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Positive integer x leaves a remainder of 2 when divided by 8.
What is the remainder when x+9 is divided by 8

\(\begin{array}{|rcll|} \hline x & \equiv & 2 \pmod{8} \\ x+{\color{red} 9 } & \equiv & 2+{\color{red} 9 } \pmod{8} \\ x+9 & \equiv & 11 \pmod{8} \\ x+9 & \equiv & 11-8 \pmod{8} \\ x+9 & \equiv & 3 \pmod{8} \\ \hline \end{array}\)

Paradox.

Ya :D Did you forgot to login? Who are you?

Check the way you are putting it in 

\(\ln(6^{719})\\ =719 \ln 6\\ \approx 1288.2750583749715456\)

\(-7/9 + 8\\ = -7/9 + 72/9\\ = (72-7)/9\\ = 65/9\\ = 7\;\;2/9\)

\(\quad 16x^2-49\\ = (4x)^2-7^2\\ = (4x-7)(4x+7)\)

\(\left(\dfrac{6x^3y^{-7}}{9xy}\right)^{-1}\\ = \dfrac{9xy}{6x^3y^{-7}}\\ = \dfrac{9xy^8}{6x^3}\\ =\dfrac{3y^8}{2x^2}\)

Put it on the calc on this site,

it shows \(\mathtt{\dfrac{6}{100}\times 16,824 = 1009.44}\)

ALAN WILL EXPLAIN IT TO YOU TOMORROW, HOPEFULLY. THE EQUATION, AS GIVEN, IS THE PROBLEM. THAT IS WHY I MODIFIED IT. IF YOU SUBSTITUTE 1/2 FOR e, THEN YOU WILL GET THE RIGHT ANSWER: SOLVE FOR t: 85/6=85 * 1/2^(t/719), WHERE t=ELAPSED NUMBER OF YEARS, 719=HALF-LIFE OF THE RADIOACTIVE SUBSTANCE. THAT IS THE END OF THIS MATTER FOR ME!.

Max is correct and gives the exact answer....which is the same answer I gave (rounded) as

.   sqrt(98) = sqrt(49 x 2) = 7  sqrt2 = 9.899 in

But your answer does not satisfy the given equation for decay...

the question CLEARLY states the amount of isotope is given by  85e^(-t/719)   where t is in years.

The ORIGINAL amount is given by t= 0      85

85e^((-1858.59)/719) = the amount at 1858.59 years = 6.4088   which is NOT 1/6 of 85 = 14.166

x^2+x^x2

\(x^2+x^{x^2}\ne x^4\\~\\ x^2\times x^2=x^4\\~\\ x^2+ x^2=1x^2+ 1x^2=2x^2\\~\\\)

I am going to use another way to solve this equation.

aftenn provided a good start, Substitute x = ln a.......

\(\dfrac{e^x+3}{4e^{-x}+5}=2\\ e^x+3=\dfrac{8}{e^x}+10\\ e^{\ln a}+3=\dfrac{8}{e^{\ln a}}+10\\ a+3 = \dfrac{8}{a}+10\\ a= \dfrac{8}{a}+7\\ a^2-7a-8=0\\ (a+1)(a-8)=0\\ (e^x+1)(e^x-8)=0\\\text{Set each factor to 0}\\ e^x=-1 \text{ have no real solutions}\\\text{PS: }x=\pi i \text{ is a solution but it's not real}\\ e^x=8\\ x = \ln 8\)

If you want imaginary solutions too x = pi(i) or x = ln 8

\(18+ \dfrac{6}3(2^2+5)\\ = 18 + 2(9)\\ = 36\)

Same as Kaleb's answer.(Kaleb? Sanders? Sorry I come from Hong Kong)

Just multiply both sides of the equation by 16 to find m !

232 m = 16 x 87

m = 16 x 87 / 232 = 6

isn't this expanded? (-3)^0 = 1

5 into expanded form is 5

1 into expanded form is 1.

in expanded form