Answer:
Total eclipse was 
minutes longer than in 2005.
Step-by-step explanation:
It is given in the question that total eclipse in 2003 lasted 
minutes and in 2005 it lasted 3/8 of a minute.
We have to calculate how much longer did it last in 2003.
So we have to subtract the duration in 2003 by duration in 2005
Total time taken in 2003- total time taken in 2005
minutes
780, or C, is correct. Since 9.75 is 1/80th of the actual measurement, we can set up an equation like this:
9.75 = 1/80x
Then multiply each side by 80:
780 = x
The actual building's height is 780 ft.
We can check this answer by plugging in 780 for x in the original equation:
9.75 = 1/80(780)
9.75 = 9.75
Check! <span>✓</span>
The rest of the question is the attached figure
========================================
solution:
========
As show in the attached figure
∠M = ∠R = 54.4°
∠N = ∠T = 71.2°
∠O = 180° - (∠M + ∠N) = 180° - (54.4°+71.2°) = 54.4°
∠S = 180° - (∠R + ∠T) = 180° - (54.4°+71.2°) = 54.4°
∠O = ∠S = 36°
∴ Δ MNO is similar to Δ RTS
So, the correct statement:
The triangles each have two given angle measures and one unknown angle measure.
Angelica is 23
Steps
Use the equation x=2j-5
Plug in jahiems age (14) for j and solve for c
Answer:
360 mi/h
Step-by-step explanation:
The speed for the outbound trip was ...
speed = distance/time = (715 mi)/(2 1/6 h) = 330 mi/h
The inbound speed was ...
(715 mi)/(1 5/6 h) = 390 mi/h
The airspeed of the plane is the average of these two ground speeds, so is ...
(330 +390)/2 = 360 . . . . mi/h
