The answer is B. because it’s 1 long line, it is “and”; if it were two separate lines, it would be “or”
Answer:
4
Step-by-step explanation:
This situation has two unknowns - the total number of half dollars and the total number of quarters. Because we have two unknowns, we will write a system of equations with two equations using the two unknowns.
- h+q=31 is an equation representing the total number of coins
- 0.50h+0.25q=11 is an equation representing the total value in money based on the number of coin. 0.50 and 0.25 come from the value of a half dollar and quarter individually.
We write the first equation in terms of q by subtracting it across the equal sign to get h=31-q. We now substitute this for h in the second equation.
0.50(31-q)+0.25q=11
15.5-0.50q+0.25q=11
15.5-0.25q=11
After simplifying, we subtract 15.5 across and divide by the coefficient of q.
-0.25q=-4.5
q=4
We now know of the 31 coins that 4 are quarters.
Answer:
Step-by-step explanation:
.20 is a rational number between -1/4 and 2/5.
It can be part of a ratio.
20:100
A fraction
20/100 or 1/5
I hope this helps!
~cupcake
Hello :
all n in N ; n(n+1)(n+2) = 3a a in N or : <span>≡ 0 (mod 3)
1 ) n </span><span>≡ 0 ( mod 3)...(1)
n+1 </span>≡ 1 ( mod 3)...(2)
n+2 ≡ 2 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 0×1×2 ( mod 3) : ≡ 0 (mod 3)
2) n ≡ 1 ( mod 3)...(1)
n+1 ≡ 2 ( mod 3)...(2)
n+2 ≡ 3 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 1×2 × 3 ( mod 3) : ≡ 0 (mod 3) , 6≡ 0 (mod)
3) n ≡ 2 ( mod 3)...(1)
n+1 ≡ 3 ( mod 3)...(2)
n+2 ≡ 4 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 2×3 × 4 ( mod 3) : ≡ 0 (mod 3) , 24≡ 0 (mod3)
