16•0.10= 1.60$
8•0.05= 0.40¢
12•0.25= 3.00$
1.60$+0.40¢+3.00$= 5.00$
3•0.10= 0.30¢
5•0.05= 0.25¢
0.30¢+0.25¢= 0.55¢
5.00$-0.55¢= 4.45$
4.45$ is left in the piggy bank.
Answer:

The graph in the attached figure
Step-by-step explanation:
Let
x-----> money that Tasha can spend on new boots
we know that
The inequality that represent the situation is equal to

The solution is the interval--------> (40,∞)
All real numbers greater than 40
In a number line the solution is the shaded area to the right of the dashed line at 40 (open circle)
see the attached figure
Answer:
(x, y) = (3, 26)
Step-by-step explanation:
I like to solve systems of equations like this by graphing. A graphing calculator easily shows the solution for both x and y.
(x, y) = (3, 26)
__
As an alternative, here, the equations can be subtracted.
y = f(x)
y = g(x)
0 = f(x) - g(x) . . . . subtract the second equation from the first
0 = 11x -7 -(3x +17)
0 = 8x -24 . . . . simplify
0 = x -3 . . . . . . .divide by 8
3 = x
Now, we can substitute into either equation:
y = g(x) = 3(3) +17 = 26
The solution is (x, y) = (3, 26).
Answer:
Step-by-step explanation:
a) A relation R is symmetric when it includes the inverse relation, for example if it includes (8,9) then it should also include (9,8), if not, then the relation is not symmetric, you can see that in this case the relation includes (3,4) but not (4,3), therefore it is not symmetric
b) A relation is antisymmetric when it never includes the inverse relation, for example if it includes (8,9) then it can not include (9,8), if it does then it is not antisymmetric. In this case you can see that it first starts with (1,2) but then it also includes (2,1) so then it is not antisymmetric
c) A relation is reflexive if for each number of the domain set it includes the pair that is two times that same number, for example if 8 is in the domain then the relation should include (8,8). if not then it is not reflexive. In this case you can see that the domain S includes 1 but (1,1) is never on the relation or for example (2,2) is also never in the relation.