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I think that is supposed to be 12^o   and  9 ^o

For one building  tan 9  = height₁ / 100  

   for the other building   tan 12 = height₂ /100       Now you should be able to find their heights and the difference.....

Both digits, 1 and 2, appear 50,000 times each between 1 and 100,000, including ALL duplicates such as 111, 2222.....etc.

If you exclude All duplicates, then both 1 and 2 appear 40,951 times each between 1 and 100,000 inclusive.

If you are wondering how to write the sqrt of i :    (this is NOT the answer to the question though    see alan's answer below))

i = sqrt (-1) =   -1^1/2

the square root of this would be       

-1^1/4         or      \(\sqrt[1/4]{-1}\)

This makes a right triangle with legs   (opposite)  = 80     and (adjacent ) = 130

arctan(80/130) = ........^o      angle of depression

The minimum value is -1, and the maximum value is 6.

When (in seconds) will the amoebas fill up half the room? 

Since the amoeba doubles every second, the room was half full one second before the room was full. 

So the answer is at mark 59 minutes 59 seconds, or 3,599 seconds. 

_.

QN = 12                PR = 14       PN = 7

F is a foot of the altitude from M to PN

Find angle QPN using QN and PN

Find QP and MP using the Pythagorean theorem

Find MF and PF using angle QPN and MP

Find angle MRF using MF and FR

And finally, find  OR = 5*sqrt(2).

PLUS     0     

its just x^28y^21

Let  O  be the center of the circle.

Let  A  be the center of the smaller left-hand circle.

Let  B  be the center of the smaller top circle.

Let  C  be the center of the smaller right-hand circle.

Let  D  be the center of the smaller bottom circle.

ABCD is a square with each side = 2.

Therefore, the area of this square is 4.

Each of the smaller circle has an area of pi·1²  =  pi

However, one-fourth of each circle is contained in the center square, so each circle contributes

another ¾·pi of area to the white portion.

Total white area  =  4 + 4(¾·pi)  =  4 + 3·pi

The distance from B to D is the diagonal of the central square. Since each side of the central square

is 2, the diagonal is  2·sqrt(2).

So, the diameter of the outside circle is  2 + 2·sqrt(2)  and its radius is 1 + sqrt(2).

The area of the outside circle is:  pi·( 1 + sqrt(2) )²,  making the area of the shaded section:

     pi·( 1 + sqrt(2) )²  -  (4 + 3·pi)

     pi

The largest will be the one with the largest NUMERATOR and smallest DENOMINATOR

the largest NUMERATOR  would be   c+d       ,     right?

the smallest denominator would then be     _  +  _               Any ideas? 

He starts at 8000 gallons      when he adds 2000 gals it will overflow

120 gal go out each hour       225 gal goes in each hour     net : 105 gal per hour IN

105 x = 2000 gal

x = 19.048 hours until overflow      

1 pm + 19 .048 hours =  8.048 AM  = ~~8 :03 AM   Since he checks every hour, he will first see the overflow at   9:00 AM

              4^x - 4^x-1  =  24

      (2²)^x - (2²)^x-1  =  24                             rewrite 4 as 2²   

           2^2x - 2^2x-2  =  24                             simplify

          2^2x - 2^2x/2²  =  24                            rewrtie 2^-2 as 1/2²

    4[ 2^2x - 2^2x/2² ]  =  4[ 24 ]                      multiply both sides by 4

           4·2^2x - 2^2x  =  96                            simplify

        4·2^2x - 1·2^2x  =  96                            rewrite

            (4 - 1)·2^2x  =  96                            factor

                   3·2^2x  =  96                            simplify

                      2^2x  =  32                            divide both sides by 3

                      2^2x  =  2⁵                             rewrite

                       2x  =  5                               rewrite

                         x  =  2.5                            divide

(2x)^x  =  (2 · 2.5)^2,5  =  5^2.5  =  5² · 5^0.5  =  25·sqrt(5)

2) 

Triangle(BPS) is similar to Triangle(BAC)     --->     Area(BPS) / Area(BAC)  =  BP2 / BA2  =  22 / 32  =  4/9

     Consequently, Area(APSC)  =  5/9 · Area(ABC)

Similarly, Area(DWT) / Area(DAC)  =  4/9     --->     Area(WACT)  =  5/9 · Area(DAC)

Therefore, Area(APSCTW)  =  5/9 · Area(ABCD)     --->     Area(APSCTW)  =  5/9 · 180  =  20

g(x) is defined only for [-4,4], so the width is 8.

What does that mean?

Thanks EP. You are absolutely right. I missed th repetitions of 9! There are 250 altogether as you calculated.

The inequality is 2x - 3y + 1 > 0.

1. 

    3x^2 + 12x + 4

=  3(x^2 + 4x) + 4

=  3(x^2 + 4x + 4) + 4 - 12

=  3(x + 2)^2 - 12

a = 3     h = -2     k = -12

2. 

The vertex of the parabola will be found half-ways between the focus and the directrix.

The vertex will be the point (4, 1.5).

A formula for the parabola is:  y - k  =  a(x - h)2  where (h, k) are the coordinates of the vertex.

For this situation:  y - 1.5  =  a(x - 4)2.

Since the point  (8, 6)  is on the parabola:     6 - 1.5  =  a(8 - 4)2

                                                      --->      4.5  =  a(4)2     --->     4.5  =  16·a     --->     a  =  0.28125

The equation is:  y - 1.5  =  0.28125(x - 4)2.

Multiplying out:   y - 1.5  =  0.28125(x2 - 8x + 16)

                           y - 1.5  =  0.28125x2 - 2.25x + 4.5

                                   y  =  0.28125x2 - 2.25x + 6

The sum of the coefficients is then 0.28125 - 2.25 + 6 = 129/32

Find the vertex of the graph of the equation y = -2x^2 + 9x - 18.

Hello Guest!

\( y = -2x^2 + 9x - 18\\ y'=-4x+9=0\\ \color{blue}x_P= \frac{9}{4}\)

\( y = -2x^2 + 9x - 18\\ y=-2\cdot (2.25)^2 + 9\cdot 2.25) - 18\\ y=-10.125+20.25-18\\ \color{blue}y_P=-7.875\)

\(P_{vertex}(2.25,-7.875)\)

  !

I think you left off some numbers   such as

9998    9992    9982   9928    etc   

If  n = 1,  (7n + 1)/(3n + 5)  =  1 

If  n = 9,  (7n + 1)/(3n + 5)  =  2.

The quotient  (7n + 1)/(3n + 5)  can never reach 3 because, as n approaches infinity,

the  lim (7n + 1)/(3n + 5)  = 7/3.

There are only two possibilities. 

(1, 2, 4) , (1, 2, 6) , (1, 2, 8) , (1, 2, 10) , (1, 3, 5) , (1, 3, 7) , (1, 3, 9) , (1, 4, 6) , (1, 4, 8) , (1, 4, 10) , (1, 5, 7) , (1, 5, 9) , (1, 6, 8) , (1, 6, 10) , (1, 7, 9) , (1, 8, 10) , (2, 3, 4) , (2, 3, 6) , (2, 3, 8) , (2, 3, 10) , (2, 4, 5) , (2, 4, 7) , (2, 4, 9) , (2, 5, 6) , (2, 5, 8) , (2, 5, 10) , (2, 6, 7) , (2, 6, 9) , (2, 7, 8) , (2, 7, 10) , (2, 8, 9) , (2, 9, 10) , (3, 4, 6) , (3, 4, 8) , (3, 4, 10) , (3, 5, 7) , (3, 5, 9) , (3, 6, 8) , (3, 6, 10) , (3, 7, 9) , (3, 8, 10) , (4, 5, 6) , (4, 5, 8) , (4, 5, 10) , (4, 6, 7) , (4, 6, 9) , (4, 7, 8) , (4, 7, 10) , (4, 8, 9) , (4, 9, 10) , (5, 6, 8) , (5, 6, 10) , (5, 7, 9) , (5, 8, 10) , (6, 7, 8) , (6, 7, 10) , (6, 8, 9) , (6, 9, 10) , (7, 8, 10) , (8, 9, 10) , Total =  60  distinct ways.

What is (n/n-1) for any positive integer n? (Answer in terms of n.)

Hello Guest!

\(\frac{n}{n-1}\) for any positive integer n is \(\frac{(2)}{(2)-1}=2\)

  !