Answer:
B)60m^2
Step-by-step explanation:
The area of the rectangle is 40m^2 and the area of the triangle is 20m^2. So 40+20 is 60.
Answer:
33°
Step-by-step explanation:
Where a transversal crosses parallel lines at right angles, all of the angles at their points of intersection are right angles.
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Angles at U are right angles, congruent to ∠UXZ. So, acute angle UXW in right triangle UXW is the complement of acute angle UWX:
∠UXW = 90° -57° = 33°
Answer:
If two lines are perpendicular and neither one is vertical, then one of the lines has a positive slope, and the other has a negative slope. Hope this helps.
Step-by-step explanation:
Answer:
482 km
63.94 degrees
Step-by-step explanation:
to solve this question we will use the cosine rule. For starters, draw your diagram. From point A, up north is 500km and 060 from there, another 300. If you join the point from the road junction back to the starting point, yoou have a triangle.
Cosine rule states that
C = 
where both A and B are the given distances, 500 and 300 respectively, C is the 3rd distance we're looking for and c is the given angle, 060
solving now, we have
C = 
C = ![\sqrt{250000 + 90000 - [215000 cos(60) }]](https://tex.z-dn.net/?f=%5Csqrt%7B250000%20%2B%2090000%20-%20%5B215000%20%20%20cos%2860%29%20%20%7D%5D)
C = ![\sqrt{340000 - [215000 * 0.5 }]](https://tex.z-dn.net/?f=%5Csqrt%7B340000%20-%20%5B215000%20%2A%200.5%20%20%7D%5D)
C = ![\sqrt{340000 - [107500 }]](https://tex.z-dn.net/?f=%5Csqrt%7B340000%20-%20%5B107500%20%20%7D%5D)
C =
C = 482 km
The bearing can be gotten by using the Sine Rule.
= 
sina/500 = sin60/482
482 sina = 500 sin60
sina = 
sina = 0.8983
a = sin^-1(0.8983)
a = 63.94 degrees
2 11/78 you just have to simply multiply 64/143 x2 because since it can't be subtracted. And after you multiply 64/143, you just have to subtract normally which is 128-21 and 286-208. Then since your answer came out to an improper fraction, so you have to simplify by dividing 107 divided by 78. Then you will get your answer which is 2 11/78.
