20 kg remaining in bag 1.
40 kg remaining in bag 2.
Answer: 1 .Thus for a graph to have an Euler circuit, all vertices must have even degree. The converse is also true: if all the vertices of a graph have even degree, then the graph has an Euler circuit, and if there are exactly two vertices with odd degree, the graph has an Euler path.
2. A graph has an Euler circuit if and only if the degree of every vertex is even. A graph has an Euler path if and only if there are at most two vertices with odd degree.
Step-by-step explanation:
Can i have branily plz
• So we know that.....
x represent bags of snack and y is bottles of water.
This equations shows the total amount and the cost of each water bottle and snack:
20.00 = 2.50x + 1.00y
Total: $20.00
Snack: $2.50
Water Bottle: $1.00
And this question shows the total items:
11 = x + y
Which there will be some snack + some water bottle = 11 items
—————————————————————
• Now I’m going to first solve for x, which is the amount of bags of snack.
I will use the equation, 11 = x + y.
(First, we’ll subtract y from both side, since we’re solving for x [UNDO])
11 = x + y
-y = - y
_______
11 - y = x —> so x is equal to 11 minus y.
—————————————————————
• Now we’re going to plug the 11 - y as x in the equation: 20.00 = 2.50x + 1.00y to solve for y.
20.00 = 2.50 (11 - y) + 1.00y
20.00 = 27.5 - 2.50y + 1.00y (Distributed)
20.00 = 27.5 - 1.50y (Combine like terms)
20.00 = 27.5 - 1.50y
-27.5 = -27.5 (Subtract -27.5 both side)
——————————
-7.5 = - 1.50y
-7.5 = -1.50y
—— ——— (Divide both side by -1.50)
- 1.50 = -1.50
5 = y
y is equals to 5, which means that there are 5 water bottles.
Now we know there are 11 items total and because there are 5 water bottles, there will be 6 bags of snacks. 11-5=6
—————————————————————
ANSWER:
They bought 6 bags of snacks! :)
Answer:
Step-by-step explanation:
Vertex A of the triangle ABC when rotated by 90° counterclockwise about the origin,
Rule to be followed,
A(x, y) → P(-y, x)
Therefore, A(1, 1) → P(-1, 1)
Similarly, B(3, 2) → Q(-2, 3)
C(2, 5) → R(-5, 2)
Triangle given in second quadrant will be the triangle PQR.
If the point P of triangle PQR is reflected across a line y = x,
Rule to be followed,
P(x, y) → X(y, x)
P(-1, 1) → X(1, -1)
Similarly, Q(-2, 3) → Y(3, -2)
R(-5, 2) → Z(2, -5)
Therefore, triangle given in fourth quadrant is triangle XYZ.
