kay oss can solve equations at the rate of 26 per hour. Her friend, Dan D. Lyons, can solve equations at a rate of 20 per hour. When they start their homework, Kay has already solve 3 equations in study hall and Dan has already solved 8. How long will it take until they have solved the same number of equations

0.83 hours

Step-by-step explanation:

Hi, to answer this question we have to write a system of equations:

Kay can solve equations at the rate of 26 per hour (x), and he already solved 3 equations.

Kay=3+26 x

Dan D can solve equations at the rate of 20 per hour (x), and he already solved 8 equations.

Dan d = 8+20x

If both equations are equal: (they have solved the same number of equations)

3+26x = 8+20x

Solving for x (hours)

26x-20x =8-3

6x= 5

x =5/6

x = 0.83 hours  

Feel free to ask for more if needed or if you did not understand something.

u bout ugly as   a mf

step-by-step explanation: lol

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752