Let's use our largest amount (quarter) to see how many can fit evenly into 1.74.
- 1.74/0.25 = 6.96
- The most quarters that can fit into $1.74 is 6.
---Multiply $0.25 by 6 = $1.50
---Subtract $1.50 from $1.74 = $0.24
---We need to use the rest of the coins to fill up $0.24.
Now let's use our second largest amount (dime) to see how many can fit evenly into 0.24.
- 0.24/0.10 = 2.4
- The most dimes that can fit into $0.24 is 2.
---Multiply $0.10 by 2 = $0.20
---Subtract $0.20 from $0.24 = $0.04
No nickels can fit into $0.04, because they are worth $0.05.
We can use our least amount (pennies) to fill in the $0.04 remaining.
Your answer is 6 quarters, 2 dimes, and 4 pennies.
Collinear=On the same line
N-U
N-T
U-T
I believe that is what this questions is asking. :)
Answer:
<h2>
The eleventh term of the sequence is 64</h2>
Step-by-step explanation:
The sequence given is an arithmetic sequence
14, 19, 24, …………., 264
The nth term of an arithmetic sequence is given as;
Tn = a+(n-1)d where;
a is the first term = 14
d is the common difference = 19-14=24-19 = 5
n is the number of terms = 11(since we are to look for the eleventh term of the sequence)
substituting the given values in the formula given;
T11 = 14+(11-1)*5
T11 = 14+10(5)
T11 = 14+50
T11 = 64
The eleventh term of the sequence is 64
#9 is
32divided(10-8)divided2-3
Answer:
A)-7
B)2
C)-2
D)-4
Step-by-step explanation:
