Answer:
Step-by-step explanation:
You can multiply -3 by 4 and then multiply the result of that by 5, or you could write -3, 4 and 5 in other orders (6 possible orders) and obtain the same result.
I would multiply (-3) by 4, obtaining -12, and then multiply this -12 by 5, obtaining -60. But I could also multiply 5 by 4, obtaining 20, and then multiply this 20 by -3, obtaining the same -60.
What I have described here are the commutative and associative properties of multiplication.
Answer:
x=2(3-y)/5
Step-by-step explanation:
5x=6-2y
x=6-2y/5
As the new mathematical operation is defined by a△b=a^2-b/b-a^2, the value of 4△3 using the same operation will be 4△3 = -1
As per the question statement, we are given a new mathematical operation a△b=a^2-b/b-a^2 and we are supposed to find the value of 4△3 using the same operation.
Given, a△b=a^2-b/b-a^2
now 4△3 = (4^2-3) / (3-4^2)
4△3 = (16-3) / (3-16)
4△3 = 13 / -13
4△3 = -1
Hence, as the new mathematical operation is defined by a△b=a^2-b/b-a^2, the value of 4△3 using the same operation will be 4△3 = -1.
- Mathematical operation: An operator in mathematics is often a mapping or function that transforms components of one space into elements of another.
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6-7n=29 would be your answer
Answer:
Step-by-step explanation:
Given a general quadratic formula given as ax²bx+c = 0
To generate the general formula to solve the quadratic equation, we can use the completing the square method as shown;
Step 1:
Bringing c to the other side
ax²+bx = -c
Dividing through by coefficient of x² which is 'a' will give:
x²+(b/a)x = -c/a
- Completing the square at the left hand side of the equation by adding the square of half the coefficient x i.e (b/2a)² and adding it to both sides of the equation we have:
x²+(b/a)x+(b/2a)² = -c/a+(b/2a)²
(x+b/2a)² = -c/a+(b/2a)²
(x+b/2a)² = -c/a + b²/4a²
- Taking the square root of both sides
√(x+b/2a)² = ±√-c/a + b²/√4a²
x+b/2a = ±√(-4ac+b²)/√4a²
x+b/2a =±√b²-4ac/2a
- Taking b/2a to the other side
x = -b/2a±√√b²-4ac/2a
Taking the LCM:
x = {-b±√b²-4ac}/2a
This gives the vertex form with how it is used to Solve a quadratic equation.
