Answer:
1 1/2 cups
Step-by-step explanation:
2/3 cup of almonds = 4 cups of trail mix
4 cups of trail mix = 2/3 cup of almond
4 cups of trail mix = 2/3 cup of almond
1 cup of trail mix = 2/3 ÷ 4 = 1/6 cup of almond
1 cup of trail mix = 1/6 cup of almond
9 cups of trail mix = 1/6 x 9 = 3/2 cup of almond
3/2 cups = 1 1/2 cups
Answer and Explanation:
Solution: The operation of concatenation for a set of string on p. and the set is
AB = {XY | X ∈ A and y ∈ B}.
We need to satisfy all these following properties to find out the standard set is closed under concatenation.
1- Union of two standard sets also belongs to the classic collection. For example, A and B are regular. AUB also belongs to a regular group.
2- Compliment of two standards set A and B are A’ and B’ also belonging to the standard set.
3- Intersection of two standards set A and B is A∩B is also a regular set member.
4- The difference between two regular sets is also standard. For example, the difference between A and B is A-B is also a standard set.
The closure of the regular set is also standard, and the concatenation of traditional sets is regular.
Answer:
The Answer is: x / y = 8 / 3
Step-by-step explanation:
3(x - 2) - 4(2y - 1) + 2=0
3x - 6 - 8y + 4 + 2 = 0
3x - 8y = 0
3x = 8y
3x/y = 8
x / y = 8 / 3
Hope this Helps! Have an Awesome Day!! (-:
Answer:
1.80
Step-by-step explanation:
trust me 100% sure
The number of dimes is 75 coins.
The number of nickels is 1 coin.
<u>Step-by-step explanation:</u>
It is given that,
A box contains 76 coins, only dimes and nickels.
- The 10 cent coin is called a dime.
- Let, the number of dimes be x.
- The 5 cent coin is called a nickel.
- Let, the number of nickels be y.
<u>This is represented by the system of equations :</u>
x + y = 76 -------(1)
10x + 5y = 4.90 ------(2)
Multiply eq (1) by 5 and subtract eq (2) from eq (1),
5x + 5y = 380
-<u>(10x + 5y = 4.90)</u>
<u>-5x = 375.1</u>
x = 375/5
x = 75 coins.
The number of dimes is 75 coins.
Substitute x=75 in eq (1),
y = 76 - 75
y = 1 coin.
The number of nickels is 1 coin.