Answer:
Part 1) The length of the longest side of ∆ABC is 4 units
Part 2) The ratio of the area of ∆ABC to the area of ∆DEF is
Step-by-step explanation:
Part 1) Find the length of the longest side of ∆ABC
we know that
If two figures are similar, then the ratio of its corresponding sides is proportional and this ratio is called the scale factor
The ratio of its perimeters is equal to the scale factor
Let
z ----> the scale factor
x ----> the length of the longest side of ∆ABC
y ----> the length of the longest side of ∆DEF
so
we have
substitute
solve for x
therefore
The length of the longest side of ∆ABC is 4 units
Part 2) Find the ratio of the area of ∆ABC to the area of ∆DEF
we know that
If two figures are similar, then the ratio of its areas is equal to the scale factor squared
Let
z ----> the scale factor
x ----> the area of ∆ABC
y ----> the area of ∆DEF
we have
so
therefore
The ratio of the area of ∆ABC to the area of ∆DEF is
Answer: 1) Vertex: (6, -2) Focus: (6, -7/4) Directrix: y = -9/4
2) Vertex: (-2, -1) Focus: (-7/4, -1) Directrix: x = -9/4
<u>Step-by-step explanation:</u>
Rewrite the equation in vertex format y = a(x - h)² + k or x = a(y - k)² + h by completing the square. Divide the b-value by 2 and square it - add that value to both sides of the equation.
- (h, k) is the vertex
- p is the distance from the vertex to the focus
- -p is the distance from the vertex to the directrix
1) y = x² - 12x + 34
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2) x = y² + 2y - 1
Answer:
Square
Step-by-step explanation:
It has four coordinates which means four vertices which is four sides. Can also be a rectangle or parallelogram.
.
Answer:
C) 78
Step-by-step explanation:
its technically 77.5714... but rounds to 78, hope that helps
Determine whether the relation is a function. {(−3,−6),(−2,−4),(−1,−2),(0,0),(1,2),(2,4),(3,6)}
Gennadij [26K]
Answer:
The relation is a function.
Step-by-step explanation:
In order for the relation to be a function, every input must only have one output. Basically, you can't have 2 outputs for 1 input but you can have 2 inputs for 1 output. Looking at all of the points in the relation, we see that no input has multiple outputs, so the answer is yes, the relation is a function.