Answer: No, she does not have enough.
Step-by-step explanation:
1. You have the following information given in the problem above:
- The measures of the three pieces are: 32 centimeters, 41.19 centimeters and 57.8 centimeters long.
- She need 200 centimeters of ribbon for the box.
2. Therefore, you must add the measures given in the problem to know if Marie has enough ribbon to decorate the gift box. Then:
3. As you can see:
130.99 cm<200 cm
Therefore, she does not have enough ribbon.
Answer:
Step-by-step explanation:
Given the explicit function as
f(n) = 15n+4
The first term of the sequence is at when n= 1
f(1) = 15(1)+4
f(1) = 19
a = 19
Common difference d = f(2)-f(1)
f(2) = 15(2)+4
f(2) = 34
d = 34-19
d = 15
Sum of nth term of an AP = n/2{2a+(n-1)d}
S20 = 20/2{2(19)+(20-1)15)
S20 = 10(38+19(15))
S20 = 10(38+285)
S20 = 10(323)
S20 = 3230.
Sum of the 20th term is 3230
For the explicit function
f(n) = 4n+15
f(1) = 4(1)+15
f(1) = 19
a = 19
Common difference d = f(2)-f(1)
f(2) = 4(2)+15
f(2) = 23
d = 23-19
d = 4
Sum of nth term of an AP = n/2{2a+(n-1)d}
S20 = 20/2{2(19)+(20-1)4)
S20 = 10(38+19(4))
S20 = 10(38+76)
S20 = 10(114)
S20 = 1140
Sum of the 20th terms is 1140
Answer:
-6s-c+1
Step-by-step explanation:
(-3s-4c+1)+(-3s+3c)
We have been given the above expression. To find the sum, we simply collect the like terms and combine them;
(-3s-4c+1)+(-3s+3c) = -3s + -3s -4c + 3c + 1
-3s + -3s -4c + 3c + 1 = -3s - 3s + 3c - 4c + 1
-3s - 3s + 3c - 4c + 1 = -6s - c + 1
Therefore;
(-3s-4c+1)+(-3s+3c) = -6s-c+1