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We have an irregular  pentagon

The sum of the interior angles   = 

(Number of Sides - 2) * 180  =

(5 - 2) * 180  =

3 *  180  =

540°

So....this implies that

(2x + 10 )    + 148  +  90   + 112  +  2x   =  540        simplify

4x  +  360  =  540      subtract  360  from both sides

4x  =  180      divide both sides by  4

x  =  45

Thank you so much!  I actually found the transcript of my class from a friend so I'm covered on that front (phew) All your answers were so helpful!

2) Note that the side of the equilateral triangle = Side of the square

So.....draw EC

And in triangle ECB,  EB  = CB    and angle CBE  =  30°

But  ....since  EB  = CB, the angles opposite those sides, ECB  and CEB will be equal

And either will be  =     (180  - 30)  /  2   =   150 / 2   =  75°

So....angle ECB  =  75°

So.....angle  ECD  will be complementary to this  =   15°

2^500 =2.73..... x 10^150,  3^375 =8.326..... x 10^178,   5^250 =5.527....x 10^174

Therefore, from least to greatest:

2^500, 5^250, 3^375

Thanks!

These will  be parallel if  4x  and 2x are supplementary...so....

2x  + 4x  = 180°     simplify

6x  =  180°     divide both sides by 6

x  = 30°

4)  This one isn't too difficult

The total area of the rectangle  is  7 x 5   =  35 units^2    (1) 

The white triangle is obtuse.....

The base of this triangle......sitting on the "bottom" of the rectangle  = 2 units

The altitude occurs outside the triangle   and is  =  5 units

So....the area of this triangle is   (1/2) base * altitude  =  (1/2) (2) (5)  =  5 units^2    (2)

So.....the red  area is just the difference of (1)  and (2) =

35  -  5    =     30 units^2

P.S.   -  hope you're feeling better  !!!

3) 

Because transversal EF cuts parallels AB, CD.....then angle  BEF  = angle GFE

And since  EF = FG.....then angles FEG  and EGF are equal

So.....angle FEG  =  [  180  - (100 +x) ] / 2   =  (80 - x)/2

And angles  BEF +  FEG + AEG  = 180

Sustituting....we have   that

(100 + x)  +  (80 - x) / 2  +  x   =   180      multiply through by 2

2(100 + x)  + 80 - x  + 2x  =  360     simplify

200 +2x  + 80 - x + 2x  =  360

280  + 3x  =  360     subtract 280  from both sides

3x  =  80       divide both sides by 3

x  = 80/3 °   ≈  26.66°

8-(2+1)+3 to the power of 2

[8 - (2+1) + 3]^2 =

[8 - 3 + 3]^2 =

[5 + 3]^2 =

8^2 = 64

Omi67:

x = 1    and    y = -2

These values balance the two equations.

Mmmmm....let's see..

1) Since BAC  = 23, then angle DAQ  will be complementary to this  = 90 - 23  = 67

And QDA  =  90 - DAQ  =  23

And since DP = PC, then in triangle DPC.....angle  DCP = angle PDC

But......since tranversal AC  cuts parallels DC and AB....then angle BAC  = angle DCP  = 23

So....PDC  also = 23

So....

QDA  + PDC   =  23 + 23   =  46°

A system of equations is shown.

−7x−8y=9

−4x+9y=−22

What are the values of x and y?

The new area  =   

Original Area *  Dilation Factor^2  =

58 * 8^2   =   

3712 ft^2

Let's convert "bearing"  into standard degrees....we can convert back at the end.

A bearing of  130°  is  320°

A wind from the SW means that the wind is blowing at a 45° angle

a)  Velocity of the bat  =   

[ 10cos (320°) i, 10 sin (320°) j  ]  =

[ 7.66 i, - 6.43 j ]

 Wind  velocity  =  [  5cos (45°) i , 5 sin (45°) j  ]  =

[ 3.54 i, 3.54 j ]

b)   Velocity of the bat =   

[ (7.66 + 3.54) i ,  ( -6.43 + 3.54 j ]  =

[ 11.2 i ,- 2.89 j ]

c)  Resultant velocity of the bat  = 

sqrt  [  11.2^2  + 2.89^2 ]  = 

11.57  m/s

Final direction  =    arctan  (-2.89/ 11.2)  = -14.47° = -14°  [to the nearest degree ]

Converting this to bearing, we have   90  + 14   =    104°

Average Speed = Total distance covered / Elapsed time

Elapsed time =10 minutes / 60 minutes =1/6 of an hour

Average speed =7 miles / (1/6) hours

                         =7 x 6

                         = 42 mph - your average speed.

If you are travelling by bus, for example, the bus will stop at various "bus stops" and then drive at various speeds until you get to your destination. All that counts here is that you must know "total distance travelled" and the "elapsed time to cover that distance". Then you can calculate your average speed using the above formula.

Where is the "diagram" ??

3 1⁄8 − 1 7⁄8 =

[8*3+1]/8  -  [8*1+7]/8 =

25/8  -  15/8 =

10/8 = 5/4 =1 1/4 - which is the answer.

Here are some pointers to the right direction. I’ll leave you to fill in the details:

(a)  b = 10*cos(130°)*i + 10*sin(130°)*j

      w = 5*cos(45°)*i + 5*sin(45°)*j

(b) v = b + w.   (add corresponding terms in i and j)

(c) If v = r*i + s*j then |v| = sqrt(r²+s²) and angle = tan^-1(s/r)

The question doesn’t say that f(x) is a polynomial; it just asks for g(x) to be a polynomial, and for the limit of the sum of the two functions to have a zero asymptote.

.

But Alan

 3(x^4+x^3+x^2+1)/(x^2+x-2) 

does not divide to give a polynomial so how can you say that this question makes sense.  ?

"Let f(x) = 3(x^4+x^3+x^2+1)/(x^2+x-2). Give a polynomial g(x) so that f(x) + g(x) has a horizontal asymptote of 0 as x approaches positive infinity."

Thanks!!

I liked the way you did this, hectictar....the "congruent" triangles approach was a nice touch...!!!

Angle ADP  = 30°   and angle APD  = 60°

So.....since we have a 30-60-90 triangle......AP  =  1/√3   = √3/3

And PD  =   2 / √3  =  2√3 / 3

And angle DPB = 120°  and angle PDB = 30° ..... so   angle  PBD = 30°

So...triangle DPB  isosceles with  PD  = PB  =  2√3 / 3

And using the Law of Sines

DB/ sin(120)  = PB / sin (30)

DB  =  sin (120) * [ 2√3 / 3 ]  /  (1/2)  =    √3 * 2√3 / 3   =  2

So...the perimeter of triangle BDP  =

PD + PB + DB  =     4√3 / 3  +  2   =   

2  +  4√3 / 3

So   w  = 2,  x = 4 , y  = 3  and z  = 3 

And their sum is  12

DB and DP trisect angle ADC,   so   m∠ADP  =  m∠PDB  =  m∠BDC  =  90° / 3  =  30°

∠ABD  and  ∠BDC  are alternate interior angles, so they have the same measure.

If we draw a line from  P to  BD  that is perpendicular to side BD, we will have congruent triangles.

The angle measures of these triangles are  30°, 60°, and 90° .

The side opposite the  60°  angle  =  1 , so

the side opposite the 90° angle  =  2 / √3

perimeter of triangle BDP   =  2  +  2/√3  +  2/√3   =   2  +  4/√3   =   2  +  4√3 / 3

2 + 4 + 3 + 3   =   12