True
<span>Vertical angles are opposite angles with the same vertex </span>
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, ![\frac{AE}{ED}=\frac{AB}{CD}](https://tex.z-dn.net/?f=%5Cfrac%7BAE%7D%7BED%7D%3D%5Cfrac%7BAB%7D%7BCD%7D)
⇒ ![\frac{2x+10}{x+3}=\frac{10}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B2x%2B10%7D%7Bx%2B3%7D%3D%5Cfrac%7B10%7D%7B4%7D)
⇒ ![8x+40=10x+30](https://tex.z-dn.net/?f=8x%2B40%3D10x%2B30)
⇒ ![x=5](https://tex.z-dn.net/?f=x%3D5)
Hence, AE=2x+10=2(5)+10=20 units
The length of AE is 20 units.
Answer:
x = 13
Step-by-step explanation:
First start with the blank equation
![a^{2} + b^{2} = c^{2}](https://tex.z-dn.net/?f=a%5E%7B2%7D%20%2B%20b%5E%7B2%7D%20%3D%20c%5E%7B2%7D)
Plug in the legs and hypotenuse. Remember c squared is always the hypotenuse.
![5^{2} + 12^{2} = c^{2}](https://tex.z-dn.net/?f=5%5E%7B2%7D%20%2B%2012%5E%7B2%7D%20%3D%20c%5E%7B2%7D)
Then solve
25 + 144 = ![c^{2}](https://tex.z-dn.net/?f=c%5E%7B2%7D)
169 =![c^{2}](https://tex.z-dn.net/?f=c%5E%7B2%7D)
![\sqrt{169} = \sqrt{c^{2} }](https://tex.z-dn.net/?f=%5Csqrt%7B169%7D%20%3D%20%5Csqrt%7Bc%5E%7B2%7D%20%7D)
c = 13
Answer:
1/3 divided by 3= 1/9
9 divided by 1/3= 27
Step-by-step explanation: