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The sum of two positive integers is 35. When the smaller integer is subtracted from twice the larger, the result is 34. Find the two integers. 

Let's call the larger number X 

Let's call the smaller number Y 

                                                                      X + Y  =  35 

                                                                    2X – Y  =  34 

Add the two equations                                       3X  =  69 

Divide both sides by 3                                         X  =  23   the larger number     

Plug 23 in place of X in the first equation      23 + Y  =  35 

Subtract 23 from both sides                                  Y  =  35 – 23 

                                                                              Y  =  12   the smaller number 

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A triangle with vertices A(3,-1) , B(-2,6) and C(5,4) is rotated 90 degrees clockwise about Vertex B. Find the co-ordinates of the image of vertex C

We have  that

( y  - k)  = a(  x - h)^2              where (x, y)  is a point on the graph and  (h, k)  is the  vertex.....solving for  "a"  we have

(-8 - 1)   = a ( 6  - 3)^2

(-9)  = a (3)^2

-9  = 9a

a = -1

So  we  have this

y = -1 ( x - 3)^2  + 1      expand and simplify

y = - (x^2  - 6x + 9)  +  1

y = -x^2 + 6x   - 9 + 1

y = -x^2  + 6x   - 8

Here's the  graph :  https://www.desmos.com/calculator/403szzeeqc

First one

Period : Correct   !!!

Midline  =  [ High Point  + Low Point ]   /  2  =  [44 + 20  ]  / 2  =   64 / 2   = 32   Correct!!!

Amplitude   [ High Point - Low Point ] /  2   =  [44  - 20 ]  /  2  =    24 / 2 =   12    Correct!!!

Second one : Correct  !!!

Good job  !!!!

C ' =   (-4, -1)

See here : 

Unit circles are arranged as shown. Find the area of the shaded part.

Let t hours be the time travelled at 7mph. Then:

7t + 11(6 - t) = 54

-4t = -12

t = 3 hours at 7mph

Hence 6 - 3 = 3 hours at 11 mph.

#2:  Let M be the midpoint of side AC. Draw BM.

Since triangle ABC is an isosceles triangle BM is the angle bisector of angle(B) and also the altitude from B to side AC.

This makes angle(AMB) a right angle.

Since AC = 16, AM = 8.

Using the Pythagorean Theorem on triangle(AMB), we can find that BM = 15.

In right triangle(AMB), cos(A)  =  8/17   --->   angle(A)  =  cos^-1(8/17)  =  61.9275^o.

Draw the bisector of angle(A). Call the point where this angle bisector intersects BM point 'X'.

Angle(MAX) = ½·angle(A)  =  ½·61.9275^o  =  30.96^o.

To find XM of triangle(MAX)  --->   tan(30.96^o)  =  XM/8   ---XM  =  8·tan(30.96^o)  =  4.8.

The radius of the incircle is  4.8.

A cone is circumscribed by a hemisphere radius of which is equal to the height of the cone. Calculate the ratio of the volume of the cone to the volume of the hemisphere.

I'll try the first one.

Since the triangle has side lengths of 10, 24, and 26, this triangle is a right triangle (10² + 24² = 26²).

The center of the circumcenter of a right triangle is the midpoint of the hypotenuse.

Since the hypotenuse has a length of 26, the distance from this point to any of the vertices is 13.

Therefore, the area of the circumcircle is:  pi·13²  =  169·pi.

The center of the incircle is the point of intersection of the angle bisectors.

Call the vertex between the 10 and 24 sides = C (it is a right angle).

Call the vertex between the 24 and 26 sides = A.

Call the center of the incircle X.

Draw the perpendicular from X to side AC; call this point Y; this is a radius of the incircle;

call its value 'h'.

Angle C = 90^o; therefore, angle(XCY) = 45^o.

I will need to find the size of angle(XAY).

I will use the sin(A) in triangle ABC) and find one-half of that value.

sin(A) = 10/26   --->   A  =  sin^-1(10/26)  =  22.61986^o   --->   angle(XAY)  =  11.31^o.

Now, to find the value of h:

Call the distance from C to Y 'x'; therefore, the distance from A to Y is '24 - x'.

In triangle(XCY), tan(45^o)  =  h/x   --->   h  =  x·tan(45^o).

In triangle (XAY), tan(11.31^o)  =  h/(24 - x)   --->  h  =  (24 - x)·tan(11.31^o). 

Setting these two equation equal to each other:  x·tan(45^o)  =  (24 - x)·tan(11.31^o)

--->   x·tan(45^o)  = 24·tan(11.31^o) - x·tan(11.31^o)

--->   x·tan(45^o) + x·tan(11.31^o)  =  24·tan(11.31^o)

--->   x·( tan(45^o) + tan(11.31^o) )  =  24·tan(11.31^o)

--->   x  =  24·tan(11.31^o) / [ tan(45^o) + tan(11.31^o) ]

--->   x  =  4

Therefore, the area of the incircle is:  pi·4²  =  16·pi.

169·pi  -  16·pi  =  153·pi 

Ohh, I see. I haven't grasped Vietas yet.

Since a line has a constant slope, we can write the equation of the line as (y+ 3)/x = (3 - 1)/(1 - 0) = 2.  Then y + 3 = 2x, so the standard form is 2x - y = 3.

Howard bought a total of  4.5 + 1  =  5.5  pounds of jellybeans.

At  $0.80  per pound, it cost him  $0.80 x 5.5  =  $4.40.

To find how high up the tower the guy wire is attached, use sine.

    sin(36^o)  =  x / 60

                x  =  60 · sin(36^o)

                x  =  35.27 ft

To find the new angle, use inverse sine.

    sin(x^o)  =  32.57 / 110

           x^o  =  sin^-1( 32.57/110)

           x^o  =  17.2^o

5 + 5*x  =  57

 5 + 5x  =  57

Subtract  5 from both sides:

        5x  =  52

Divide both sides by 5

          x  =  52/5  =  10.4

Ratio of children to adults  =  2:3

Set up the equation:  2:3  =  80:x

                                  2/3  =  80/x

Cross-multiply:          2·x  =  80·3

                                   2x  =  240

Divide by 2:                  x  =  120

In the diagram, O is lacated at (0,0), U is located at (7,6),
and V is located at (11,6).
Assume that R, S, T, U, V, and W are midpoints.

\(1)\\ \begin{array}{|rcl|rcl|rcl|} \hline R &=& U + \dfrac{TR}{2} & \dfrac{TR}{2}+T &=& U & T &=& \dfrac{Q}{2} \\ && & \dfrac{TR}{2}+\dfrac{Q}{2} &=& U \\ && & \dfrac{TR}{2} &=& U-\dfrac{Q}{2} \\ R &=& U + U-\dfrac{Q}{2} \\ \mathbf{R} &=& \mathbf{2U-\dfrac{Q}{2}} \\ \hline S &=& T+TS & T+\dfrac{TS}{2} &=& V & T &=& \dfrac{Q}{2} \\ & & & \dfrac{Q}{2}+\dfrac{TS}{2} &=& V \quad | \quad *2 \\ & & & Q+TS &=& 2V \\ & & & TS &=& 2V-Q \\ S &=& \dfrac{Q}{2} + 2V-Q \\ \mathbf{S} &=& \mathbf{2V-\dfrac{Q}{2}} \\ \hline \end{array}\)

\(2)\\ \begin{array}{|rcl|rcl|rcl|} \hline && &P &=& S + \dfrac{QP}{2} & QP &=& 2(S-Q) \\ && &P &=& S + (S-Q) \\ && &P &=& 2S-Q & S&=&2V-\dfrac{Q}{2} \\ && &P &=& 4V-Q-Q \\ && &\mathbf{P} &=& \mathbf{4V-2Q} \\ \hline && &P &=& 2R & R&=&2U-\dfrac{Q}{2} \\ && &\mathbf{P} &=& \mathbf{4U-Q} \\ P= 4U-Q &=& 4V-2Q \\ 4U-Q &=& 4V-2Q \\ \mathbf{ Q }&=& \mathbf{4V-4U} \\ \hline \end{array} \\ 3)\\ \begin{array}{|rcl|rcl|rcl|} \hline W&=& \dfrac{1}{2}( R+S )& R&=&2U-\dfrac{Q}{2} & S&=&2V-\dfrac{Q}{2} \\ W&=& \dfrac{1}{2}( 2U-\dfrac{Q}{2}+2V-\dfrac{Q}{2} ) \\ W&=& \dfrac{1}{2}( 2U+2V-Q ) \\ \mathbf{ W }&=& \mathbf{U+V-\dfrac{Q}{2}} \\ \hline \end{array}\)

\(\text{Summary:}\quad Q=4V-4U \\ \begin{array}{|rcll|} \hline T&=& \dfrac{Q}{2} \\ R&=& 2U-\dfrac{Q}{2} \\ S&=& 2V-\dfrac{Q}{2} \\ W&=& U+V-\dfrac{Q}{2} \\ P&=& 4U-Q \\ \hline \end{array}\)

Substitute \(Q\):

\(\begin{array}{|rcll|} \hline T &=& -2U+2V \\ Q &=& -4U+4V \\ R &=& 4U-2V \\ S &=& 2U \\ W &=& 3U-V \\ P &=& 8U-4V \\ \hline \end{array}\)

Substitute \(U(7,6)\) and \(V(11,6)\):

\(\begin{array}{|rcll|} \hline \mathbf{T} &=& -2(7,6)+2(11,6) = (-14+22, -12+12) =\mathbf{(8,0)}\\ \mathbf{Q} &=& -4(7,6)+4(11,6) = (-28+44, -24+24) =\mathbf{(16,0)}\\ \mathbf{R} &=& 4(7,6)-2(11,6) = (28-22, 24-12) =\mathbf{(6,12)}\\ \mathbf{S} &=& 2(7,6) = \mathbf{(14, 12)} \\ \mathbf{W} &=& 3(7,6)-(11,6) = (21-11, 18-6) =\mathbf{(10,12)}\\ \mathbf{P} &=& 8(7,6)-4(11,6) = (56-44, 48-24) =\mathbf{(12,24)}\\ \hline \end{array}\)

Let  x  represent the county's population in 1990.

In the year 2000, the county's population was 15,922,543.

This represents an increase of  25.325.3%.

To find the original population: 

  add the original population to the increase in population to get the final population

  -  the original population  =  x

  -  the increase in population is  25.325.3% · x   =  0.253253 · x

  -  the final population is 15,922,543

                                                  x  +  0.253253·x  =  15,922.543

                                                          1.253253·x  =  15,922.543

                                                                          x  =  15,922.543 / 1.253253

                                                                          x  =  12,704,971

Let the roots be p, q and r, and show that p + q + r = 7.

If p, q and r are distinct positive integers, they have to be 1, 2 and 4.

First question

---------------

\(cos(B)=sqrt(5)/5\)

Steps:

-Draw triangle ACB, where C is the right angle,

-Given 2 sides: BC=4 and AC=8, Thus the third side AB (Which is the hypotenuse)= 16+64=AB^2, \(AB=\sqrt{80}\)

Notice \(\sqrt{80}=4\sqrt{5}\)

\(cos(B)=\frac{Adjacent side}{hypotenuseside}\)

\(cos(B)=\frac{4}{4\sqrt{5}}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}\)

\(tan(A)=\frac{Opposite}{Adjacent}=\frac{4}{8}=\frac{1}{2}\)

\(csc(B)=\frac{4\sqrt{5}}{8}=\frac{\sqrt{5}}{2}\) Notice csc is the reverse of sin (I.e. sin = opposite/hypotenuse,  csc (cosecant) is hypotenuse/opposite)

Keep in mind that sec is reverse of cos (cos=adjacent/hypotenuse, while sec=hypotenuse/adjacent)

-------------------------------

Second question,

-Draw triangle ACB, where C is the right angle,

-Now 2 choices, either recall what a 30-60-90 triangle is. or using trigonometry (Let's use trigonometry)

\(sin(60)=\frac{7\sqrt{5}}{x}\) where x is the hypotenuse

\(x=\frac{7\sqrt{5}}{sin(60)}=\frac{14\sqrt{15}}{3}=18.07\)

Now we have 2 sides so let's find the third side by Pythagoras theorem.

\(BC^2=(\frac{14\sqrt{15}}{3})^2-(7\sqrt{5})^2\) Use the calculator

\(BC=\frac{7\sqrt{15}}{3}\)

Perimeter is the sum of the 3 sides of the triangle.

\(7\sqrt{5}+\frac{14\sqrt{15}}{3}+\frac{7\sqrt{15}}{3}\)

Simplifying:

\(7\sqrt{15}+7\sqrt{5}\) (Choice 3)

=42.76335926...

Let the number of calls in the 3rd evening=C
The number of calls in the first evening    =C - 5
The number of calls in the 2nd evening     =3C

C + C - 5 + 3C =95, solve for C
C = 20 calls in the 3rd eveningg
20 - 5 = 15 calls in the first evening
3 x 20 = 60 calls in the 2nd evening.
20 + 15 + 60 = 95 calls received in 3 evenings.

Hello Melody,

Thank you sooo much for your esponse.I do see my error. my problem is as with sooo many other sums, that I tend to just do the sum, like I did in my example, and do not really comprehend, or see the restrictions. Also, I hardly ever test the result back in the original equation, to test if the answer does indeed make sense or not...something i should really start learning to do...you agree?..

A right prism has a base that is an equilateral triangle. The height of the prism is equal to the height of the base. If the volume of the prism is 81, what are the lengths of the sides of the base?

Let Noel's salary = N
Robert's salary    = N - 512.75
John's salary       =3[N - 512.75]

N + N - 512.75 + 3[N - 512.75] =47645.25, solve for N
N = $9,939.25 - Noel's salary
$9,939.25 - $512.75 = $9,426.50 - Robert's salary
3 x $9,426.50 = $28,279.5 - John's salary
$9,939.25 + $9,426.50 + $28,279.50 = $47,645.25 - Their total salaries.

A chef plans to mix 100% vinegar with Italian dressing. The Italian dressing contains 12% vinegar. The chef wants to make 160 milliliters of a mixture that contains 23% vinegar. How much vinegar and how much Italian dressing should she use?