Answer:
6+ 45 = 500 #!_ djdjbdkeknf
Third term = t3 = ar^2 = 444 eq. (1)
Seventh term = t7 = ar^6 = 7104 eq. (2)
By solving (1) and (2) we get,
ar^2 = 444
=> a = 444 / r^2 eq. (3)
And ar^6 = 7104
(444/r^2)r^6 = 7104
444 r^4 = 7104
r^4 = 7104/444
= 16
r2 = 4
r = 2
Substitute r value in (3)
a = 444 / r^2
= 444 / 2^2
= 444 / 4
= 111
Therefore a = 111 and r = 2
Therefore t6 = ar^5
= 111(2)^5
= 111(32)
= 3552.
<span>Therefore the 6th term in the geometric series is 3552.</span>
Let's start by assuming Armando's house is between Joey's and the park.
Let

be the distance Joey walked to Armando's house.
<span>The park is 9/10 mile from Joey's home. Joey leaves home and walks to Armando's home. Then Joey and Armando walk 3/5 mile to the park.
</span>


That's probably the answer they're looking for. But what if the park is between Joey and Armando's houses or Joey is between the park and Armando? (The latter isn't really possible with the given distances.)
Let

be the distances between three collinear points like we have here. Our equation is really a few equations in one, something like

Let's get rid of the plus/minuses. Squaring,



For us, that's a quadratic equation for


I'll skip right to the solutions,


We could have gotten the 3/2 just by adding 9/10+3/5 but this was more fun.
Answer:
C.
Since the square root of 20 is around 4.5, we know that ≈4.5 is almost at the half mark between 4 and 5 on the number line.