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How do you solve the B in Ax plus By equals C?
To solve for B in the equation ( Ax + By = C ), you first isolate the term involving B. Rearranging gives ( By = C - Ax ). Then, divide both sides by y (assuming y is not zero) to solve for B: ( B = \frac{C - Ax}{y} ).
How do you solve a AND b OR c in logical operators?
There is nothing to "solve". You can evaluate the expression when each of a, b and c are TRUE or FALSE. But that is not solving.
Prove if a union c equals b union c and a intersect c equals b intersect c then a equals b?
suppose x is in B. there are two cases you have to consider. 1. x is in A. 2. x is not in A Case 1: x is in A. x is also in B. then x is in A intersection B. Since A intersection B = A intersection C, then this means x is in A intersection C. this implies that x is in C. Case 2: x is not in A. then x is in B. We know that x is in A union B. Since A union B = A union C, this means that x is in A or x is in C. since x is not in A, it follows that x is in C. We have shown that B is a subset of C. To show that C is subset of B, we do the same as above.
P equals b plus 3a plus c solve for b?
p=b+3a+c p-3a-c=b+3a-3a+c-c p-3a-c=b b=p-3a-c
How do you solve ax bx-c and solve for x?
To solve the equation ( ax + bx - c = 0 ) for ( x ), first combine like terms on the left side to get ( (a + b)x - c = 0 ). Next, isolate the term involving ( x ) by adding ( c ) to both sides, resulting in ( (a + b)x = c ). Finally, divide both sides by ( (a + b) ) to find ( x = \frac{c}{a + b} ), assuming ( a + b \neq 0 ).
How do you solve a equals b plus c plus d solve for c?
If: a = b+c+d Then: c = a-b-d
What is the complement of the union B and C?
not (b or c) = (not b) and (not c)
Is it true that if a union c equals b union c then a equals b?
No- this is not true in general. Counterexample: Let a = {1,2}, b = {1} and c ={2}. a union c = [1,2} and b union c = {1,2} but a does not equal b. The statement be made true by putting additional restrictions on the sets.
Solve the equation ax-b equals c for x?
ax - b = c ax = b + c x = (b + c)/a
How do you solve the B in Ax plus By equals C?
To solve for B in the equation ( Ax + By = C ), you first isolate the term involving B. Rearranging gives ( By = C - Ax ). Then, divide both sides by y (assuming y is not zero) to solve for B: ( B = \frac{C - Ax}{y} ).
How do you solve a AND b OR c in logical operators?
There is nothing to "solve". You can evaluate the expression when each of a, b and c are TRUE or FALSE. But that is not solving.
If j and k are intersecting lines A and B are points on j and C and D are points on k how many planes contain points A B C and D?
exactly one
Prove if a union c equals b union c and a intersect c equals b intersect c then a equals b?
suppose x is in B. there are two cases you have to consider. 1. x is in A. 2. x is not in A Case 1: x is in A. x is also in B. then x is in A intersection B. Since A intersection B = A intersection C, then this means x is in A intersection C. this implies that x is in C. Case 2: x is not in A. then x is in B. We know that x is in A union B. Since A union B = A union C, this means that x is in A or x is in C. since x is not in A, it follows that x is in C. We have shown that B is a subset of C. To show that C is subset of B, we do the same as above.
What is the formula to solve for b a equals 2b plus c?
If: a = 2b+c Then: a-c = 2b And: b = (a-c)/2
Prove A union B minus A intersection B equals A minus B union B minus A?
complement of c
Solve for indicated letter b equals 8c for c?
c equals b over 8
What A union C minus B equals A minus B union C minus B?
The expression ( (A \cup C) - B = (A - B) \cup (C - B) ) represents the set of elements that are in either ( A ) or ( C ) but not in ( B ). On the left side, ( (A \cup C) - B ) includes all elements from ( A ) and ( C ) excluding those in ( B ). The right side, ( (A - B) \cup (C - B) ), combines the elements in ( A ) without ( B ) and those in ( C ) without ( B ), which captures the same set of elements. Thus, both sides are equal, demonstrating a property of set difference and union.
