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So the inverse for f(x) is
f−1(x)=√x+9+3

ok let me do the math is that ok

There are 24 permutations which meet all your criteria: They are > 500,000 and multiples of 5 and multiples of 3 as follows:

(612345, 612435, 613245, 613425, 614235, 614325, 621345, 621435, 623145, 623415, 624135, 624315, 631245, 631425, 632145, 632415, 634125, 634215, 641235, 641325, 642135, 642315, 643125, 643215) >>Total = 24 such numbers.

The probability is: 24 / 6! ==24 / 720==1 / 30

1 / a       1 / b

3660     61

1860     62

1260     63

960      64

780      65

660      66

510      68

460      69

420      70

360      72

300      75

285      76

260      78

240      80

210      84

204      85

180      90

160      96

150      100

135      108

120      120

110      132

105      140

Yo have 23 pairs that==1 / 60

Initial amount in wallet is $77.

2/7 of initial amount of spent on Friday.

∴ The amount spent on Friday = 2/7 × 77

                                                 = 2 × 11 = 22

∴ $22 spent on Friday.

∴ The remaining amount in wallet after spending

  2/7 on Friday is $77 - $22 = $55.

Now, 2/5 of the remainder $55 spent on

Saturday

∴ The amount spent on Saturday = 2/5 × 55

                                                     = 2 × 11 = 22

∴ The amount spent on Saturday is $22.

∴ The total amount spent is

= Amount spent on Friday + Amount spent on Saturday

= 22 + 22 = 44

∴ The amount left = 77 - 44 = 33

∴ 33 left in the wallet.

Awesome thanks, I had trouble typing it into the calculator, but that notation works.

2/7 of 77 is 22

77-22=55.

2/5 of 55 is 22

55-22=33.

(1+ .10/12)¹² = 1.104713         Your balance will go up by   'balance'  *  .104713

Your new balance at the end of a year will be         ' original balance'   *   1.104713

abcde...bs wrote:

I just played with the numbers until the circles overlapped, and I got a=5, b=3, c=15, and d=15

This is the correct solution set ... But me thinks you are serving a big pile of  BULLSHIT as an early Christmas present...

Your graph loads with A = 5; B = 1; C = 3; D = 9 producing two non-intersecting circles that are tangent to different points on the x-axis when y = 0. Setting A, B, C, D to the solution numbers also produces two non-intersecting circles ... meaning your equations are worthless.

For sure, I’d like to see your solution method... using your presented equations. If it is doable, this would be analogous to traveling from New Jersey to Florida via the North Pole.  

...While you’re at the North Poll, you should ask Father Christmas for a muckrake for Christmas, because you will have to do a massive amount of muckraking to present a valid solution with this bullshit. 

GA

--. .-

This was already answered by heureka: https://web2.0calc.com/questions/please-help-me_118 

The answer is 55

a = -3/2.

The points of intersection are (-1,3), (2,2).

a = 15/4.

y = -8/5*x + 26

Why are you trolling?

Everyone's time is valuable please do not post trivial questions. Not only does this take up time, it takes away space from problems from people that actually might need help.

If you really don't know (I doubt it), then sorry, and the answer is

1+1=2

I just played with the numbers until the circles overlapped, and I got a=5, b=3, c=15, and d=15

Then I actually did the problem 

Do you want to see my work?

I also ask, what do you want to do with it?

\(\sqrt{160-64\sqrt3}\\ =4\sqrt{10-4\sqrt3}\)

Let the numbers be a and b

\(10^5 \le a,b \le 10^6-1\\ so\\ 10^{10} \le ab \le (10^6-1)^2\\ 10^{10} \le ab \le 9.99998*10^{11}\\ \frac{10^{10}}{10} \le\frac{ ab}{10}\qquad \qquad \frac{ ab}{100} \le \frac{9.99998*10^{11}}{100}\\ \text{I only care about the smallest one so ...}\\ 10^{9} \le\frac{ ab}{ \textcolor{red}{10}} \qquad \qquad \frac{ ab}{100} \le 9.99998*10^{9}\)

\(\frac{ab}{10 }\ge \text{10 digits}\qquad \frac{ab}{100}< 11\;\;digits\)

The smallest common divisor must be between 10 and 100 inclusive.

The smallest common divisor is bigger or equal to 10  and 10 has  2 digits.

ok, so what did you figure out?

DC = 4√3 + 5 - 2√3 + 4

f^1 (1) = 1/sqrt(2)

f^2 (1) = 1/sqrt(3)

f^3 (1) = 1/sqrt(4)

Do you see a pattern?

I hope this helped. :))

=^._.^=

1 + 2 + 3 + 4  + ... + 10 = 55

Let the number she didn't include be called x. 

55 - x = a square number. 

x has to be in the interval [1, 10]

The largest square number less than 55 is 49. 

55 - x = 49

x = 6

I hope this helped. :))

=^._.^=

I figured it out using my graph !!!!! 

Which line?

1)  Find the slope of the given line ... 2x - 7y  =  11     --->     -7y  =  -2x + 11

                                                                                     --->        y  =  (2/7)x - 11/7

      The slope of this line is 2/7.

2)  The slope of all lines parallel to this line will also be 2/7.

3)  Use the point-slope form to find the equation of the wanted line:

             y - y₁  =  m(x - x₁)     --->     y - -1  =  (2/7)(x - 3)

4)  Simplify the answer to step 3 and place it into the form  Ax + By  =  C