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Lera25 [3.4K]
3 years ago
5

If the vertices of triangle bcd are B(-3,3) and c(3,5) and d(-1,0). what is the perimeter?

Mathematics
2 answers:
Klio2033 [76]3 years ago
8 0
D ( BD ) = √( 2² + 3² ) = √(4+9) = √13 ≈ 3.61
d ( DC ) = √( 4² + 5² ) = √(16+25) = √41 ≈6.40
d ( CB ) = √( 6² + 2² ) = √(36+4) = √40 ≈ 6.32
The perimeter:  3.61 + 6.40 + 6.32 = 16.33
Answer:
C ) 16.33   
Olin [163]3 years ago
4 0

Sorry i hit one star instead of five on helpful and now it wont let me change it! However, you are correct...the answer was C. Just took test. Thanks!

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This is for algebra 2 can someone tell me if I’m right and can someone explain how do you get a= ?
algol13

Answers:

a = 2

h = 3

k = -3

You have the correct h and k values. Nice work so far.

==========================================================

Explanation:

(h,k) is the vertex. So (h,k) = (3,-3) meaning h = 3 and k = -3

To get the value of 'a', we need to plug in some point on the red graph. Each point is of the form (x,y). We can pick any point we want that isn't the vertex.

Let's pick (0,3) which is the y intercept. I'm picking this because 0 is an easy number to work with

------------

y = a|x-h| + k

y = a|x-3| + (-3) .... plug in the h,k values

y = a|x-3| - 3

3 = a|0-3| - 3 .... plug in (x,y) = (0,3)

3 = a|-3| - 3

3 = a*3 - 3

3 = 3a - 3

3a-3 = 3

3a = 3+3 ... adding 3 to both sides

3a = 6

a = 6/3 ... divide both sides by 3

a = 2

The value a = 2 indicates that the parent function y = |x| has been stretched vertically by a factor of 2. So it is twice as tall as before. Then it has been shifted to place the vertex at (3,-3) as shown in the graph.

------------

You may be wondering why you can't pick on the vertex for (x,y)

Let's see what happens if we use (x,y) = (3,-3)

y = a|x-3| - 3

-3 = a|3-3| - 3 ... plug in (x,y) = (3,-3)

-3 = a|0| - 3 ... uh oh, we get 0 here

-3 = a*0 - 3

-3 = 0 - 3

-3 = -3

We get a true statement, which is nice, but it doesn't tell us anything about what the value of 'a' is. That 'a' term goes away entirely. So I avoided using x = 3 to prevent x-3 from being 0.

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