Whether dividing constant terms or polynomials, we always have definitive terms when it comes to division. Suppose we say, 10x divided by 2. The dividend is the 10x and the divisor is the 2. In other words, the dividend is the number to be divided by the divisor, to obtain the answer called the quotient.
When dividing polynomials, your main goal is to be able to divide the dividend evenly into the <em>divisor</em>. For example, we divide x²+2x+1 by x+1. The first thing you're going to focus is, what term will completely divide the first term of the polynomial? That would be x. Why? Because when you multiply x with x+1, the product is x²+x. When you subtract this from the polynomial, the x² will cancel out. All you have to do is subtract x from 2x, yielding x. Then, you carry down the last term of the equation: +1. You do the steps again. The term that will completely divide x+1 by x+1 is 1. When you subtract the two, you will come up with zero. That means there is no remainder. The polynomial is divisible by the divisor.
x + 1
------------------------------------
x+1| x²+2x+1
- x²+x
----------------------
x +1
- x +
------------
0
Answer:
You have to multiply the denorminator to both sides in order to make x the subject :
He should start cooking at 15:00
Answer:
An ellipse and a rectangle.
Step-by-step explanation:
If Jamal cuts the right circular cylinder anywhere but its extremities, the resulting shapes on both pieces will be an ellipse.
If he cuts precisely in a perpendicular way in relation to the ends, he will then form two new right circular cylinders, then the ellipses obtained would be circles.
If Jamal cuts the right circular cylinder lengthwise, going from one end to the other, even if it's not perpendicular to the base, he will obtain a rectangular shape.
Answer:
i) (0, 2) and (1, 2), ii) (0.333, 1.333) and (1, 2).
Step-by-step explanation:
i) Let be 
, if 
, which is equivalent to the following system of equations:
Now, this system is now represented by means of a graphing tool and whose outcome is attached below. There are two solutions: (0, 2) and (1, 2)
ii) Let be , if 
, which is equivalent to the following system of equations:
Now, this system is now represented by means of a graphing tool and whose outcome is attached below. There are two solutions: (0.333, 1.333) and (1, 2)
