Do they know how to divide?
If so, tell them to divide the number in as many ways until you go past the number
For example, if your seeing if 9 is a prime number divide 9 by ,1,2,3,4,5,6,7,8,9 and only count the whole numbers.
This sounds complicated so just teach them simple time tables
1/3x - 2 = 1/4x + 11....it looks to me like one y was subbed in for another...
ur 2 equations are :
y = 1/3x - 2 and y = 1/4x + 11...but we need them in standard form...
y = 1/3x - 2
-1/3x + y = -2 ..multiply by 3
-x + 3y = -6.....or 3y - x = -6 <==
y = 1/4x + 11
-1/4x + y = 11 ...multiply by 4
-x + 4y = 44 or 4y - x = 44 <==
Answer:
The quadratic x2 − 5x + 6 factors as (x − 2)(x − 3). Hence the equation x2 − 5x + 6 = 0
has solutions x = 2 and x = 3.
Similarly we can factor the cubic x3 − 6x2 + 11x − 6 as (x − 1)(x − 2)(x − 3), which enables us to show that the solutions of x3 − 6x2 + 11x − 6 = 0 are x = 1, x = 2 or x = 3. In this module we will see how to arrive at this factorisation.
Polynomials in many respects behave like whole numbers or the integers. We can add, subtract and multiply two or more polynomials together to obtain another polynomial. Just as we can divide one whole number by another, producing a quotient and remainder, we can divide one polynomial by another and obtain a quotient and remainder, which are also polynomials.
A quadratic equation of the form ax2 + bx + c has either 0, 1 or 2 solutions, depending on whether the discriminant is negative, zero or positive. The number of solutions of the this equation assisted us in drawing the graph of the quadratic function y = ax2 + bx + c. Similarly, information about the roots of a polynomial equation enables us to give a rough sketch of the corresponding polynomial function.
As well as being intrinsically interesting objects, polynomials have important applications in the real world. One such application to error-correcting codes is discussed in the Appendix to this module.
Answer:
cm
Step-by-step explanation:
The volume of the box is:
V = height * length * width
V = x*(66 - 2*x)*(90 - 2*x)
V = (66*x - 2*x^2)*(90 - 2*x)
V = 5940*x - 132*x^2 - 180*x^2 + 4*x^3
V = 4*x^3 - 312*x^2 + 5940*x
where x is the length of the sides of the squares, in cm.
The mathematical problem is :
Maximize: V = 4*x^3 - 312*x^2 + 5940*x
subject to:
x > 0
2*x < 66 <=> x < 33
In the maximum, the first derivative of V, dV/dx, is equal to zero
dV/dx = 12*x^2 - 624*x + 5940
From quadratic formula
But 
, then is not the correct answer.
56,000=5.6x10^4
Hope this helps :)
