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1)
In rectangle ABCD, we have AD = 3 and AB = 4.
Let M be the midpoint of AB, and let X be the point such that MD = MX, angle MDX = 77 degrees,
and A and X lie on opposite sides of DM. Find angle XCD, in degrees.

Picture: https://latex.artofproblemsolving.com/1/b/3/1b39a7de7d5064e27d8ac7d0e00bb8b0db806199.png

$\text{Let $\angle XCD = x$ } \\ \text{Let $\angle MDX = \angle DXM = 77^{\circ}$ } \\ \text{Let $\angle MDC=\angle MCD = \beta $ } \\ \text{Let $\angle MXC=\angle MCX = \gamma $ } \\ \text{Let $\angle DMC = 180^{\circ} - 2 \times \beta $} \\ \text{Let $\angle XMC = 180^{\circ} - 2 \times \gamma $}$

$\begin{array}{|rcll|} \hline \angle DMX &=& 180^{\circ} - 2 \times 77^{\circ} \\ &=& 26^{\circ} \\\\ \angle XMC &=& \angle DMC - 26^{\circ} \\ &=& 180^{\circ} - 2 \times \beta - 26^{\circ} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \angle XMC = 180^{\circ} - 2 \times \beta - 26^{\circ} &=& 180^{\circ} - 2 \times \gamma \\ 180^{\circ} - 2 \times \beta - 26^{\circ} &=& 180^{\circ} - 2 \times \gamma \quad & | \quad \gamma = \beta + x \\ 180^{\circ} - 2 \times \beta - 26^{\circ} &=& 180^{\circ} - 2 \times (\beta + x) \\ 180^{\circ} - 2 \times \beta - 26^{\circ} &=& 180^{\circ} - 2 \times \beta -2\times x \\ - 26^{\circ} &=& -2\times x \\ 26^{\circ} &=& 2\times x \quad & | \quad :2 \\ \mathbf{13^{\circ}} &=& \mathbf{ x } \\ \hline \end{array}$

The angle XCD is $\mathbf{13^{\circ}}$

thanks so much this is really helping me with my study guide before the exam starts

Last one before I head off to bed.....

(a)  They are similar because the corresponding sides are in the same ratio

36/30   = 48/40

6/5  = 6/5             

This is also the scale factor of the larger prism base to the smaller prism base  

(b)  If the prisms are the same height......the prism with base DEF will have a greater  volume

It will be (Scale Factor)^2   greater than the smaller prism  =  (6/5)^2  =  36/25  times as great 

1) 

a)   A =   f(n)  =  5 + .4 ( n )

b)   f^-1A   =  f^-1(n) 

Write  y   for  f(n)....and we have that

y  = 5 + .4(n)      subtract 5 from both sides

y - 5 = .400(n)        divide both sides by .4

[ y - 5] /. 4=  n     "swap"  y  and  n

[ n - 5] / .4= y       for y, write  f^-1 (n)

[ n - 5] / .4  = f-1(n)

c)  f(10)   is the amount of water present in the tank at 6PM  =

5 + .4(10)  =  5  + 4  =  9    ⇒  9000 gallons

d)  f^-1(10)   =   [ 10 - 5 ] / .4  =  12.5     this is the number of hours after 8AM  when the tank will contain 10,000 gallons of water   ⇒   8:30 PM

$0.12\,=\,\frac{100(0.05)+x(0.2)}{100+x} \\~\\ 0.12(100+x)\,=\,100(0.05)+x(0.2) \\~\\ 12+0.12x\,=\,5+0.2x \\~\\ 12-5=0.2x-0.12x \\~\\ 7\,=\,0.08x \\~\\ 87.5=x$

the amount of  20% dye to add  is  87.5 mL .

2.   

The perimeter  is 

2 ( 12 + 15)  =  2 (27)  =  54ft

The length of one diagonal  forms a hypotenuse of a right triangle  with legs of 12 and 15...its length is

sqrt  ( 12^2  + 15^2)  =   sqrt (144 + 225)  =  sqrt (289)  =  17 ft

And there are two diagonals so their combined length is  2 * 17  =  34 ft

So   .....  the total rolls she needs  is

[ 54 + 34 ]  / 20    =   88/ 20  = 4.4 rolls

But we can't purchase partial rolls...so.....she needs  5 rolls

Notice that...

3 * 3  =  9

9 * 3  =  27

13 * 3  =  39

20 * 3  =  60

Each value of  y  is  3  times the corresponding  x  value.

y  =  3x

This is a direct variation.

In a  cyclic quadrilateral   [ a rectangle inscribed in a circle]  opposite angles sum to 180°

Therefore  ...this implies that

(x - 2)  + (x + 2)   =  180      simplify

2x  =  180      divide both sides by 2

x  =  90°

So   

Angle A  =  x + 2   =  90 + 2  =  92° 

Angle B  =  x - 2  = 90 - 2  =  88°

Angle D  =  x - 10  =  90 - 10  =  80° 

Note that  the sine of an angle  is just  opposite side / hypotenuse

Here.....the opposite side  = 11.8  and the hypotenuse  =12

So...we need to use the sine inverse  (arcsin) function to determine the angle this makes with the ground

arcsin ( 11.8 / 12)  =  angle measure  =  79.5°

So....the ladder is not safe with this angle

Here's a pic :

The angle across from the 5cm side  is 30°

In a 30-60-90  triangle like this, the side opposite the 60° will be √3  times as long as the side opposite the 30° angle

So  BC  =  height of the triangle  =  5√3

So....the area  =  (1/2) base * height  =  (1/2) (5) (5√3)  =

(25/2)* √3  cm^2   ⇒   21.7 cm^2

1)

a)  This is known as  the  Tangent-Secant Theorem....it says that :

"When a tangent and a secant are drawn from one single external point to a circle, the square of the length of tangent segment must be equal to the product of lengths of whole secant segment and the exterior portion of secant segment."

In plain English....it says that    TU^2  = TR* TS

b)  RT  =  9      ST  =  4

We can find  TU  thusly :

TU^2  = TR * TS

TU^2  = RT * ST

TU^2  =  (9) * (4)

TU^2   =  36       take the square root of both sides

TU  =   6

MIRB, The counts (not the actual sequence) correspond to row #8 of Pascal’s triangle (with zero starting the count, so it’s actually the ninth row). If zero were included in the count then row #9 would give the counts for each number.

Again, despite Mr. BB’s insistence, there is nothing coincidental about these numeric relationships. All of these are mathematically related.  The set theory gives the solution to the sum, which is what the question asks for. The binomial distribution gives the leading coefficient for the count of each sequence (via a generating function). Still, further logic and mathematics are required to produce the numeric sequences. There are 362,880 ways to arrange those numbers, but only 256 of them fall within the criteria of the question.   

GA

Well, Mr. BB, like usual... you’re an arrogant idiot. 

Heureka’s post details the 256 individual solutions to this problem, and they link unambiguously and decidedly to the count obtained via the set theory solution—that is, they are NOT incidental nor are they coincidental.   The counts relate to the binomial distribution in Pascal’s triangle. (See below.)

log_x a  -  log_1/x a^3

Using the change-of -base  rule, we have that

log a  / log x    -    log a  /  log (1/x)   =

log a / log x  -  log a^3  / log x^-1   =

log a / log x  -  3log a / [ - log x]   =

log a / log x   +  3log a   /log x  =

4 [log a  /  log x ]  =      [ reverse the change-of-base rule ]

4 log_x a

2)

a)  They are similar......each side of Rectangle 2 will be  k times as long as the corresponding side of Rectangle 1

b )   Let  the Perimeter of Rectangle 1 be    2 ( L + W) = (2L + 2W)......so  the  perimeter of Rectangle 2 will be   2( kL + kW )  =   2k (L + W)  =   k (2L + 2W).....note that the perimeter of  Rectangle 2 is k times that of Rectangle 1

c)  The area of Rectangle 1 is  ( L x W).....the area of Rectangle 2 is  kL x kW   since each side is k times that of Rectangle 1.....so   kL x kW  =  k * k * L * W  =  k^2 (L * W)....so Rectangle 2 has  an area k^2 times that of Rectangle 1

6)  In a triangle. a segment drawn parallel to a base always splits the sides in equal proportions

So we have that

48/ 40   = ? / 85   

6/ 5  =  ? / 85      cross-multiply

85 * 6   =  5 * ?

510  =  5 * ?      divide both sides by  5

102   =  ?

We  can check this

48 / 40  =  102 / 85

6 / 5  =   6/ 5

Whew !   Finally got one correct....time for Brewski !   Thanx  GA !

7)  The bisected angle  in the first triangle will always set up this relationship

10 / 24  =  ?  / 36

5/12  = ? / 36        cross - multiply

36 * 5  =  12 * ?       

180   =  12 * ?      divide both sides by 12

15  =  ? 

8)  Similarly

[ 5x - 6 ]  / 42   =  28 / 49

[ 5x - 6 ] / 42  =   4 / 7        cross-multiply

7[ 5x - 6 ]  =  42 * 4

35x - 42  =  168       add 42  to  both sides

35x  =  210   divide both sides by 35

x  =  6

4. similar triangles

so   14 is to 8   as  FM is to 9......ratio:

    14/8 = FM/9     so FM = 14/8 * 9 = 15.75 feet

Now for the other side

14/8 =HM/12     HM = 14/8 *  12 =21 feet

Total = 14 + 15.75 +21 = 50.75 ft

5. Use pythagoream theorem to find the diagonal       then take 1/2

   BD^2 = 8^2 + 14^2 = 260

BD= 16.125        1/2 BD = BE = 8.062 ft

6.  I THINK    40 is to 85    as   (40+48) is to (?+85)

      as a ratio:   40/85 = (40+48) / (?+85)     solve for ?

              (?+85) = (40+48) / (40/85)

               ?+85  = 187

                  ? =102

Heureka,

thank you very much, I honestly appreciate your time!!

1)

△ABC ~ △HGF    and    AB = 72    and    BC = 48    and    HG = 12

GF / BC  =  HG / AB

GF / 48   =   12 / 72

GF   =   48 * 12 / 72

GF   =   8

2)

Since  $\frac{21}{42}\,=\,\frac{9}{18}$    , the sides are proportional.

And the included angle is the same.

So by   SAS  ,   △FGH ~ △UVW

3)

By the AA theorem, △ABC ~ △EDC . So

AB / AC  =  ED / EC

15 / 18   =   31.5 / x

x * 15 / 18  =  31.5

x   =   31.5 * 18 / 15

x   =   37.8

hallo, please help with this:

$\large{ \log_x(a)-\log_{\frac1x}(a^3) }$

$\begin{array}{|rcll|} \hline && \mathbf{ \large{ \log_x(a)-\log_{\frac1x}(a^3) } } \\ &\large{=}& \large{ \log_x(a)-3\times \log_{\frac1x}(a) } \qquad \boxed{ \log_{\frac1x}(a) = \dfrac{\log_x(a)}{\log_x(\frac{1}{x})} } \\ &\large{=}& \large{ \log_x(a)-3\times \dfrac{\log_x(a)}{\log_x(\frac{1}{x})} } \qquad \boxed{ \log_x\left(\frac{1}{x}\right) = \log_x(1) - \log_x(x) } \\ &\large{=}& \large{ \log_x(a)-3\times \dfrac{\log_x(a)}{ \log_x(1) - \log_x(x) } } \\ &\large{=}& \large{ \log_x(a)-3\times \dfrac{\log_x(a)}{ \log_x(1) - \log_x(x^1) } } \\ && \qquad | \quad \log_x(1) =\log_x(x^0)= 0 \\ && \qquad | \quad \log_x(x^1) = 1 \\ &\large{=}& \large{ \log_x(a)-3\times \dfrac{\log_x(a)}{ 0 - 1 } } \\ &\large{=}& \large{ \log_x(a)+3\times \log_x(a) } \\ &\mathbf{\large{=}}& \mathbf{\large{ 4\times \log_x(a) } }\\ \hline \end{array}$

Second one

Scale factor  =  1/2      Dilation Center  = (8, -6)

A'  =

[(-4, 4) - (8, - 6) ] * (1/2)  + (8, - 6) =

[ (-12, 10] * (1/2) + (8, -6)  =

(-6, 5) + (8, -6)  =

(2, -1)

B' = 

[ (4, 2)  -  (8, -6) ] *(1/2) + (8, -6) =

[ (-4, 8) ] * (1/2) +  ( 8, -6)  =

(-2, 4)  + ( 8 , - 6)

(6, -2)

C'  =

[ (-6, -4) - ( 8, -6) ] * (1/2)  +  (8, -6)  =

[ (-14, 2) ] * (1/2) + (8, -6)  =

(-7, 1)  +  ( 8, - 6)

(1, -5)

Here's a pic :

Volume of a square pyramid = a^2  h/3      where a=base length   h= height (I looked it up)

yields  270 cm^3   of wax needed

Each cube is  4 cm x 4 cm x 4cm = 64 cm^3

270 cm^3 / 64 cm^3/cube = 4.218 cubes......he will need 5 of the little cubes of wax.

Sweet,

thanx a million Hectictar!!