Answer:
The value is 
Step-by-step explanation:
The diagram illustrating the question is shown on the first uploaded image
From the question we are told that
The distance from city A to B is AB = 467.3 miles
The bearing from B to C is 
The bearing from B to A is 
The bearing from A to B is 
The bearing from A to C is 
Generally from the diagram
=> 
Also
and
=> 
=> 
Generally according to Sine Rule
=> 
So
=> 
Also
Generally the additional flyer miles that Adam will receive if he takes the connecting flight rather than the direct flight is mathematically represented as
![k = [CA +BC] - AB](https://tex.z-dn.net/?f=k%20%3D%20%5BCA%20%2BBC%5D%20%20-%20AB%20)
=> ![k = [260 .1 +316.8]- 467.3](https://tex.z-dn.net/?f=%20k%20%3D%20%5B260%20.1%20%2B316.8%5D-%20467.3%20)
=>
Cº b<span>. </span>Points<span> on the </span>x<span>-axis ( </span>Y. 0)-7<span> (6 </span>2C<span>) are mapped to </span>points<span>. --IN- on the </span>y<span>-axis. ... </span>Describe<span> the transformation: 'Reflect A ALT if A(-5,-1), L(-</span>3,-2), T(-3,2<span>) by the </span>rule<span> (</span>x<span>, </span>y) → (x<span> + </span>3<span>, </span>y<span> + </span>2<span>), then reflect over the </span>y-axis, (x,-1) → (−x,−y<span>). A </span>C-2. L (<span>0.0 tº CD + ... </span>translation<span> of (</span>x,y) → (x–4,y-3)? and moves from (3,-6) to (6,3<span>), by how.</span>
Based on the elements and charges in Copper (II) Oxalate, CuC₂O₄(s), the solubility in pure water is 1.7 x 10⁻⁴ M.
<h3>What is the solubility of Copper (II) Oxalate in pure water?</h3>
The solubility equilibrium (Ksp) is 2.9 x 10⁻⁸ so the solubility can be found as:
Ksp = [Cu²⁺] [C₂O₄²⁻]
Solving gives:
2.9 x 10⁻⁸ = S x S
S² = 2.9 x 10⁻⁸
S = 1.7 x 10⁻⁴ M
Find out more on solubility at brainly.com/question/23659342.
#SPJ1
12 is 20% of 60 because 12/60 = 0.2.
Answer:
6.338
this is the same because thousandths means 3 decimal places and it already has only 3 places, so it is already rounded to the thousandths. Rounded to hundredths place would be 6.34 and tenths would be 6.3.
