Answer:
B
27
Explanation:
Step One
The very first step is to show that triangles ΔABC, ΔFDC and ΔGEC are similar to one another.
1. All three triangles have a right angle in them.
- ΔABC has a right angle at A
- ΔFDE has a right angle at <FDE
- ΔGEC has a right angle at <GEC
2. All three triangles have a common angle at C
3. All three triangles are similar by AA
Step Two
Find the ratios of the heights to one another.
AC / DC = 3/2
AB / DF = 3/2 The sides of similar triangles are in the same ratio.
Step Three
Find the area of ΔABC
Area ΔABC = 1/2 AB * AC
Area ΔABC = 81
Step Four
Find the Area of ΔDFC
Area of ΔDFC = 1/2 DF * DC
But DF and DC are known in terms of AB and AC
Area of ADC = 1/2 * 2/3 * AB * 2/3 * AC
Area of ADC = 1/2 * 4/9 * AB * AC
However 1/2 AB * AC = 81 so
Area ADC = 4/9 * 81 = 36
That's a very long complex step. Make sure you follow it through.
Step Five
Find the area of ΔGEC
By a similar process
EC = 1/3 AC
EG = 1/3 AB
Area ΔGEC = 1/2 * EC * EG
Area ΔGEC = 1/2 * 1/3 AC * 1/3 AB
Area ΔGEC = 1/2 * 1/9 * AB * AC
But 1/2 AB * AC = 81
Area ΔGEC = 1/9 * 81 = 9
Last Step
Find the area of the shaded area
ΔFDC - ΔGEC = Shaded area
36 -9 = 27 = shaded area
The two passages have the theme "time erases everything," as shown in the first answer option.
We can arrive at this answer because:
- The passage from the poem "Ozymandias," shows how all the glory, wealth, and magnificence of a great leader fades over time.
- In this poem, we can see that someone seen as the king of kings, can have all his power terminated and buried in the sand.
- In "On Seeing the Elgin Marbles," we can see how memorable moments pass and become memories of "the old days," even though it seems like it will last forever.
In this case, we can see that the two poems reinforce the idea that everything is fleeting, and that time brings the end of all things.
The two passages the question refers to can be seen in the image below.
More information:
brainly.com/question/11255346?referrer=searchResults
The bearing and distance is from the starting position is 111° and 225.7 miles respectively.
<h3>What is the distance and bearing from the starting position?</h3>
The distance from A to B = 1.4 * 110 = 154 miles
The distance from B to C = 1.5 * 110 = 165 miles
The distance from A to C = x
Using Pythagoras theorem:
x = √(165² + 154²)
x = 225.7 miles
The bearing will be 64° + α
α = tan⁻¹(165/154) = 47°
Bearing from starting position = 64 + 47 = 111°
In conclusion, the bearing and distance is determined using Pythagoras theorem.
<em>Note that the complete question is:</em>
<em>From A to B a private plane flies 1.4 hours at 110 mph on a bearing of 64o. It turns at point B and continues another 1.5 hours at the same speed, but on a bearing of 154o to point C.</em>
<em>At the end of this time, how far is the plane from its starting point?</em>
<em>On what bearing(from due North) is the from its original location?</em>
Learn more about bearing and distance at: brainly.com/question/24142612
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Answer:
(2) Molecules of I2 are polar, and molecules of Br2 are nonpolar.
(3) Molecules of Br2 have stronger intermolecular forces than molecules of I2.
(4) Molecules of I2 have stronger intermolecular forces than molecules of Br2.
None
Explanation:
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