n = # of days
45 = 2.5n + 2n
45 = 4.5n
n = 10
He can paint 45 structures in 10 days
The simplified ratio would be 4 to 1! hope this helps :)
Answer:
p = -1 q = -4
Step-by-step explanation:
a system of eq and solve for p and q ??? can do :)
Eq. 1) 8p + 2q = - 16
Eq. 2) 2p - q = 2
use Eq .2 and solve for q
2p - 2 = q
plug into Eq.1 with q
8p +2(2p - 2) = - 16
8p +4p -4 = -16
12p = - 12
p = -1
plug -1 into Eq. 1 for p and solve for q
8(-1) + 2q = - 16
-8 + 2q = - 16
2q = -8
q = -4
Time=(distance)/(speed)
speed=55 mi/hr
Total distance=440 miles
Distance covered=275 miles
remaining distance=440-275=165 miles
Time taken to cover the remaining distance will be:
time=165/55=3 hours
Total time taken to travel=440/55=8 hours
if we are to use the equation given by:
55h+275=440
time taken will be found by solving for h, thus we shall have:
55h=440-275
55h=165
h=165/55
h=3 hours
Answer: 3 hours
Answer:
8 days
Step-by-step explanation:
On day 8, Isabella will save 256 nickels, bringing her total to 510.
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The number of nickels saved on day n is 2^n. The total is 2^(n+1)-2.
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The above can be written down from your knowledge of binary sequences. If you want a more formal development, read on.
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The number of nickels saved on day n is a geometric sequence with first term 2 and common ratio 2. The n-th term of the sequence is ...
an = a1·r^(n-1) = 2·2^(n-1) = 2^n
The sum of n terms of the sequence is ...
S = a1(r^n -1)/(r -1) = 2(2^n -1)/(2-1)
S = 2^(n+1) -2
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We want S > 500, so ...
500 < 2^(n+1) -2
502 < 2^(n+1)
251 < 2^n
log(251) < n·log(2)
n > log(251)/log(2)
n > 7.97 . . . . . . . . 8 days or more to save more than 500 nickels
