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\(f^{-1}(0) + f^{-1}(6) = 4\)

gcd(100, 997) = 1. The congruence is solvable and it has unique solution.

100x≡11(mod997)

⟹1000x≡110(mod997)

⟹3x≡110(mod997)

⟹999x≡36630(mod997)

⟹2x≡738(mod997)

⟹998x≡499×738(mod997)

⟹x≡369(mod997)

369 is the required unique 3-digit positive integer.

Use the definition of \(\star\) to get \(5x + 2x - 5 = 37\). Can you take it from here?

\(1_6 + 2_6 + 3_6 + \cdots + 45_6 + 50_6 + 51_6 \)

Probably the most straight forward way to takle this problem is to work out what 51 base 6 is in base 10,

do the sum and then convert back.

51base6 = 5*6+1 = 31 base 10

1+2+...+31 = 31/2(2+30)= 31*16 = 496

496/6 = 82R4

82/6=13R4

13/6=2R1

2/6=0R2

496 base 10 = 2144 base6

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I just want to play with another method  (all working is in base 6)

(0+ 1+2+3+4+5)   +   (  10+ .... 15)     +  (  20 .....+25 )      + ( 30 ....+35 )    + ( 40+....45 )+  50 +51

=   5(0+1+2+3+4+5) +0+1   +     (10*10)   + (20*10)  +(30*10)  + (40*10) +50*2

=    5* (10+10+3)  +1            +       100     +      200  +   300  +   400  + 140

=      5* 23  +1                       +      1540

=          204  +1540

=   2144 base 6

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See solution under your original posting here:  https://web2.0calc.com/questions/arithmetic-sequence_51

39==3/2*[20 + 2D], solve for D
178==4/2 * [2F + 3*3 ], solve for F

D ==3 - this is the common difference
F ==40 - this is the first term of the last 4 terms.
Sequence: 10, 13, 16.............40, 43, 46, 49...

Your full sequence looks like this:
10 , 13 , 16 , 19 , 22 , 25 , 28 , 31 , 34 , 37 , 40 , 43 , 46 , 49 , >>Number of terms = 14>>Total Sum == 413

The average of all 14 terms==413 / 14 ==29.5

187,829, it is 31*73*83

[250 / 0.3048]  x  [12 / 0.3048] ==~32,292 sq. feet in the wall

32,292 / 200 ==161.46 ==162 gallons of paint needed to paint the entire wall.

N mod 4 ==1

N mod 3 ==2

N mod 5 ==2

LCM[3, 4, 5]==60

N ==60m + 17, where m=0, 1, 2, 3.......etc.

When m=0,  N ==17 - which is the smallest whole number that satisfies the 3 congruences.

𝐇𝐞 𝐢𝐬 𝐫𝐞𝐭𝐢𝐫𝐞𝐝! 𝐒𝐨, 𝐡𝐞 𝐜𝐨𝐦𝐞𝐬 𝐢𝐧 𝐰𝐡𝐞𝐧𝐞𝐯𝐞𝐫 𝐡𝐞 𝐰𝐚𝐧𝐭𝐬. 𝐓𝐡𝐢𝐬 𝐢𝐬 𝐧𝐨𝐭 𝐚 𝐦𝐚𝐧𝐝𝐚𝐭𝐨𝐫𝐲 𝐛𝐮𝐭 𝐯𝐨𝐥𝐮𝐧𝐭𝐚𝐫𝐲 𝐣𝐨𝐛 𝐡𝐞 𝐭𝐨𝐨𝐤 𝐨𝐧 𝐚𝐬 𝐚 𝐫𝐞𝐜𝐫𝐞𝐚𝐭𝐢𝐨𝐧 𝐨𝐫 𝐚 𝐡𝐨𝐛𝐛𝐲 𝐭𝐨 𝐤𝐞𝐞𝐩 𝐡𝐢𝐬 𝐦𝐢𝐧𝐝 𝐚𝐜𝐭𝐢𝐯𝐞.

You have already solved half of the problem !

Let the distance between A and B ==D

2 x 0.8 ==1.6 km still to jog by the slower jogger.

D - 0.90D ==0.10D km still to jog

0.10D ==1.6

D ==1.6 / 0.10 ==16 km - the distance between A and B

Check: 16 / 2 ==8 km/h - speed of the first jogger. 8 - 0.8 ==7.2 km/h speed of the 2nd jogger. 16/7.2 ==2 2/9 - hours - time of the 2nd jogger.   8 km/h x 2 hours ==7.2 km/h x 2 2/9 hours ==16 km.

I did

maybe you have to put in the exact answer, since 3.215 is just an approximation

Here's a hit: since you know that chickens have 2 legs and rabbits have 4, you can write the system of equations : c + r = 80, 2c+4r=250, where c and r are chickens and rabbits, respectively. 

Michael first determines which line the angle of rotation places the point on. This line tells Michael that the point must lie in __quadrant II__ area if r is positive or ___quadrant IV__ area if r is negative.

The radius of 2 tells Michael that the point lies on the second circle of the polar plane. The value of r is negative. Therefore, the point will lie __in the opposite quadrant as__ the angle of rotation.

Explanation: -4pi/3 is the angle, and it is in the 2nd quadrant. That means that r, the x value of the coordinate, will extend in that direction and also lay on the second quadrant. If it is negative, it will "de-extend" and go in the opposite direction, which is a 180 degrees turn to quadrant 4 which is the opposite quadrant. 

Hope this helps, 95% sure this is correct.

Thanks for helping, asinus! However, I checked the answer, and it said that is was incorrect. Furthermore, I do not know where you were wrong.

-Nathan

2^4*3^8*5^2*7^2==2 , 2 , 2 , 2 , 3 , 3 , 3 , 3 , 3 , 3 , 3 , 3 , 5 , 5 , 7 , 7>> Total Prime Factors = 16

[4+1] x [8+1] x [2+1] x [2+1]==405 divisors.

Factor   ab^3 - a^3b + bc^3 - b^3c + ca^3 - c^3a

-(a - b) (a - c) (b - c) (a + b + c)

https://web2.0calc.com/questions/math-problem_32546

Note that \({ 1 \over a} + {1 \over b} = {a \over ab} + {b \over ab} ={ {a + b} \over ab}\)

Now, we can use Vieta's formula (\({-b \over a} = {11 \over 5}\)) to find a+b.

We can also use \({c \over a} = {4 \over 5}\) to find ab.

This means \({1 \over a} + {1 \over b} = {{11 \over 5} \over {4 \over 5}} = \color{brown}\boxed{11 \over 4}\)

Not enough info...

The only ways to get 24 is if we have \(1, 2, 3, 4\), \(1,1,3,8\), or \(1,1,4,6\)

For the first case, because all the digits are distinct, there are \(4! = 24 \) ways to factor. 

For the second case, we have to account for overcounting by dividing by \(2!\)(the amount of ways to order 2 1s), giving us: \({4 ! \over 2!} = 12\)

LIkewise, the third case has 12, by the same logic. 

Now, can you take it from here?

Simplifying into a quadratic, we have: \(x^2 - 12x - 30 = 0 \).

Now, to solve, use the quadratic formula: \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\), where \(a = 1\), \(b = -12\), and \(c = -30\). 

Can you take it from here?

We have the following equation: \(xy + 1 = (x-4)(3y-1)\)

SImplifying the right hand side and isolating the constant gives us: \(-3 = 2xy - x - 12y\)

From here, we can factor out \(2y\) from the right-hand side of the equation. This gives us: \(-3 = 2y(x-6) - x\).

Now, notice how if we add 6 to both sides of the equation, we can factor in terms of \((x - 6)\): \(3 = (2y - 1)(x-6)\). 

Now, we need 1 term to equal 3, and the other to equal 1. There are 2 ways to do this, but only 1 will satisfy the question. 

Can you take it from here?

If \(f(x) = 2/x + 1,\) then we can plug in -1/5 as x into the equation to get 2/(-1/5)+1 = -10+1 = -9