7.5
X= (b-c)/a
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.
ax-9 = b x = (b+9)/a
x(-b)=m(x-c)
7.5
Ax + B = Bx + C Ax - Bx = (C - B) x (A - B) = (C - B) x = (C - B) / (A - B)
x + y + z = 0 x = a - b, y = b - c, z = c - a, therefore a - b + b - c + c - a = ? a - a + b - b + c - c = 0
X= (b-c)/a
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.
ax - b = c ax = b + c x = (b + c)/a
Use the quadratic formula, with a = 1, b = 7, c = 9.
x = b/(a + c)
For this, you need the Pythagorean theorem. Let A equal x + 9 Let C equal 26 A^2 + B^2 = C^2 (x + 9)^2 + B^2 = 26^2 x^2 + 18x + 81 + B^2 = 676 x^2 + 18x + B^2 = 595 This doesn't help you find x, but I hope this did something!
ax-9 = b x = (b+9)/a
x(-b)=m(x-c)
The product of the means equals the product of the extremes. In other words, if A is to B as C is to D, then B times C equals A times D, so... A = B x C ÷ D B = A x D ÷ C C = A x D ÷ B D = B x C ÷ A