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x(2x - 7) = 3  simplify

2x^2 - 7x - 3  = 0

2 ( x^2 - (7/2)x - 3/2)  = 0     divide through by 2

x^2 - (7/2)x - 3/2   = 0      complete the square on x

x^2 - (7/2)x  = 3/2

x^2 - (7/2)x  + 49/16   =  3/2 + 49/16

(x - 7/4)^2  =  (24 + 49) / 16

( x - 7/4)^2 = ( 73) / 16           take both roots

x - 7/4  =  ±√(73) / 4

x =   [ 7 ±√(73)  ] / 4

m = 7   n = 73   and p =4

So    m + n + p  =   84

x^2 + (a - (1/a))x - 1 = 0

x^2 +[ (a^2 - 1) / a ]x - 1  = 0   complete the square on x

x^2 +  [ (a^2 - 1) / a ] x  =  1

x^2 + [ (a^2 - 1) / a ] x  +  (a^2 - 1)^2/ [4a^2]  =   1 + (a^2 - 1)^2 / [4a^2 ]

(x + (a^2 - 1)/[2a] )^2   = [ 4a^2 + a^4 - 2a^2 + 1 ] / [ 4a^2]

(x + (a^2 - 1) / [2a] )^2  =  [ a^4 + 2a^2 + 1] / [ 4a^2]

(x + (a^2 - 1) / [2a])^2 =  [ a^2 + 1]^2 / [ 4a]^2      take both roots

( x + (a^2 - 1) / [2a] )  =  ±√ [( a^2 + 1)^2 / [4a]^2]

x + (a^2-1) / [2a]  = ± ( [ a^2 + 1] / [2a] )

x =  ± ([a^2 + 1] / [2a]) - (a^2-1) /[2a ]

The area (in km^2)  = (3.14) (.75)^2

So...the density is

32 / [ (3.14) (.75)^2 ]  ≈    18 antelopes/km^2

\((2,3)\)  is at the polar coordinates of  \((r,\theta)\)  so:

\(2\ =\ r\cos\theta\\~\\ 3\ =\ r\sin\theta\)

\((x_1,y_1)\)  is at the polar coordinates of  \((2r,\theta+\frac{\pi}{2})\)  so:

\(\begin{array}{} x_1\ =\ (2r)\cos(\theta+\frac{\pi}{2})&=&(2r)(-\sin\theta)&=&(-2)(r\sin\theta)&=&(-2)(3)&=&-6\\~\\ y_1\ =\ (2r)\sin(\theta+\frac{\pi}{2})&=&(2r)(\cos\theta)&=&(2)(r\cos\theta)&=&(2)(2)&=&4 \end{array}\)

\((x_2,y_2)\)  is at the polar coordinates of  \((-r,-\theta)\)  so:

\(\begin{array}{c} x_2\ =\ (-r)\cos(-\theta)&=&(-r)\cos\theta&=&-(r\cos\theta)&=&-2\\~\\ y_2\ =\ (-r)\sin(-\theta)&=&(-r)(-\sin\theta)&=&r\sin\theta&=&3 \end{array}\)_

The sum of 3 real numbers is known to be zero.
If the sum of their cubes is \(\pi^e \), what is their product equal to?

\(x+y+z = 0 \\ x^3+y^3+z^3 = \pi^e \)

1.)

\(\begin{array}{rcll} 0&=&(x+y+z)^3 \\ &=& x^3+y^3+z^3 +6xyz+3\left( x^2(y+z)+y^2(x+z)+z^2(x+y)\right) \end{array} \)

2.)

\(\begin{array}{rcll} 0&=&(x+y+z)(x^2+y^2+z^2) \\ &=& x^3+y^3+z^3 + x^2(y+z)+y^2(x+z)+z^2(x+y) \\ x^2(y+z)+y^2(x+z)+z^2(x+y) &=& -(x^3+y^3+z^3) \\ \end{array} \)

\( \begin{array}{rcll} x^3+y^3+z^3 +6xyz+3\left( x^2(y+z)+y^2(x+z)+z^2(x+y)\right) &=& 0 \\ x^3+y^3+z^3 +6xyz+3\left( -(x^3+y^3+z^3) \right) &=& 0 \\ -2(x^3+y^3+z^3) +6xyz &=& 0 \\ 6xyz &=& 2(x^3+y^3+z^3) \\ xyz &=& \dfrac{1}{3}(x^3+y^3+z^3) \\ \mathbf{xyz} &=& \mathbf {\dfrac{ \pi^e}{3}} \\ \end{array}\)

As follows:

1 + 2 + 3 - 4 + 5 + 6

We want to subtract as much as possible. So the minimum possible value this expression can obtain is

1 + 2 + 3 - (4 + 5 + 6)   =   6 - 15   =   -9

n₁  =  0

n₂  =  1

n₃  =  6

n₄  =  2

n_k  =  n_{(k mod 4)}     and     n₀  =  n₄

We need the tangent here

The shadow length is the adjacent side  to the angle and we are  trying to  find the opposite side....so...

tan (43)  = Tree  Height

                 ___________             multiply bothsides by 9.3

                        9.3

9.3 * tan (43)  = Tree Height  ≈   8.67 m

In triangle ABC, B = 90 degrees, and AC = 13 cm.

Determine:

A) BC if C = 17 degrees. B) A if BC = 6cm .

A) If C = 17°  then  A = 90 -17  = 73°

By the Law of Sines

BC / sin 73  =  AC / sin90

BC / sin 73 = 13/1

BC = 13sin(73)    ≈  12.4

B)  A if BC = 6

Again by the Law of Sines

sin A / BC  = sin90 / AC

sin A / 6 = 1 /13

sin A = 6/13      take the sine inverse to find angle A

arcsin (6/13) = A ≈ 27.5°

Subtraction Property of Equality

Division Property of Equality

2.

For how many positive integers x is x^2+4x+4 between 10 and 50?

Note   x^2 + 4x + 4  =  (x + 2)^2

Whe x = 1        (x + 2)^2  = 9 

When x = 2      (x + 2)^2  =  16

When x = 5       (x + 2)^2 = 49

When x = 6       (x + 2)^2   =  64

So....  four integers....2, 3 , 4 , 5

1.

Solve for x, where x>0 and 0=-9x^2-3x+2 Express your answer as a simplified common fraction.

Multiply through by - 1  and we have

9x^2 + 3x - 2  = 0

(3x + 2) (3x - 1)  = 0

Only the second factor produces a positive solution of  1/3

Let me check!!!

It says the answer is 5, do you know where you went wrong?

The solutions are    1/2 + √5/2      and    1/2 - √5/2

So we have

[  (1/2 + √5/2) - (1/2 - √5/2) ]^2  =

[ √5/2 + √5/2 ]^2  =

[ 2√5 / 2 ]^2

[ √5 ] ^2  =

5

EDITED

3y^2 + 5y + 2  = 4

3y^2 + 5y - 2  = 0

(3y - 1) (y + 2)  = 0

Set each factor to 0...solve for y

3y - 1 = 0             y + 2 = 0

3y =1                    y = - 2

y =1/3

y = -2  is the least  value

Ah, tertre....you beat me to it....sneaky  !!!!

( x + y)^2  = 78

x^2 + 2xy + y^2  = 78

If xy = 13 .....then

x^2 + y^2 + 2(13) = 78

x^2 + y^2 + 26 = 78

x^2 + y^2  = 52

So

(x - y)^2  =  

x^2 - 2xy + y^2  =

(x^2 + y^2) - 2xy =

(52) - 2(13)  =

26

(x+y)^2=x^2+y^2+2xy=78

x^2+y^2+2(13)=78

x^2+y^2=52

(x-y)^2=x^2+y^2-2xy, 52-2(13)=52-26=26,

-tertre

For what values of k does the equation 3x^2 + 4x + k = 0 have no real roots? One real root? Two real roots?

No real roots  if the discriminant is <  0   ....so....

(-4)^2 - 4(3)(k) < 0

16 - 12k < 0

16 < 12k

4/3 < k

k > 4/3

One real root if the disriminant = 0

(-4)^2 - 4(3)k = 0

16 - 12k = 0

16 = 12k

4/3 =  k 

It's obvious that we must have two real roots if  k < 4/3

4x^2 - 4x + m

If this has one zero.....then the discriminant  = 0    ....so....

(-4)^2 - 4(4)m  = 0

16 - 16m  = 0

16 = 16m     divide both sides by 16

1  = m

If we have only one solution, the discriminant must  = 0

So

32^2  - 4ac  = 0             ⇒  c  = 130 - a

1024 - 4  (a) (130 - a)  = 0

1024 - 520a + 4a^2  = 0

4a^2 -520a + 1024  = 0       divide through by 4

a^2 - 130a + 256  = 0      factor as

(a -128) (a - 2)  = 0

Set each factor to 0  and solve for a

a - 128  = 0           a - 2 =0

a = 128                  a = 2

But...since  a > c

Then a must = 128    and c = 130 - 128  = 2  

(a, c )  =  (128, 2)

By trial and error....I get that x  = 5

(2 * 5)^2 + 2*37*2(5) + 37^2  = 2209

2209 / 47   = 47