Answer:
57m^2
Step-by-step explanation:
First we need to find the area of both shaded and not shaded regions. Since we are solving area for 2-D images we will use the equation Area=length(with)
Shaded Region: 11(9)=99
Unshaded Region: 8(6)=42
Now we subtract the shaded region with the unshaded region to get an area of 57 Meters. Therefore the area of the shaded region is 57m^2
Answer:
90 km, N 46° E
Step-by-step explanation:
<em>A jet flies due North for a distance of 50 km and then on a bearing of N 70° E for a further 60 km. Find the distance and bearing of the jet from its starting point.</em>
Look at the diagram I drew of this scenario. You can see the jet flies North for 50 km, and then turns at a 70° angle to fly another 60 km. We want to find the distance from the starting point, SP, to angle C (labeled).
This will be the jet's distance from its starting point.
In order to find the bearing of the jet from its starting point, we will need to find the angle formed between distances b and c, labeled angle A.
The <u>Law of Cosines</u> will allow us to use two known sides and one known angle to solve for the sides opposite of the known angle.
In this case, the known angle is 110° (angle B) so we will use the <u>Law of Cosines</u> respective to B.
Substitute the known values into the equation and solve for b, the distance from the starting point (A) to the endpoint (C).
- b² = (60)² + (50)² - 2(60)(50) cos(110°)
- b² = 6100 -(-2052.12086)
- b² = 8152.12086
- b = 90.28909602
- b ≈ 90 km
The distance of the jet from its starting point is 90 km. Now we can use this b value in order to calculate angle A, the bearing of the jet.
The <u>Law of Cosines</u> with respect to A:
Substitute the known values into the equation and solve for A, the bearing from the starting point (clockwise of North).
- (60)² = (90.28909602)² + (50)² - 2(90.28909602)(50) cosA
- 3600 = 8152.12086 - 6528.909602 cosA
- -4552.12086 = -6528.909602 cosA
- 0.6972252853 = cosA
- A = cos⁻¹(0.6972252853)
- A = 45.79519
- A ≈ 46°
The bearing of the jet from its starting point is N 46° E. This means that it is facing northeast at an angle of 46° clockwise from the North.
Each number in the sum is even, so we can remove a factor of 2.
2 + 4 + 6 + 8 + ... + 78 + 80 = 2 (1 + 2 + 3 + 4 + ... + 39 + 40)
Use whatever technique you used in (a) and (b) to compute the sum
1 + 2 + 3 + 4 + ... + 39 + 40
With Gauss's method, for instance, we have
S = 1 + 2 + 3 + ... + 38 + 39 + 40
S = 40 + 39 + 38 + ... + 3 + 2 + 1
2S = (1 + 40) + (2 + 39) + ... + (39 + 2) + (40 + 1) = 40×41
S = 20×21 = 420
Then the sum you want is 2×420 = 840.
27-2b > -6
-2b > -6-27
-2b > -33
b < -33/-2 (you have to change the sign when you divide with a negative number)
b < 16.5
(all the signs are supposed to have a line under them to make them greater or equal or less than or equal to)
Since opposite exterior angles are congruent I would plug in numbers for X till you find a number that goes into both and gets you the same product.