What is the slope of line A? A. 3 B.-3 C.-1/3 D.1/3
1 answer:
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Answer:
Part A)
1) The graph in the attached figure N 1
2) The coordinate rule is (x,y) -----> (x,y+10)
3) The translation of the function up 10 units means that the initial deposit is $60 instead of $50
Part B)
1) The graph in the attached figure N2
2) The coordinate rule is (x,y) -----> (x-10,y)
3) The translation of the function right 10 units means that the initial deposit is equal to $10
Part C)
1) In each translation, the slope is the same (m=5) are parallel lines
2) The vertical translation would be up 40 units
Step-by-step explanation:
we have
where
f(x) --> represents Jeremy's account balance
x ---> the time in years
Part A)
The translation of the function is up 10 units.
The rule of the translation is equal to
(x,y) -----> (x,y+10)
so
The new function will be
The graph in the attached figure N 1
The translation of the function up 10 units means that the initial deposit is $60 instead of $50
Part B)
The translation of the function is right 10 units.
The rule of the translation is equal to
(x,y) -----> (x-10,y)
so
we have
----> function Part A
The new function will be
The graph in the attached figure N 2
The translation of the function right 10 units means that the initial deposit is equal to $10
Part C)
1. Look at the translations, what characteristic of the graph stayed the same in each translation?
In each translation, the slope is the same
The slope m is equal to m=5
Are parallel lines
2. Look at the original graph and the graph of the translation right 10 units. What vertical translation of the graph in Part B would put the graph back to its original position?
we have
The vertical translation would be up 40 units
The rule of the translation is equal to
(x,y) -----> (x,y+40)
so
The new function will be
Answer:
Step-by-step explanation:
Given: ∠H ≅∠L and line HJ congruent to Line JL.
Since, In triangles HIJ and LKJ,
∠H ≅ ∠L
HJ ≅ JL
And, ∠HJI ≅ ∠ LJK
Therefore, By ASA postulate,
Δ HIJ ≅ Δ LKJ
This is why Option first is correct.
Note: Option third ΔHIJ≅ΔJKL is not correct,
Because, HI≠JK