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2.

We can find angle BAC as follows

6^2  = 2(5)^2  - 2(5^2) cos(BAC)

[36 - 50]  / (-50)  = cos(BAC)

-14/-50  = cos(BAC)

7/25 = cos (BAC)

arccos ( 7/25) = BAC

And since BAC is an inscribed angle, the BOC has twice its measure =  2arccos(7/25)

And   cos [ 2 arccos (7/25) ]  =    - 527/ 625

And OC  = OB

So......using the Law of Cosines

6^2  = 2(OC)^2 - 2(OC)^2 cos [2arccos(7/25) ]

36 = 2(OC)^2 - 2(OC)^2 [ -527/ 625]

36 = 2(OC)^2 + 2(OC)^2 [ 527/625]

18 = OC^2 [ 1 + 527/625]

18 = OC^2 [ 1152/ 625]

18 * 625 / 1152  = OC^2  =  11250/ 1152  =  625/ 64

So.....the area of triangle OBC  =

(1/2) OC^2 * sin (BOC)

(1/2) (625/64) * sin [ 2 arccos(7/25) ]  = 

(625/ 128) * sin [ 2 arccos (7/25) ]  =

(625 / 128) * (336/625) =

336/128    =

( 21 /  8  )   units^2

Altitudes AX and  BY of acute triangle ABC  intersect at H. If angle AHB=132, then what is angle ACB?

 \(\begin{array}{|rcll|} \hline \angle ACB &=& 180^{\circ} - 132^{\circ} \\ &=& 48^{\circ} \\ \hline \end{array} \)

The quadratic $-x^2+2x-4$ can be written in the form $a(x+b)^2+c$,
where $a$, $b$, and $c$ are constants.
What is $a+b+c$?

\(\begin{array}{|rcll|} \hline && -x^2+2x-4 \\ &=& -(x^2-2x)-4\\ &=& -\left((x-1)^2-1 \right) - 4 \\ &=& -(x-1)^2 +1 - 4 \\ &=& -(x-1)^2 -3 \quad & | \quad a(x+b)^2+c \\ &&& |\quad a=-1 \\ &&& |\quad b=-1 \\ &&& |\quad c=-3 \\ &&& |\mathbf{a+b+c} = -1-1-3 \mathbf{= - 5 } \\ \hline \end{array} \)

For what value of $m$ does the equation $(x+4)(x+1) = m + 2x$ have exactly one real solution?

Express your answer as a common fraction.

\(\begin{array}{|rcll|} \hline (x+4)(x+1) &=& m + 2x \\ x^2+5x+4 &=& m+2x \\ x^2+3x+4-m &=& 0 \\\\ x &=& \dfrac{-3\pm \sqrt{9-4(4-m)} }{2} \\ &=& \dfrac{-3\pm \sqrt{9-16+4m} }{2} \\ &=& \dfrac{-3\pm \sqrt{4m-7} }{2} \\\\ && \text{one real solution: $ \sqrt{4m-7}=0$} \\ \sqrt{4m-7} &=& 0 \\ 4m-7 &=& 0 \\ 4m &=& 7 \\ \mathbf{ m } & \mathbf{=} & \mathbf{ \dfrac{7}{4} } \\ \hline \end{array}\)

It was actually 7, thank you though

"C"  is correct

We have a cone topped by a hemisphere.....the total volume is

pi * r^2 * h / 3   +  (2/3) pi*r^3   =

pi  [ 4^2 * 10 / 3   + (2/3)* 4^3 ]  =

pi  [ 160/ 3  + 128/3 ] =

pi [ 288/3 ]  =

96 pi   cm^3

Volume of a sphere  =

(4/3) pi * radius^3  =

(4/3) * 3.14 * 5^3  ≈

523 cm^3

We need to find the radius of the sphere

Surface area  =  4 * pi * r^2

113.04  = 4 * pi * r^2       divide both sides by 4 pi

113.04 / [ 4 pi]  =  r^2

√ [ 113.04/ [ 4pi] ]  = r

So...the volume is given by

(4/3) pi *  r^3

(4/3)pi * (√ [ 113.04/ [ 4pi] ])^3  =  

35.97 pi   in^3

The circumference  of Cylinder A   = 3m

So...the radius  is given as

(3) / (2 * 3.14)   = r

The circumference of Cylinder B  = 5m

So..... the radius is given as

(5) / (2 * 3.14)  = r

So.....the volume of Cylinder A   =

(3.14) * [ 3 / (2*3.14)]^2  * 5  =  3.58 m^3

The volume of Cylinder B  =

(3.14) [ (5) / (2 * 3.14) ]^2 * 3  =  5.97 m^3

Cylinder B  has the greater volume

1)

The  area of an equilateral triangle   is  given by

√3/4 side^2

So

64√3  =  √3 /4  side^2

256 = side^2

16 cm  = side  of original equilateral triangle

Decreasing each side by 4 cm  produces an area of  

√3/ 4 * 12^2   =  

36√3  cm^2

So  the area decrease  is    64√3  - 36√3   =  28√3 cm^2

The pylon  cosists of a cylinder with a radius of 1 ft  and a height of 3 ft topped by a cone with a radius of 1ft  and a height of 2 ft

So.....the total volume is

pi * r^2 * h  +  pi * r^2 * h  /3  =

pi  [ 1^2 * 3   +  1^2 * 2 / 3 ]  =

pi [ 3 + 2/3 ]  =

[ 3 + 2/3] pi   ft^3

\bump

\(\frac{1}{4}<\frac{x}{7}<\frac{1}{3}\)

Multiply through by 7

7/4 < x < 7/3

1 3/4  < x <  2 1/3

The integer that satisfies this is   x  = 2

3x^2  + x  - 4

3(x^2 + (1/3)x  - 4/3)       complete the square on x

Take 1/2  of (1/3)  =   1/6

Square this.....=  1/36......add and subtract it within the parentheses

3(x^2  + (1/3)x + 1/36  - 4/3 - 1/36 )

3  [ (x + 1/6)^2   -  48/36 - 1/36]

3 [ (x + 1/6)^2  -  49/36 ]

3(x + 1/6)^2  - 49/12

So......k  =  -49/12

Notice  that we can  write  (5, 2, 4)  as the linear combination

1(5, 0, 0)  + 1(0, 2 , 0)  + 1(0, 0, 4)   =   vector (OE)  +vector( ED)   + vector (DP)

So......the length of  OE   =    5

Midpoint  of  (1,2) , (19,4)   =  ( 10, 3)

The line containing points  (0,7) and (4, - 3)  has a slope of    [ -3 - 7] / [4 - 0]  =  -10/4  = -5/2

So....a line perpendicular to this will have a slope of   2/5

So....the equation of this line is

y  = (2/5)(x - 10) + 3

y = (2/5)x - 4 + 3

y = (2/5)x - 1

So....when x  = 20

y =  (2/5)(20) - 1

y  = 8 - 1

y  = 7

EDITED TO CORRECT A SIGN ERROR

Let the height of the monument  = h

Let the distance from the base of the monument to a point directly under the apex  = x

So   we have these equations

tan(42)  =  h / x     rearrange as

x  =   h / tan (42)      (1)

We also have

tan (37.8)  =  h / (100 + x)

(100 + x) tan (37.8)   = h

100*tan(37.8)  + x *tan (37.8)  =  h     (2)

Sub  (1)  into (2)

100* tan(37.8)  + (h /tan(42)* tan (37.8)  = h

100* tan(37.8)  =  h  -  h [ tan (37.8) / tan(42) ]

100* tan (37.8)  =  h  [  1  - tan(37.8)/tan(42) ]

h  =     [ 100 * tan (37.8) ]  /  [  1 - tan(37.8)/tan(42) ] ≈  560 ft

\( 3x^2 + \left(3 + \frac 13\right)x + \frac 13\)

We can write this as

3x^2  + (10/3)x + 1/3

The discriminant  is

(10/3)^2  -  4(3)(1/3)  =

100/ 9  - 4  =

[ 100 - 36 ] / 9  =

64 / 9

x^2 - 16x + k

Take  (1/2)  of 16  = 8

Square this  =  64  =  k

\(-\frac{1}{2}-2kx = 5y \)

\( \left(\frac{1}{4},-6\right)\)

So    we can solve this  for   "k'

-1/2  - 2k(1/4)  = 5(-6)

-1/2 - (1/2)k  = - 30         multiply through by 2

-1  - k  =  -60

-k =  - 59

k   = 59

x^2 + 25x    + c

Take  (1/2)  of 25   =  25/2

Square this  =    625/4   =   c

\(\frac{-5\pm i\sqrt{11}}{6}\)

3x^2 + 5x + k

The product  of the two  roots  is

(-5 + i√11) / 6  *    (-5 - i√11) / 6    =    (25  - i^2* 11)/ 36 =   (25 - (-1)(11) ) / 36  =

[25 + 11] / 36    =   36 / 36   =    1

And this product will  equal  k/3

So

k/3  =  1         multiply both sides by 3

k  = 3

-4 < 2(x - 1) < 8          divide through by  2

-2 <  x  - 1 <  4           add 1 to everything

-1 < x  < 5

So

a + b  =

-1  +  5  =

4