Let us find the co ordinates of each vertex of the triangle .
Vertex A ( in firs second quadrant) = ( -5 ,3)
vertex B in third quadrant = ( -5, -5)
vertex C in fourth quadrant = ( 4, -2)
let us use distance formula AB^2 = ( -5 - 3)^2 + (-5 - -5 )^2 = 64 + 0
AB= 8
BC^2 = ( -2 - -5 )^2 + ( 4 - - 5)^2 = 9 + 81 = 90
BC = 9.48
AC^2 = ( -2 -3)^2 +( 4- -5)^2 = 25 + 81 = 106
AC= 10.29
Perimeter = sum of length of AB+ BC+ Ac = 8 + 10.29 + 9.48= 22.77
Yes, (3,-1) is a solution to the graphed system of inequalities.
<h3>What are system of inequalities?</h3>
When mathematical expressions are compared, with non-strict equality, then such mathematical statements are called mathematical inequalties.
A collection of inequalities for which we consider a common solution for all inequalities is called a system of inequalities.
As we can see in the graph that the two lines are interest in the second quadrant at point such as(x, -y).
So, the point will be a solution also because it lies in the shaded region.
Therefore, Yes, (3,-1) is a solution to the graphed system of inequalities.
Learn more about graphing inequalities here:
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Answer:
a
Step-by-step explanation:
We are looking for an exponential function 
where r is less than 1.
a works because it is in the form where r is less than 1.
b is in the form but r is more than 1.
c. and d. are not even exponential functions.
c. is a polynomial
d. is a constant polynomial (no variable)
Answer;
4.215
Step-by-step explanation:
there you go
First you have to factor the equation because there are technically 2 x’s in an x^2. You will get x(x+6)=0 because x*x=x^2 and 6*x=6x. Now you can come up with possible values. If you make the first x, 0, you are multiplying everything by 0 and will get the product of 0. So, one of the and coordinates is x=0. You can also substitute -6 for the second x in the equation so that in the parentheses it would be x(-6+6)=0. You’re basically multiplying 0 by x which will guarantee you to get 0. So, the answer is (0,-6). If you have any questions feel free to comment.
