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Suppose that the common root is x1, then x1^3 - 3x1 + b = 0 ...... (1), and x1^2 + bx1 - 3 = 0 ..... (2).

Multiply equation (2) by x1 and subtract the resulting equation from equation (1).

That gets you b(1 - x1^2) = 0, so either b = 0 or x1 = + or - 1.

Substitute the x1 values into (2) and calculate the corresponding values of b.

You should find that he sum of the three possible values for b is zero.

Find 2^{-1} mod{185}, as a residue modulo 185. (Give an answer between 0 and 184, inclusive.)

\(\begin{array}{rcll} \text{Let} \\ & 2\cdot 2^{-1} \equiv 1 \pmod{185} \\ \end{array} \)

\(\begin{array}{rcll} \text{Let} \\ & 2\cdot 92 = 184 \equiv -1 \pmod {185} \\ \end{array} \)

\(\begin{array}{llcll} \text{square this equation: } \\ & (2\cdot 92)(2\cdot 92) &\equiv& (-1)(-1) \pmod{185} \\ & (2) (92^2\cdot 2) &\equiv& 1 \pmod{185} \\ & (2) (16928) &\equiv& 1 \pmod{185} \quad | \quad 16928 \equiv 93 \pmod{185} \\ & (2)\underbrace{(93)}_{=(2)^{-1}} &\equiv& 1 \pmod{185} \\ \end{array}\)

\(\text{So $2^{-1}=\boxed{93}$ is the multiplicative inverse to $2$ modulo $185$}.\)

Let A=111111 and B=142857. Find a positive integer N with six or fewer digits such that 

N is the multiplicative inverse of AB modulo 1,000,000.

\(\begin{array}{rcll} \text{Let} \\ & A=111111 \\ \text{and} \\ & B=142857 \\ \end{array}\)

\(\begin{array}{rcll} A\text{ and } B \text{ are factors of } 999999. \\ & 9A=999999 \\ \text{and} \\ & 7B=999999 \\ \end{array}\)

\(\begin{array}{rcll} A\text{ and } B \text{ modulo } 1000000. \\ & 9A \equiv -1 \pmod{1000000} \\ \text{and} \\ & 7B \equiv -1 \pmod{1000000} \\ \end{array}\)

\(\begin{array}{rcll} \text{Multiply these equations: } \\ & (9A)(7B) &\equiv& (-1)(-1) \pmod{1000000} \\ & (AB) (9\cdot 7) &\equiv& 1 \pmod{1000000} \\ & (AB)\underbrace{(63)}_{=(AB)^{-1}} &\equiv& 1 \pmod{1000000} \\ \end{array}\)

\(\text{So $N=\boxed{63}$ is the multiplicative inverse to $AB$ modulo $1000000$.} \)

If f(x)=x^5-1/1, find f^-1(-31/96).

If the permutations of the first 5 letters, a, b, c, d, e,

are arranged in alphabetical order from the beginning to the end,

then what is the alphabetical order of the 77th term? How about the 100th term?

sorted...

\(\begin{array}{|rcll|} \hline 1.& abcde \\ 2.& abced \\ 3.& abdce \\ 4.& abdec \\ 5.& abecd \\ 6.& abedc \\ 7.& acbde \\ 8.& acbed \\ 9.& acdbe \\ 10.& acdeb \\ 11.& acebd \\ 12.& acedb \\ 13.& adbce \\ 14.& adbec \\ 15.& adcbe \\ 16.& adceb \\ 17.& adebc \\ 18.& adecb \\ 19.& aebcd \\ 20.& aebdc \\ 21.& aecbd \\ 22.& aecdb \\ 23.& aedbc \\ 24.& aedcb \\ 25.& bacde \\ 26.& baced \\ 27.& badce \\ 28.& badec \\ 29.& baecd \\ 30.& baedc \\ \hline \end{array} \begin{array}{|rcll|} \hline 31.& bcade \\ 32.& bcaed \\ 33.& bcdae \\ 34.& bcdea \\ 35.& bcead \\ 36.& bceda \\ 37.& bdace \\ 38.& bdaec \\ 39.& bdcae \\ 40.& bdcea \\ 41.& bdeac \\ 42.& bdeca \\ 43.& beacd \\ 44.& beadc \\ 45.& becad \\ 46.& becda \\ 47.& bedac \\ 48.& bedca \\ 49.& cabde \\ 50.& cabed \\ 51.& cadbe \\ 52.& cadeb \\ 53.& caebd \\ 54.& caedb \\ 55.& cbade \\ 56.& cbaed \\ 57.& cbdae \\ 58.& cbdea \\ 59.& cbead \\ 60.& cbeda \\ \hline \end{array} \begin{array}{|rcll|} \hline 61.& cdabe \\ 62.& cdaeb \\ 63.& cdbae \\ 64.& cdbea \\ 65.& cdeab \\ 66.& cdeba \\ 67.& ceabd \\ 68.& ceadb \\ 69.& cebad \\ 70.& cebda \\ 71.& cedab \\ 72.& cedba \\ 73.& dabce \\ 74.& dabec \\ 75.& dacbe \\ 76.& daceb \\ \color{red}77.& \huge{\color{red}daebc} \\ 78.& daecb \\ 79.& dbace \\ 80.& dbaec \\ 81.& dbcae \\ 82.& dbcea \\ 83.& dbeac \\ 84.& dbeca \\ 85.& dcabe \\ 86.& dcaeb \\ 87.& dcbae \\ 88.& dcbea \\ 89.& dceab \\ 90.& dceba \\ \hline \end{array} \begin{array}{|rcll|} \hline 91.& deabc \\ 92.& deacb \\ 93.& debac \\ 94.& debca \\ 95.& decab \\ 96.& decba \\ 97.& eabcd \\ 98.& eabdc \\ 99.& eacbd \\ \color{red}100.& \huge{\color{red}eacdb } \\ 101.& eadbc \\ 102.& eadcb \\ 103.& ebacd \\ 104.& ebadc \\ 105.& ebcad \\ 106.& ebcda \\ 107.& ebdac \\ 108.& ebdca \\ 109.& ecabd \\ 110.& ecadb \\ 111.& ecbad \\ 112.& ecbda \\ 113.& ecdab \\ 114.& ecdba \\ 115.& edabc \\ 116.& edacb \\ 117.& edbac \\ 118.& edbca \\ 119.& edcab \\ 120.& edcba \\ \hline \end{array} % % // ======================================================================= \begin{array}{|rcll|} \hline \hline \end{array} % % // ======================================================================= \begin{array}{|rcll|} \hline \hline \end{array} % % // ======================================================================= \begin{array}{|rcll|} \hline \hline \end{array}\)

We need to know the gravitational acceleration, which is approximately 9.8 metres per second per second in magnitude.

There are four positive integers a,b,c,d less than 8 which are invertible modulo 8.

Find the remainder when (abc+abd+acd+bcd)(abcd)^{-1} is divided by 8.

\(\begin{array}{|rcll|} \hline && (abc+abd+acd+bcd)(abcd)^{-1} \\ &=& \dfrac{abc}{abcd} +\dfrac{abd}{abcd}+\dfrac{acd}{abcd}+\dfrac{bcd}{abcd} \\ &=& \dfrac{1}{d} +\dfrac{1}{c}+\dfrac{1}{b}+\dfrac{1}{a} \\ &=& a^{-1}+b^{-1}+c^{-1}+d^{-1} \\ \hline \end{array} \)

\(\begin{array}{|lrcl|} \hline \gcd(1,8)=\gcd(3,8)=\gcd(5,8)=\gcd(7,8)=1 \quad & | \quad a=1\quad b=3\quad c=3\quad d=5 \\\\ \begin{array}{|rcl|} \hline \phi(8)&=& 8\cdot\left(1-\dfrac12 \right) \\ &=& 4 \\ \hline \end{array} \\ \hline \end{array}\)

\(\small{ \begin{array}{|rcll|} \hline && \Big( 1^{-1} \pmod {8} + 3^{-1} \pmod {8} + 5^{-1} \pmod {8} + 7^{-1}\pmod {8} \Big)\pmod {8} \\\\ &=& \Big( 1^{\phi(8)-1} \pmod {8}+ 3^{\phi(8)-1} \pmod {8} + 5^{\phi(8)-1} \pmod {8} + 7^{\phi(8)-1}\pmod {8} \Big) \pmod {8} \\\\ &=& \Big( 1^{4-1} \pmod {8}+ 3^{4-1} \pmod {8} + 5^{4-1} \pmod {8} + 7^{4-1}\pmod {8} \Big) \pmod {8} \\\\ &=& \Big( 1^{3} \pmod {8}+ 3^{3} \pmod {8} + 5^{3} \pmod {8} + 7^{3}\pmod {8} \Big) \pmod {8} \\\\ &=& ( 1^{3}+ 3^{3} + 5^{3} + 7^{3} ) \pmod {8} \\\\ &=& ( 1+ 27 + 125 + 343 ) \pmod {8} \\\\ &=& 496 \pmod {8} \\\\ &=& 62\cdot 8 \pmod {8} \\\\ &\mathbf{=}&\mathbf{ 0 \pmod {8} } \\ \hline \end{array} }\)

What is the residue modulo 16 of the sum of the modulo 16 inverses of the first 8 positive odd integers?

Express your answer as an integer from 0 to 15, inclusive.

\(\begin{array}{|lrcl|} \hline \gcd(1,16)=\gcd(3,16)=\gcd(5,16)=\gcd(7,16) \\ =\gcd(9,16)=\gcd(11,16)=\gcd(13,16)=\gcd(15,16)=1 \\\\ \begin{array}{|rcl|} \hline \phi(16)&=& 16\cdot\left(1-\dfrac12 \right) \\ &=& 8 \\ \hline \end{array} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && \Big( 1^{-1} \pmod {16} + 3^{-1} \pmod {16} + 5^{-1} \pmod {16} \\ && + 7^{-1} \pmod {16} + 9^{-1} \pmod {16} + 11^{-1} \pmod {16} \\ && + 13^{-1} \pmod {16} + 15^{-1} \pmod {16} \Big) \pmod {16} \\\\ &=& \Big( 1^{\phi(16)-1} \pmod {16} + 3^{\phi(16)-1} \pmod {16} + 5^{\phi(16)-1} \pmod {16} \\ && + 7^{\phi(16)-1} \pmod {16} + 9^{\phi(16)-1} \pmod {16} + 11^{\phi(16)-1} \pmod {16} \\ && + 13^{\phi(16)-1} \pmod {16} + 15^{\phi(16)-1} \pmod {16} \Big) \pmod {16} \\\\ &=& \Big( 1^{8-1} \pmod {16} + 3^{8-1} \pmod {16} + 5^{8-1} \pmod {16} \\ && + 7^{8-1} \pmod {16} + 9^{8-1} \pmod {16} + 11^{8-1} \pmod {16} \\ && + 13^{8-1} \pmod {16} + 15^{8-1} \pmod {16} \Big) \pmod {16} \\\\ &=& \Big( 1^{7} \pmod {16} + 3^{7} \pmod {16} + 5^{7} \pmod {16} \\ && + 7^{7} \pmod {16} + 9^{7} \pmod {16} + 11^{7} \pmod {16} \\ && + 13^{7} \pmod {16} + 15^{7} \pmod {16} \Big) \pmod {16} \\\\ &=& ( 1^{7}+ 3^{7} + 5^{7} + 7^{7} + 9^{7}+ 11^{7} + 13^{7}+ 15^{7} ) \pmod {16} \\\\ &=& ( 1+ 2187 + 78125 + 823543 + 4782969 \\ && + 19487171 + 62748517+ 170859375 ) \pmod {16} \\\\ &=& 258781888 \pmod {16} \\\\ &=& 16173868\cdot 16 \pmod {16} \\\\ &\mathbf{=}&\mathbf{ 0 \pmod {16} } \\ \hline \end{array}\)

​Find x. Round to the nearest tenth if necessary.

\(\begin{array}{|rcll|} \hline x\cdot (x+1) &=& 10 \cdot 3 \\ x^2+x &=& 30 \\ x^2+x-30 &=& 0 \\ (x-5)(x+6) &=& 0 \\\\ x_1 &=& 5 \\ x_2 &=& -6 \qquad \text{no solution, because }x \lt 0 \\ \hline \end{array} \)

d. 5

Not free,I'll cash app, I've got all the way to the end and now I'm stuck, the tutor I had was no help at all.

Don't just try to get free answers, try to understand the concepts as well

LOL I'm kind of talking about my benchmark.

Solving the quadratic, we get x=-6 or 5. x must be positive, so x=5

I have no clue what I'm doing, I have until AUG 2nd to pass this benchmark online. I'll cash app you if you help me plzz plzz 

Again, using power of a point, x^2+x=30. Solve from there.

We can use power of a point to solve this problem. Given that two chords BC and DE which intersect at A, (BA)(AC)=(DA)(AE). So we know that 10x=32, so x=3.2.

Let the larger angle = x, then:

The 2nd angle          = x, and:

The 3rd angle           = 1/2x, so that:

x + x + 1/2x =180

2.5x = 180

x =180 / 2.5

x = 72 degrees, and:

72 / 2 =36 degrees - the smallest angle. So that:

72 + 72 + 36 = 180

Well, young person!! The reason you are getting different answers is because there is NO DIRECT solution to your equation!!. I can readily see that this a combined financial equation for a future value of $1,307,997.74, a present value of $100,000, a periodic payment of $5,000 and 40 periods, which appear to be "years". You can only solve for "i" using iteration, or trial and error method. You can easily solve it by using a dedicated financial calculator, if you have one. 

Go online and use this financial calculator: https://arachnoid.com/finance/

1- Enter $100,000 as negative under "pv"

2- Enter $5,000 as negative under "pmt"

3-Enter 40 under "np"

4- Try any interest rate and enter it under "ir"

5- Press "fv" and compare the amount to your $1,307,997.74

6- Change the interest rate in "4" above until you get as close as possible to "fv" of $1,307,997.74.

Note: I tried it on your behalf and got 5%, which gives the exact "fv" of $1,307,997.74 !!!.

First, we find the four numbers, which are 1, 3, 5 and 7. We then expand (abc+abd+acd+bcd)(abcd)^(-1) to get a^(-1)+b^(-1)+c^(-1)+d^(-1). Plugging in 1, 3, 5, and 7, we get 1^(-1)+3^(-1)+5^(-1)+7^(-1) which is congruent to 1+3+5+7 mod 8. Adding them together, we see that 1+3+5+7=16, which is congruent to 0 mod 8.\( \)

Um...

We would need to know the force of gravity to determine when it hits its peak point...

-MathCuber 

P.S I may be wrong i'm only 10...

If a number is congruent to 2^(-1), it basically means that when multiplied by 2, it is congruent to 1 mod 185. So the answer is 93.

I can't understand your question

Thanks so much! I would've never thought of using that way... Thanks again

3162277660.17 - This is in "standard notation"

3.162277660.17 - put a decimal point in front of the first number, which is a "3"

Count the number of digits BETWEEN the two decimal points and you should get 9. This 9 is the exponent or power of 10. Then write the whole number like this: 3.16227766017 x 10^9. You remove the first decimal point. So, your number 3.16227766017 x 10^9 is now in "scientific notation"

Oops! You are right!

I should have q1 =  25/51 and q2 = 24/50.