I’m pretty sure you can first do 10% at a time which you can round to 13 which is 3 so 3 plus 3 then add tax
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.
Directly proportional, both increase/decrease, Multiple k(constant you needed to solve for)..
Inversely --one increases as the other decreases, increase the denominator & fraction gets smaller.... So
Problem:
Dirextly to X (× by X)
Inversely to Y, (÷ by Y)
Z= k X/Y
Solve, working backwards: Z =5 becomes
5 = k (X) ÷ (Y). And
X=15 and Y =9 becomes
5= k(15)/(9). Undo ÷9, ×9 both sides
45= k(15). Now undo ×15 by ÷15 both sides
45/15 = 3. So k=3, (also called your constant of variation)....and your Equation/Proportion becomes...
Z = 3X/Y
Use the new #s.... X=23, Y=14
Z= 3×23÷14 = 69/14. If they want spirochete answer divide and round to corresponding # if decimals
Answer:
There is no diagram or any information of what you are dilating but a simple tip is to DIVIDE THE TARGET BY 2
Step-by-step explanation:
You would divide by two because you're dilating by 1/2
Answer:
B=-2+5 collecting like terms
B=3 answer