Answer:
CD = two square root of 10 end square root
Step-by-step explanation:
To find the length of a segment, use the distance formula. Substitute the order pairs for the endpoints of the segment. CD has the end points (-7, -4) and (-1, -2).

Answer:
Option d. $22154 is the right answer.
Step-by-step explanation:
To solve this question we will use the formula 
In this formula A = amount after time t
P = principal amount
r = rate of interest
n = number of times interest gets compounded in a year
t = time
Now Lou has principal amount on the starting of first year = 10000+5000 = $15000
So for one year 

= $15900
After one year Lou added $5000 in this amount and we have to calculate the final amount he got
Now principal amount becomes $15900 + $ 5000 = $20900
Then putting the values again in the formula



So the final amount will be $22154.
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Answer:
#3.) The initial value of 16 gal at x = 5 minutes means that 16 gallons of water was present 5 minutes after the barrel started leaking.
#4.) Find how many minutes until the barrel is empty of water.
Let y = 0, to solve for time x.
0 = (-2/5)*x + 18
(2/5)*x = 18
x = (5/2)* 18 = 5*9 = 45 minutes
Up to and after 45 minutes, the barrel is empty of water.
Step-by-step explanation:
#2.)
minutes: 5, 10, 15, 20
water(gal): 16, 14, 12, 10
Find slope: slope m = (14 - 16)/(10 - 5) = -2/5
y - 10 = (-2/5)*(x - 20)
y - 10 = (-2/5)* x + 8
y = (-2/5)*x + 18
rate of change slope means that for every minute 2/5 gallons of water is lost
#3.) The initial value of 16 gal at x = 5 minutes means that 16 gallons of water was present 5 minutes after the barrel started leaking.
#4.) Find how many minutes until the barrel is empty of water.
Let y = 0, to solve for time x.
0 = (-2/5)*x + 18
(2/5)*x = 18
x = (5/2)* 18 = 5*9 = 45 minutes