Lets find that out writing equations and solving them, a multiple of 1/6 is something like this (1/6)x, so we have
(1/6)x > 3/6
and
<span>(1/6)y < 4/6
</span>lets solve both equations:
<span>(1/6)x > 3/6
</span>x > 6(3/6)
x > 3
<span>(1/6)y < 4/6
</span>y < 6(4/6)
y < 4
So the number must be between 3 and 4, which is obvious, lets try with 3.5 then, that is 3 5/10 or 35/10 = 7/2 in fractional form, and lets try it out:
(1/6)(7/2) = 7/12
finally we compare with the original fractions:
1/6 < 7/12
2/12 < 7/12
So, it comply with being greater than 1/6, now lets compare with 4/6
7/12 < 4/6
<span>7/12 < 8/12
</span>therefore is also smaller than 4/6 and hence 7/12 is a multiple of 1/6 between 3/6 and 4/6
Answer:
y<-2 or y>= 2
Step-by-step explanation:
y + 3 < 1
y - 3 + 3 < 1 -3
y < -2
2y +1 >= 5
2y - 1 + 1 <= 5 - 1
2y <= 4
2y/2 <= 4/2
y = 2
To solve for x we proceed as follows:
from the laws of logarithm, given that:
log_a b=c
then
a^c=b
applying the rationale to our question we shall have:
log_5 x=4
hence
5^4=x
x=625
Answer: x=625
