Vas happening?
You break it down
3 ^5 is 3 x 3 x 3 x 3 x 3 = 729
4 ^ 5 is 4 x 4 x 4 x 4 x 4 = 1024
729 times 1024 = 746,496
12 ^ 10 is 12 x 12 x 12 x 12 x 12 x 12 x 12 x 12 x 12 x 12 = 6.19174e10
So the answer is incorrect
No, the total cost will be less than $1000.
There will be 7 buses with 44 students each and one bus with 39 students. So in there will be 8 buses in all.
8*$95=$760
Answer:
13r²(2rs + 4r³ - 3s⁴)
Step-by-step explanation:
In equation 26r³s + 52r⁵ - 39r²s⁴;
The GCF of 26, 52, and 39 = 13
The GCF of r³, r⁵ and r² = r²
The GCF of s, (no "s"), and s⁴ = no "s" (Since one of the number doesn't have "s")
Now we can factor out 13r² from all three expressions;
26r³s + 52r⁵ - 39r²s⁴
=> <em>13r²(2rs) + 13r²(4r³) - 13r²(3s⁴)</em>
To factor it all together;
<u>13r²(2rs + 4r³ - 3s⁴)</u>
Hope this helps!
Answer: A. 9567m
i hope i'm correct let me know if i'm wrong
<span>it depends how the interest is calculated, but there's not much of a difference
assuming its continuously compouned, you use this formula: A(t)=Pe^(rt), where A is the final amount, P is the initial investment, r is the interest, and t is the time in years
you want to find t such that A(t)=18,600 so 18,600=1000e^(.0675t)
you need to use logarithm to figure it out, take the natural log of both sides
the following properties will come into use:
ln(a*b)=ln(a)+ln(b)
ln(a^b)=bln(a)
ln(e)=1
taking the natural log
ln(18,600)=ln(1000e^(.0675t))
ln(18,600)=ln(1000)+ln(e^.0675t)
ln(18600)=ln(1000) + .0675t
now solve for t: t= (ln(18600)-ln(1000))/.0675
t=43.31</span>
