Answer: Angle is 3
Step-by-step explanation:
16x - 27 = 5x + 6
-5x -5x
______________
11x - 27 = 6
+ 27 +27
___________
11x = 33
__ __
11 11
x = 3
The vertex of the function f(x) exists (1, 5), the vertex of the function g(x) exists (-2, -3), and the vertex of the function f(x) exists maximum and the vertex of the function g(x) exists minimum.
<h3>How to determine the vertex for each function is a minimum or a maximum? </h3>
Given:
and

The generalized equation of a parabola in the vertex form exists

Vertex of the function f(x) exists (1, 5).
Vertex of the function g(x) exists (-2, -3).
Now, if (a > 0) then the vertex of the function exists minimum, and if (a < 0) then the vertex of the function exists maximum.
The vertex of the function f(x) exists at a maximum and the vertex of the function g(x) exists at a minimum.
To learn more about the vertex of the function refer to:
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Answer:
1. 1/8 ÷ 3/4 = 1/6
2. 3/5 ÷ 3/2 = 2/5
3. 4/1 ÷ 2/3 = 6
4.9/4 ÷ 6/5 = 15/8
5.22/4 ÷ 2/5 = 261/80
Step-by-step explanation:
Hope this Helped
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.
Answer:
Ok, we have a system of equations:
6*x + 3*y = 6*x*y
2*x + 4*y = 5*x*y
First, we want to isolate one of the variables,
As we have almost the same expression (x*y) in the right side of both equations, we can see the quotient between the two equations:
(6*x + 3*y)/(2*x + 4*y) = 6/5
now we isolate one off the variables:
6*x + 3*y = (6/5)*(2*x + 4*y) = (12/5)*x + (24/5)*y
x*(6 - 12/5) = y*(24/5 - 3)
x = y*(24/5 - 3)/(6 - 12/5) = 0.5*y
Now we can replace it in the first equation:
6*x + 3*y = 6*x*y
6*(0.5*y) + 3*y = 6*(0.5*y)*y
3*y + 3*y = 3*y^2
3*y^2 - 6*y = 0
Now we can find the solutions of that quadratic equation as:

So we have two solutions
y = 0
y = 2.
Suppose that we select the solution y = 0
Then, using one of the equations we can find the value of x:
2*x + 4*0 = 5*x*0
2*x = 0
x = 0
(0, 0) is a solution
if we select the other solution, y = 2.
2*x + 4*2 = 5*x*2
2*x + 8 = 10*x
8 = (10 - 2)*x = 8x
x = 1.
(1, 2) is other solution