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Let triangle ABC have side lengths AB=13, AC=14, and BC=15.

There are two circles located inside angle BAC which are tangent to rays AB, AC, and segment BC.

Compute the distance between the centers of these two circles.

\(\begin{array}{|rcll|} \hline s &=& \dfrac{a+b+c}{2} \\ s &=& \dfrac{13+15+14}{2} \\ \mathbf{s} &=& \mathbf{ 21 } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline r &=& \sqrt{\dfrac{(s-a)(s-b)(s-c)}{s}} \\ &=& \sqrt{\dfrac{(21-13)(21-15)(21-14)}{21}} \\ &=& \sqrt{\dfrac{(8)(6)(7)}{21}} \\ &=& \sqrt{\dfrac{336}{21}} \\ &=& \sqrt{16} \\ \mathbf{r} &=& \mathbf{ 4 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline r_b &=& \sqrt{\dfrac{s(s-a)(s-c)}{s-b}} \\ &=& \sqrt{\dfrac{21(21-13)(21-14)}{21-15}} \\ &=& \sqrt{\dfrac{21(8)(7)}{6}} \\ &=& \sqrt{ 196 } \\ \mathbf{r_b} &=& \mathbf{ 14 } \\ \hline \end{array}\)

\(\mathbf{AU=\ ?}\)

\(\begin{array}{|rcll|} \hline AU &=& AV \\ AU+AV &=& (a+BU)+(c+CV) \\ &=& a+c+BU+CV \quad | \quad BU = BU',\ CV=CV' \\ &=& a+c+BU'+CV' \quad | \quad BU'+CV' = b \\ &=& a+c+ b \\ AU+ AV&=& 2s \quad | \quad AV = AU \\ 2AU &=& 2s \\ \mathbf{ AU } &=& \mathbf{ s } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline x &=& \sqrt{\Big(s-(s-b)\Big)^2 + (r_b-r)^2 } \\ &=& \sqrt{b^2 + (14-4)^2 } \\ &=& \sqrt{15^2 + 10^2 } \\ &=& \sqrt{(3*5)^2 + (2*5)^2 } \\ &=& \sqrt{5^2*(3^2+2^2) } \\ &=& \sqrt{5^2*13 } \\ \mathbf{x } &=& \mathbf{ 5\sqrt{ 13 } } \\ x &\approx& 18 \\ \hline \end{array} \)

The distance between the centers of these two circles is \(\mathbf{\approx 18}\)

A.

volume of rectangular prism  =  length * width * height

volume of rectangular prism  =  20 cm * 9 cm * 8 cm

volume of rectangular prism  =  1440 cm²

B.

volume of circular cylinder  =  area of circular base * height

volume of circular cylinder  =  π * radius² * height

radius  =  diameter / 2  =  10/2 cm  =  5 cm

volume of circular cylinder  =  π * (5 cm)² * 15 cm

volume of circular cylinder  =  π * 25 * 15 cm³

volume of circular cylinder  =  375π  cm³

volume of circular cylinder  ≈  1178  cm³        (That is to the nearest cubic centimeter.)

C.

volume of remaining wax   =   volume of rectangular prism - volume of circular cylinder

volume of remaining wax   =   1440 cm³  -  375π  cm³

volume of remaining wax   =   ( 1440 - 375π ) cm³

volume of cube  =  volume of remaining wax  =  ( 1440 - 375π ) cm³

volume of cube  =  side * side * side  =  ( side )³

                                                        Equate both representations of volume of cube.

( side )³  =  ( 1440 - 375π ) cm³

                                                        Take the cube root of both sides of the equation.

side  =  \(\sqrt[3]{1440-375\pi}\)  cm

                                                        Plug  \(\sqrt[3]{1440-375\pi}\)  into a calculator.

side  ≈  6 cm

This is what I found....

Let's label the figure like this:

A line from the center of a circle to a point of tangency is perpendicular to its tangent line, so  CD  is perpendicular to  AB .

And so  △CDA  is a right triangle  and  △CDB  is a right triangle.

By the Pythagorean theorem:

\(r^2+s^2\,=\,16^2\)

and

\(r^2+t^2\,=\,(\frac{15}{2})^2\)

and

\(16^2+(\frac{15}{2})^2=(s+t)^2\)

Now we have 3 equations and 3 unknowns.

Now I am really confused....because when I used WolframAlpha to solve the system of equations, it found that

\(r\,=\,\frac{240}{\sqrt{1249}}\,=\,\frac{240\sqrt{1249}}{1249}\,\approx\,6.791\)

Label the drawing along with me.

Label the bottom left vertex of the triangle A

Label the top vertex of the triangle B

Label the bottom right vertex of the triangle C

Label the point of tangency on the right D

Bisect the base of the triangle and label the point E - this is also the center of the circle

A line from the center of a circle to the point of tangency of a line is a right angle

So the little triangle CDE is a right triangle

The large triangle ABC is almost an equilateral triangle.  If it were equilateral, angle DCE would be 60^o and DEC 30^o 

The sides are slightly longer than the base, so DCE is slightly larger than 60^o but for now let's call it 60^o - we'll fix it later.

Calling DCE 60^o makes DEC 30^o (remember, it's just for now) 

Recalling our trig, we know that the cosine of 30^o is the (square root of 3) over 2

The cosine of our small triangle, CDE, is DE over CE

The cosine of our small triangle, CDE, is (sqrt 3) over 2

Set them equal                            DE / CE = (sqrt 3) / 2

Multiply both sides by CE            DE = CE * (sqrt 3) / 2

But CE is half of 15 so                 DE = 7.5 * (sqrt 3) / 2

                                                    DE = 3.75 * (sqrt 3)

But remember angle CED is a little smaller than 30^o so that makes the cosine a little larger

Looking at the multiple choices, 4  * (sqrt 3) is a little larger than 3.75 * (sqrt 3)

None of the other choices is even close, therefore A has to be the right answer .... 4 * (sqrt 3) 

.

Option 2 is correct, I just took the test.

There are 8 numbers between 100-159 divisible by 7

9 between 200-259, and again

9 between 300-359

So 26/180 = 13/90

I went 7A + B = 5B + A

thus    6A = 4B

From this I concluded that A and B are 4 and 6. 

(base 7) 46  =  (base 10) 28 + 6 = 34

(base 5) 64  =  (base 10) 30 + 4 = 34

The question is what is the integer in decimal.  It's 34. 

.

im sorry but part 2 was wrong

thanks. 1 was right. :)

Thank you soooooooo much! I get it now!!!!!!!!!

Partial Fraction Decomposition does the job. 

I might as well post a solution to the first problem...

a = 2 (mod 3)

a = 4 (mod 5)
a = 6 (mod 7)

a = 8 (mod 9)

The above is the system of congruences.

The answer is 314, by using the general Chinese Remainder Theorem. https://www.youtube.com/watch?v=xjtx4CaQa_4

Maybe not the most simple way....but

Let A(--2, 4)    B = (2, 4)    C  = (2, 0)    D = (-2, 0)

The height of the equilateral triangle will  be    side/ 2 * sqrt (3)  = 4/2 * sqrt (3) = 2 sqrt (3)

So point E will be located at  (0, 4 - 2sqrt (3) )

And using points B and E we can write the equation for the line containg the segment BE

The slope is    [ 4 -(4- 2sqrt (3)) ] / [ 2 - 0]  =  2sqrt(3) / 2  =  sqrt (3)

So....the equation of this line  is

y =  sqrt (3) ( x - 2) + 4

y = sqrt (3)x - 2sqrt(3) + 4

y = sqrt (3)x + (4 - 2sqrt (3) )       (1)

And we can find the equation of the line that joins AC

The slope is   [ 4 - 0 ] / [ -2 - 2 ]  =   4/-4 = - 1

So....the equation of this line is

y = -(x - 0) + 2

y = -x + 2       (2)

Set (1)  = (2) to find the x coordinate of P

sqrt (3) x + (4 - 2sqrt (3) )  = - x + 2        rearrange

x ( sqrt (3) + 1)  =  2 - 4 + 2sqrt (3)

x ( sqrt (3) + 1)  =  2sqrt (3) - 2

x  =  [ 2sqrt ( 3) - 2 ]

       _____________        rationalize the denominator

            sqrt (3) + 1

        [ 2sqrt (3) - 2 ] [ sqrt (3) - 1]             6 - 2sqrt (3) - 2sqrt (3) + 2

x =  ________________________  =   _______________________  =

        [sqrt (3)+ 1] [ sqrt (3) - 1 ]                    3 -  1

8 - 4sqrt (3)                4 - 2sqrt (3)

_________       =

    2

So the y coordinate of P  =

y = -( 4- 2sqrt (3))+ 2  =   2sqrt (3) - 2

So  the  area  of triangle APE  =   Area of triangle ABE - Area of triangle APB

Area of triangle  AEB  = (1/2)base * (height)  =  (1/2) (4) (  2sqrt (3))  = 2 (2sqrt (3))  = 

4sqrt (3)  units^2

And the area of triangle  APB  = (1/2) base (height) = (1/2)(4) [4 - (2sqrt (3) - 2 ) ]  =

2 ( 6 - 2sqrt (3) )  =  12 - 4sqrt (3)   units^2

So...the area of triangle APE  =  [ 4sqrt (3)]  - [ 12 - 4sqrt (3) ]  = [8sqrt (3) - 12] units^2 =

4 [ 2sqrt (3) - 3 ]  units^2  

\(\text{The constant term of a polynomial is the product of it's roots}\\ \text{So if $\sqrt{2}+\sqrt{5}$ is a root and $c_0$ is rational it must be that$\\ \sqrt{2}-\sqrt{5}$ is also a root}\\ \text{This is where you probably got your idea that $x^2-2x-3$ is part of it$\\$ You made a bit of an error. The actual factor is $x^2-2\sqrt{2}x-3$}\\ \text{Now given that is a factor, and we've got rational coefficients, it must be that $\\x^2 +2\sqrt{2}x-3$ is also a factor}\\ \text{That gets us $\\P(x) = (x^2+2\sqrt{2}x-3)(x^2-2\sqrt{2}x-3) = x^4-14 x^2+9$}\\ P(1)=1-14+9=-4\)

Maximum displacement is at the peak and at the valley floor   +- 14 cm      I suppose maximum would be + 14 cm

1.5 to 4.5 is one cycle as described   = 3 sec

Period = 3 sec      Freq = 1/per =   1/3 hz    Indicates the number of cycles per second

From 6 to 7.5 the weight is moving up from resting position to its peak.....then it begins moving downward back to the resting position at t = 7.5

\(\text{Does order matter? Assuming it does}\\ a)~\dfrac 1 4 \dfrac 1 2 \dfrac 1 4=\dfrac{1}{32} \\~\\ b)~\dfrac 1 2 \dfrac 1 2 \dfrac 3 4 = \dfrac{3}{16} \\~\\ c)~\left(\dfrac 1 2\right)^3=\dfrac 1 8\)

\(\text{If order doesn't matter} \\~\\ a)~\dfrac{1}{32} \cdot 3! = \dfrac{3}{16} \\~\\ b)~\left(\dfrac 1 2\right)^3 + 3\left(\dfrac 1 2\right)^2 \dfrac 1 4 = \dfrac{5}{16} \\~\\ c)~\text{unchanged}\)

x^3 - 3x^2 + 8  = 0

We have the form

ax^3 - bx^2+ cx + d  = 0

Using Vieta's Theorem, we have that

The sum of the roots  =  -( -b)/a =  r1 + r2 + r3  = --(-3)/1  = 3

And the product of the roots r1 * r2 * r3  = -d/a =  -8/1  =  - 8

So....the sum of the new roots

2r1 + 2r2 + 2r3 =   2 (r1 + r2 + r3) = 2(3)  =  6 ⇒  -b/1 ⇒  b = -6

And the product of the new roots  = 2r1 * 2r2 * 2r3  = 8(r1*r2*r3)  = 8(-8)  = -64  ⇒ -d / 1 ⇒ d = 64  

So the new polynomial  is

x^2 - 6x^2 + 64

1.  A possible formula is

P = 400(2.5)^(N )     where P is the population after N weeks

[ Note  150%  actually means that the population increases by 2.5 times every week ]

2.  After 10 weeks we have

P  = 400(2.5)^ (10)  =  400 (2.5)^10  ≈ 3,814,697

3.  We need to solve this

1,000,000,000  = 400 (2.5)^(N)     divide both sides by 400

2,500,000 = (2,5)^(N )         take the log of both sides

log 2500000 = log 2.5^(N )      and by a log property we can write

log 2500000 = (N ) log2.5        divide both sides by log 2.5

16.077 = N ≈ 16 weeks       

Present population =PP
Future population   =FP

FP = PP x [1 + R]^N
FP = 400 x [1 + 1.5]^10
FP = 3,814,697 - populations of weevils after 10 weeks.
1,000,000 =400 x 2.5^N
2,500 = 2.5^N
N =log(2,500) / log(2.5)
N = 8.54 - Number of weeks when the population hits 1,000,000.

(100, 120, 140, 160, 180, 200, 240, 260, 280, 300, 320, 340, 360, 380, 400, 420, 460, 480, 500, 520, 540, 560, 580, 600, 620, 640, 680, 700, 720, 740, 760, 780, 800, 820, 840, 860, 900, 920, 940, 960, 980)=41 Numbers.

220, 440, 660, 880  are multiples of both 20 and 55, which are excluded from the above count.

Let the dollar value of his Internet shares =S
At the end of the year, his Internet shares =1.1S
At the end of the year, his copper shares  =$9,000
1.1S + $9,000 =1.06[S + $9,000], solve for S 
S = $13,500 - dollar value of his Internet shares at beginning of the year.
$13,500 x 1.1 = $14,850 - dollar value at the end of the year.