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Wow that's actually the answer

\(f^{-1}(3) = 4\\ f^{-1}(f^{-1}(3)) = f^{-1}(4) = 7\)

Let L be the side length of square.

By Law of Cosines,

\(L = \sqrt{1^2 + 1^2 - 2(1)(2)\cos 150^\circ} = \sqrt{2 + 2\sqrt 3}\)

Area of large square = \(L^2 = 2 + 2\sqrt 3\)

Substitute numbers,

\(\quad2f(2) - f(8) \\= 2(2^2 + 2\sqrt{2}) - (8^2 + 2\sqrt{8})\\=8 + 4\sqrt 2 - 64 - 4\sqrt 2\\ =-56\)

Hint:  divide the area into an equilateral triangle and three congruent segments.

The ten's digit can be any odd number.

There are 7 choices for the first digit:  3, 4, 5, 6, 7, 8, and 9; only 0, 1, and 2 are not acceptable.

After the first digit, there are 8 choices for the second, third, and fourth digits (only 1 and 2 are not acceptable).

This means that there are  7 x 8 x 8 x 8  =  3584  possibilities.

My computer says there are 3584 [Thanks geno3141 - I only counted the first 3000!!!] 4-digit numbers that have no 1 or 2 in them. In other words, they must start with:

3000  3003  3004  3005  3006  3007  3008  3009  3033  3034.......etc. and end with .....  9993  9994  9995  9996  9997  9998  9999  Total =  3584 such numbers.

My computer says there are 138 such 4-digit numbers as follows:

3012  3021  3102  3120  3201  3210  4013  4031  4103  4130  4301  4310  5014  5023  5032  5041  5104  5140  5203  5230  5302  5320  5401  5410  6015  6024  6042  6051  6105  6123  6132  6150  6204  6213  6231  6240  6312  6321  6402  6420  6501  6510  7016  7025  7034  7043  7052  7061  7106  7124  7142  7160  7205  7214  7241  7250  7304  7340  7403  7412  7421  7430  7502  7520  7601  7610  8017  8026  8035  8053  8062  8071  8107  8125  8134  8143  8152  8170  8206  8215  8251  8260  8305  8314  8341  8350  8413  8431  8503  8512  8521  8530  8602  8620  8701  8710  9018  9027  9036  9045  9054  9063  9072  9081  9108  9126  9135  9153  9162  9180  9207  9216  9234  9243  9261  9270  9306  9315  9324  9342  9351  9360  9405  9423  9432  9450  9504  9513  9531  9540  9603  9612  9621  9630  9702  9720  9801  9810  Total =  138 such numbers.

This is a circle centered at 6,0   with aradius of 8

The line goes right through the center ...and has a slope of -1   

     so it forms a 135^o angle with the x axis

Total area = pi 8^2

    for the 135 degree portion    135/360 * pi * 8² = 75.4  units²

Here  is a picture:

https://www.desmos.com/calculator/kusprk1stg

10 years

original pop 21000   decreases    continuously

21000 e^-kt     when t = 10 years  pop = 20000

20000 = 21000 e^-10k

^{k = .004879}

 2010    t = 30               21000 e^{-.004879 (30)}  = ~~ 18141 population in 2010

That would be a number witha 7 in the same position as the two is in 1273

  x7xx   <========  this seven is 10 times larger than the seven in 1273 because it is

                                                         one postion further to the left

Honestly I don't blame the guest for assuming that because many people dump their homework here, but then again we shouldn't be quick to make assumptions and judge.

If you use the rules of the sequence, you will get the first six elements to be unique elements.

When you continue the sequence, you will find that:

     a₆  =  -a_o     a₇  =  -a₁     a₈  =  - a₂     a₉  =  -a₃     a₁₀  =  -  a₄     a₁₁  =  -a₅ 

Continuing, you find that:

     a₁₂  =  a₀     a₁₃  =  a₁     a₁₄  = a₂     ...

Etc.

Knowing this, can you calculate the value of a₁₀₀?

There are 24 girls and their average score is 89. Their total score is 24*89=2136. Let x be the number of boys. We form an equation for the average score.

(2136+92.5x)/(24+x)=91

2136+92.5x=2184+91x

1.5x=48

x=32 boys.

q = # quarters         value  = 25q

d = #dimes              value = 10d

q+ d = 25

25q + 10d = 385      from first equation   d = 25-q    use this in the second equation

25q  + 10(25-q) = 385         now you can solve for  q      then d   = 25-q

25q+250-10q = 385

15q= 135

q = 9   quarters       d = 25 - q = .......

Thank you! I had somewhat something like that but I missed one step so I got it wrong...Thanks!

1/2 of the angle at O     arcos (7/12) = 54.31^o

    so O is 108.6 ^o

        the arc you are looking for is   360 - 108.6 = 251.37^o

^{Pi (2*7)  x   251.37/360 = .....}

This is very simple!!! Your 'stuckness' will disappear in a minute.

∠AOB = ∠COB = 90º

1st      Find an angle  AOC          ∠AOC = ∠AOB + ∠COB

                                                     ∠AOC = 2[arccos( AO / OB )] = 108.6293306º

2nd     Find the central angle of a major arc ADC

                 That angle is    360º - ∠AOC = 251.3706694º

3rd       Find the length of ADC

                                         ADC = 2 * 7 * pi * ( 251.3706694 / 360 ) = 30.711

Go backwards from her result of 7

   multiply by 3  =21

     subtract 2    = 20

        divide by 4  = 5   is the original number....now proceed with the steps in the second half of the question

I know it's easy but I just need explanation...not everyone is as smart as you

f(x) = -7        soooo        -7 =  4x-3

                                       -4  = 4x

                                         x = -1

I think you can now do the others ! 

First one

   just put  '2' in to the equation for 'x'

       4(2) - 3 = 5

now put -10 in for x

the last one....put 0 in for x .......