Answer:
Quadratic equations have the solution
x = [-b +-sqr root(b^2 -4ac)] / 2a
In the given equation a = l; b = m and c = n So,
x = [-m +-sqr root(m^2 -4*l*n)] / 2*l
Step-by-step explanation:
Answer:
The steps are numbered below
Step-by-step explanation:
To solve a maximum/minimum problem, the steps are as follows.
1. Make a drawing.
2. Assign variables to quantities that change.
3. Identify and write down a formula for the quantity that is being optimized.
4. Identify the endpoints, that is, the domain of the function being optimized.
5. Identify the constraint equation.
6. Use the constraint equation to write a new formula for the quantity being optimized that is a function of one variable.
7. Find the derivative and then the critical points of the function being optimized.
8. Evaluate the y-values of the critical points and endpoints by plugging them into the function being optimized. The largest y- value is the global maximum, and the smallest y-value is the global minimum.
You would get 2 (2x+4). The two factors are the 2, and the (2x+4)
Since, both the events are independent.
So, P(A+B) = P(A)×P(B)
a) P( Head & Getting a 2 ) = P( head ) × P( Getting a 2 )
b) P( Even number & Tail ) = P( Even number ) × P( Tail )
Hence, this is the required solution.
Answer: There are no solution
Step 1: Simplify both sides of the equation.
3(x+5)−3=2(3x+1)−3x
(3)(x)+(3)(5)+−3=(2)(3x)+(2)(1)+−3x(Distribute)
3x+15+−3=6x+2+−3x
(3x)+(15+−3)=(6x+−3x)+(2)(Combine Like Terms)
3x+12=3x+2
3x+12=3x+2
Step 2: Subtract 3x from both sides.
3x+12−3x=3x+2−3x
12=2
Step 3: Subtract 12 from both sides.
12−12=2−12
0=−10
