Set V is all the unique three-digit numbers using the digits 1, 2, 3, 4, and 5. Which set would be considered a subset of set V? Set W is all the unique three-digit numbers using the digits 1, 3, and 5. Set X is all the unique three-digit numbers using the digits 2, 4, and 6. Set Y is all the unique three-digit numbers using the digits 0, 2, and 4. Set Z is all the unique three-digit numbers using the digits 3, 5, and 7.

Choice A) Set W is all the unique three-digit numbers using the digits 1, 3, and 5

Explanation:

The original set uses the digits {1,2,3,4,5}. We can only use those digits. Nothing else. So that means we can't use 0, 2, 4, 6, or 7. This rules out choices B through D.

Only choice A is left. This answer choice involves 1,3 and 5 which is the correct subset of digits to use.

Put another way, if set B was a subset of set A, then everything in B must also be found in A. Having B = {2,4,6} and A = {1,2,3,4,5} shows B is not a subset of A. We see a counter example has 6 in B, but not in set A.

This allows us to rule out choice B. Similar ideas apply to the answer choices C and D as well.

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions