If the letters are considered distinct, then the number of permutations is
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.
If we count either C as the same character, then we would be double-counting - to correct this, we would simply divided by the number of ways we can choose C from the available characters, or
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.
It’s add positive 2 every time I believe
if I’m wrong I’m so sorry and please tell me I’m wrong for other that have to use this.
If I’m right ya!! XD
Hope this helps you!
-Pam Pam
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.
The statement "<span>The rate of change of y with respect to x is inversely proportional to y^4" can be written mathematically as dy/dx = k/y^4
To solve the differential equation, we use variable saparable method.
y^4 dy = kdx
Integrating both sides gives,
y^5 / 5 = kx + A
y^5 = 5kx + 5A = Bx + C; where B = 5k and C = 5A
![y= \sqrt[5]{Bx+C}](https://tex.z-dn.net/?f=y%3D%20%5Csqrt%5B5%5D%7BBx%2BC%7D%20)
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