In this case, we can summarize that Sam had a bad debt.
<h3>What is a bad debt?</h3>
Bad debt is a liability that one incurs over which they become unable to meet the required financial obligations.
When the ability to make repayments is lacking, Sam is most likely to lose the mortgage (home loan) with the consequential forfeiture of his home.
Thus, we can conclude that Sam did not incur a good debt but a bad debt.
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Answer:
umm... there is no problem
Answer:
Not congruent; not congruent; congruent; congruent; not congruent; not congruent
Explanation:
In order to be congruent, the side lengths of both triangles must be equal.
The base of triangle ABC is 2 units. The base of triangle MNP is also 2 units. The height of triangle ABC is 3 units; but the base of triangle MNP is not 3 units, it is 4. The triangles are not congruent.
Similarly, both ABC and EFG have a base of 2 units. However, the height of ABC is 3 while the height of EFG is 4; the triangles are not congruent.
The bases of ABC and STU are both 2 units, and the height of each is 3 units. The triangles are congruent.
The base of EFG is 2, as is the base of MNP. Both triangles also have a height of 4; the triangles are congruent.
The base of EFG is 2 units, as is the base of STU. However, the height of EFG is 4 while the height of STU is 3; the triangles are not congruent.
The base of STU is 2 units, as is the base of MNP. However, the height of STU is 3 while the height of MNP is 4; the triangles are not congruent.
Answer:
x=1, y=6. (1, 6).
Step-by-step explanation:
-8x+4y=16
y=5x+1
----------------
-8x+4(5x+1)=16
-8x+20x+4=16
12x+4=16
12x=16-4
12x=12
x=12/12
x=1
y=5(1)+1=5+1=6
The third graph represents a function.
In a function, every input (x value) has <em>exactly</em> one output (y value). If even a single input has zero or two outputs, the graph does not represent a function.
A good way of testing this is using a vertical line. As you move a vertical line from left to right across a graph, it should always be touching exactly one point on the graphed line.
In this case, every graph fails this vertical line test except for the third graph, so the third graph represents a function.
