Answer:
The length of AE is 20 units.
Step-by-step explanation:
Given two segments AD and BC intersect at point E to form two triangles ABE and DCE. Side AB is parallel to side DC. A E is labeled 2x+10. ED is labeled x+3. AB is 10 units long and DC is 4 units long.
we have to find the length of AE
AB||CD ⇒ ∠EAB=∠EDC and ∠EBA=∠ECD
In ΔABE and ΔDCE
∠EAB=∠EDC (∵Alternate angles)
∠EBA=∠ECD (∵Alternate angles)
By AA similarity, ΔABE ≈ ΔDCE
therefore, 
⇒ 
⇒ 
⇒ 
Hence, AE=2x+10=2(5)+10=20 units
The length of AE is 20 units.
I am pertty sure its 0.6 but if i am wrong sorry
<h2>
Answer: −2.4f − 15</h2><h2>
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Honey, just simplify the expression to get your answer.</h3><h3>
−2.4f − 15</h3><h2>
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Hope you have a good day, Loves!~ <3
The sum of the 8 terms of the series 1-1-3-5- ... -13 is -48
The given sequence is:
1,-1,-3, . . -13
and there are 8 terms.
The related series of this sequence is:
1-1-3-5- ... -13
Notice that the series is an arithmetic series with:
first term, a(1) = 1
common difference, d = -1 - (1) = -2
last term, a(8) = -13
To find the sum of the series, use the sum formula:
S(n) = n/2 [(a(1) + a(n)]
Substitute n = 8, a(1) = 1, a(n) = a(8) = -13 into the formula:
S(8) = 8/2 [1 + (-13)]
S(9) = 4 . (-12) = -48
Learn more about sum of a series here:
brainly.com/question/14203928
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Hi there
-5i + 5-1 + 6
= -5i + 11- 1
= 5i + 10
That's the answer :0
I hope that's help... Do you have any question ?
