = 

sqrt (13)

For any perimeter,P,  the largest area that can be enclosed is a square with a side of P/4

So....the max area is

(100/4)^2 =  25^2   =   625 ft^2

Floor area of 1 drum =

 pi * ( 1.25)^2  =    1.5625 pi  ft^2     (1)

Total Floor Area = 7 * 7.5 = 52.5 ft^2   (2)

Number of drums that can be placed =  (2) / (1) =  

52.5 / [ 1.5625 pi ]  ≈  10 drums