The requirement is that every element in the domain must be connected to one - and one only - element in the codomain.
A classic visualization consists of two sets, filled with dots. Each dot in the domain must be the start of an arrow, pointing to a dot in the codomain.
So, the two things can't can't happen is that you don't have any arrow starting from a point in the domain, i.e. the function is not defined for that element, or that multiple arrows start from the same points.
But as long as an arrow start from each element in the domain, you have a function. It may happen that two different arrow point to the same element in the codomain - that's ok, the relation is still a function, but it's not injective; or it can happen that some points in the codomain aren't pointed by any arrow - you still have a function, except it's not surjective.
The answer is C :108! have a good night
Answer:
3) 8^10
Step-by-step explanation:
Use formula: (same base)
a^m × a^n = a^m+n
So,
8^3 × 8^7 = 8^3+7
= 8^10
For this problem, we are given the graph of an ellipse, and we need to determine its expression in the standard form.
The standard equation of an ellipse is given below:
Where (h,k) is the center of the ellipse, a is the horizontal radius and b is the vertical radius.
The center of the ellipse on our problem is (-2,2), the vertical radius is 2 and the horizontal radius is 3. We have:
In order to calculate the Foci, we need to first find the eccentricity of the ellipse, which is given by the following formula:
The coordinates of the foci are given by:
The coordinates for the foci are: (-2-sqrt(5), 2) and (-2+sqrt(5), 2).
When you multiply powers, (add) the exponents.
When you divide powers, (subtract) the exponents.
When you raise a power to a power, (multiply) the exponents.
Anything to the zero power equals (1).
When you have a negative exponent, (flip it) and make it (positive).
33) A
34) 3/(x^2)
35) 192 * x^5 * y^11