1 Which sequence of transformations takes AA to its image, AB ?
1 answer:
A rigid transformation preserves the dimensions of the image
The option that gives the sequence of transformation that takes ΔB to ΔB is the option B.
<u>(B) Reflection across y-axis and a translation 2 units down </u>
Reason:
Question: The sequence of transformation that takes ΔA to image ΔB
<em>From a similar question, we have, ;</em>
<em>The vertices of triangle ΔA are; (-8, 9), (-9, 4), (-3, 4)</em>
<em>The vertices of triangle ΔB are; (8, 7), (3, 2), (9, 2)</em>
<em>The options are; </em>
<em>(A) Reflection over the x-axis and 2 units translation down</em>
<em>(B) Reflection over the y-axis and 2 units translation down</em>
<em>(C) 90° rotation after a translation of 2 units down</em>
<em>(D) 90° rotation after a translation of 12units right</em>
Solution;
The length of the base of ΔA = -3 - (-9) = 6
The length of the base of ΔB = 9 - 3 = 6
Therefore, the length s of triangle A and triangle B are equal
The orientation of the base of ΔA and ΔB are both horizontal
The location of the the side that traverses four cells are both furthest from the y-axis, or the vertical
Therefore, ΔB is a reflection of ΔA across the y-axis
We also have;
The base of ΔB is 2 units lower than the base ΔA
Therefore, the correct option is option B, reflection across y-axis and a translation 2 units down
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Answer:
The Sum Of The Integers From -6 To 58 is <u>1690.</u>
Step-by-step explanation:
Given,
We have to find out the sum of integers from -6 To 58.
Firstly we will find out the total number of terms that is 'n'.
Here
Now we use the formula of A.P.
On substituting the values, we get;
So there are 65 terms in between -6 To 58.
That means we have to find the sum of 65 terms in between -6 To 58.
Now we use the formula of Sum of n_terms.
On substituting the values, we get;
Hence The Sum Of The Integers From -6 To 58 is <u>1690.</u>
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