It seems like you want to find the sum of 38 to 115:
If we notice, this is arithmetic series or the sum of arithmetic sequences.
To find the sum of the sequences, there are three types of formulas but I will demonstrate only one and the best for this problem.
This formula only applies to the sequences that have the common difference = 1.
Given that a1 = first term of sequence/series, n = number of terms and a_n = last term
We know the first term which is 38 and the last term is 115. The problem here is the number of sequences.
To find the n, you can use the following formula.
Substitute an = 115 and a1 = 38 in the formula of finding n.
Therefore the number of sequences is 78.
Then we substitute an = 115, a1 = 38 and n = 78 in the sum formula.
Hence, the sum is 5967.
The coefficient would be 1
before every variable, there is a one
So, x is equal to 1x
x+x is 2x
hope this helps:)
48,254 im not sure what your asking though..
The preimage is the solid triangle while the image is the dashed triangle. Recall that any transformation takes a preimage and maps it to an image. The "preimage" is the "before"; while the "image" is the "after"
Preimage ---> Image
Before ---> After
The "pre" is one way to help remember it means "before".
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The preimage has a horizontal segment length of 15 units (count from x=-9 to x=6 and you should count out 15 spaces). The corresponding horizontal piece on the dashed triangle is only 3 units long. Divide the image length over the preimage length
(image length)/(preimage length) = 3/15 = 1/5 = 0.2
Now if the figure wasn't reflected over the point (6,6), then the final answer would be either 1/5 or 0.2; however, this reflection does happen so we need to make the scale factor negative. So we go from 1/5 to -1/5, and we go from 0.2 to -0.2
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Final Answer as a fraction: -1/5
Final Answer in decimal form: -0.2
Answer:
Given: In triangle ABC and triangle DBE where DE is parallel to AC.
In ΔABC and ΔDBE
[Given]
As we know, a line that cuts across two or more parallel lines. In the given figure, the line AB is a transversal.
Line segment AB is transversal that intersects two parallel lines. [Conclusion from statement 1.]
Corresponding angles theorem: two parallel lines are cut by a transversal, then the corresponding angles are congruent.
then;
and
Reflexive property of equality states that if angles in geometric figures can be congruent to themselves.
by Reflexive property of equality:
By AAA (Angle Angle Angle) similarity postulates states that all three pairs of corresponding angles are the same then, the triangles are similar
therefore, by AAA similarity postulates theorem
Similar triangles are triangles with equal corresponding angles and proportionate side.
then, we have;
[By definition of similar triangles]
therefore, the missing statement and the reasons are
Statement Reason
3. Corresponding angles theorem
and
5. AAA similarity postulates
6. BD over BA Definition of similar triangle
