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.
Answer:
The roots of the equation is real and repeated
Step-by-step explanation:
Here, we want to describe the nature of the roots of the given quadratic equation
To get the nature of the roots, we find the discriminant of the equation
The discriminant is;
b^2 - 4ac
In this case, b = -28 , a = 49 and c = 4
The discriminant is thus;
-28^2 - 4(49)(4)
= 784 - 784 = 0
Since the discriminant is zero, this means that the quadratic equation has real roots which are the same
26% of 50
26/100 * 50
13
50 - 13 = 37
Answer:
9/50 or 0.18
Step-by-step explanation:
Hello!
You first cross multiply
4 * 7 = 28
5 * r = 5 * r
r * 5 = 28
Divide both sides by 5
r = 28/5 = 5.6
The answer is 28/5 or 5.6
Hope this helps!
