If
denotes the sum of the first
terms of a geometric series with first term
and common ratio
, then
Using summation notation, you have
In this case, you have
,
, and
. So the value of the sum is
Rounded to the nearest whole number, the answer would be 3.
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.
He didn't follow the order the instructions were given to him
The Right expression is:
x-3x10-13
Answer:
45
Step-by-step explanation:
AEF is a similar triangle to ABC. that means it has the same angles, and the sides (and all other lines in the triangle) are scaled from the ABC length to the AEF length by the same factor f.
now, what is f ?
we know this from the relation of AC to FA.
FA = 12 mm
AC = 12 + 28 = 40 mm
so, going from AC to FA we multiply AC by f so that
AC × f = FA
40 × f = 12
f = 12/40 = 3/10
all other sides, heights, ... if ABC translate to their smaller counterparts in AEF by that multiplication with f (= 3/10).
the area of a triangle is
baseline × height / 2
aABC = 500
and because of the similarity we don't need to calculate the side and height in absolute numbers. we can use the relative sizes by referring to the original dimensions and the scaling factor f.
baseline small = baseline large × f
height small = height large × f
we know that
baseline large × height large / 2 = 500
baseline large × height large = 1000
aAEF = baseline small × height small / 2 =
= baseline large × f × height large × f / 2 =
= baseline large × height large × f² / 2 =
= 1000 × f² / 2 = 500 × f² = 500 ×(3/10)² =
= 500 × 9/100 = 5 × 9 = 45 mm²
