The domain are all valid values for x (the independent variable) that can be used in an equation.
We have to look at any potential values of x which won't work. Easily put: in algebra, just look for values of x which cause either division by zero, or the square root of negative numbers.
A couple of examples:
y=2x+4
You can insert any negative or positive value, or zero, for x and get a valid equation. Therefore the domain is the set of all real numbers. Answers are usually written as:
x: {R}, or simply 'all real numbers'.
what about y=2/(x-1)
In this equation, x appears in the denominator. If x-1=0, then division by zero would occur.
Solve: x-1≠0
x≠1
In set notation:
x: (-∞,1)∪(1,∞)
Parentheses are next to the 1, as the domain comes up to 1, but does not include 1.
Read left to right, the domain is "negative infinity to 1, exclusive, in union with 1 to positive infinity"
19) 5c+c=-6+2
6c=-4
c=-4/6
c=-2/3
20) -20-8=-2e
-28=-2e
28/2=e
14=e
21)p-2w=2l
(p-2w)/2=l
The given statement is:
An integer is divisible by 100 if and only if its last two digits are zeros
The two conditional statements that can be made are:
1) If an integer is divisible by 100 its last two digits are zeros.
This is a true statement. If a number is divisible by 100, it means 100 must be a factor of that number. When 100 will be multiplied by the remaining factors, the number will have last two digits zeros.
2) If the last two digits of an integer are zeros, it is divisible by 100.
This is also true. If last two digits are zeros, this means 100 is a factor of the integer. So the number will be divisible by 100.
Therefore, the two conditional statements that are formed are both true.
So, the option A is the correct answer.
Yes, it is. When the definition is separated into two conditional statements, both of the statements are true
Answer:
Step-by-step explanation:
a = 3
b = 3
c = 1
Discriminate = sqrt(b^2 - 4ac)
Discriminate = sqrt(9 - 4*3*1)
Discriminate = sqrt(9 - 12)
Discriminate = sqrt(-3)
What negative value means that there are no real roots and the y intercept = 1. You should always graph this kind of question, which I am doing for you. I use Desmos which is a very handy program.
You should notice that the minimum is above the x axis. That means there are no real roots, just as the discriminate tells you.
The line y = x + 3 has slope 1, so we look for points on the curve where the tangent line, whose slope is dy/dx, is equal to 1.
y² = x
Take the derivative of both sides with respect to x, assuming y = y(x) :
2y dy/dx = 1
dy/dx = 1/(2y)
Solve for y when dy/dx = 1 :
1 = 1/(2y)
2y = 1
y = 1/2
When y = 1/2, we have x = y² = (1/2)² = 1/4. However, for the given line, when y = 1/2, we have x = y - 3 = 1/2 - 3 = -5/2.
This means the line y = x + 3 is not a tangent to the curve y² = x. In fact, the line never even touches y² = x :
x = y² ⇒ y = y² + 3 ⇒ y² - y + 3 = 0
has no real solution for y.
