Solve for y: -2(5y-1)-y=-4(y-3)
1 answer:
Answer:
<h2>Step-by-step explanation:
- 2(5y - 1) - y = - 4(y - 3)
Expand the terms in the bracket
That's
-10y + 2 - y = - 4y + 12
Group like terms
Send the constants to the right side of the equation and those with variables to the left side
That's
- 10y - y + 4y = 12 - 2
- 7y = 10
Divide both sides by - 7
We have the final answer as
<h3>Hope this helps you
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Answer:
a) False b) False c) True d) True e) True
Step-by-step explanation:
a) If A is a subset of B, and x∈B, then x∈A. False
Suppose A is Z (Set of Integers), and B is R (Set of Real Numbers). Then A is a subset of B. x ∈ B, let's say x equals π. If x ∈ B (B = Real Numbers) and x=π then x ∉ A (Z).
We could call A, any other subset of Real numbers (Q, I,..) and we both would come up to the same conclusion when it comes to Real numbers the Set.
So this conclusion is False. Not always an element of a subset is an element of a set.
b) False
For this one, I've drawn some lines, and it is useful to work with them.
Given the set {(x,y) ∈ R²| 0<x<0} then all negative and positive numbers but zero belong to this set.
c) If A and B are square matrices, then AB is also square. True
Taking into account the rules for multiplying matrices. The number of columns of A must be the same number of lines of B to there can be a matrix product.
Whenever we multiply square matrices, we'll always get square matrices. Then this conclusion is true.
d) A and B are subsets of a set S, then A∩B and A∪B are also subsets of S. True
Suppose A= Z (Integer Numbers) and B=Q (Rational Numbers) and S= R (Real Numbers)
A∩B = Z∩Q=∅ and A∪B =Z∪Q = subset
Since the ∅ empty set ⊂ in every set and ZUQ is another Subset of R this is a True conclusion.
e) True. For a matrix A, we define A²= AA
For any Power of Matrices, all we have to do is multiply any given matrix by itself for a given number of times.
M²=M*M
M³=M*M*M
