I think the correct answer is B. It is the triangle case SSA that may have one, two, or zero solutions. This case can have either number of solutions but it depends on the sides of the triangle given. Having one solution can be all of the cases except SSS, having 2 solutions can only be applied to SSA.
Answer:
Step-by-step explanation:
The given expression is 26.48. We need to write the equivalent expression for it. There are two numbers after decimal. It means that we can divide 2648 by 100 i.e.
We know that, 2648 = 662 × 4 and 100 = 25 × 4
Hence, is the equivalent expression.
It's difficult to make out what the force and displacement vectors are supposed to be, so I'll generalize.
Let <em>θ</em> be the angle between the force vector <em>F</em> and the displacement vector <em>r</em>. The work <em>W</em> done by <em>F</em> in the direction of <em>r</em> is
<em>W</em> = <em>F</em> • <em>r</em> cos(<em>θ</em>)
The cosine of the angle between the vectors can be obtained from the dot product identity,
<em>a</em> • <em>b</em> = ||<em>a</em>|| ||<em>b</em>|| cos(<em>θ</em>) ==> cos(<em>θ</em>) = (<em>a</em> • <em>b</em>) / (||<em>a</em>|| ||<em>b</em>||)
so that
<em>W</em> = (<em>F</em> • <em>r</em>)² / (||<em>F</em>|| ||<em>r</em>||)
For instance, if <em>F</em> = 3<em>i</em> + <em>j</em> + <em>k</em> and <em>r</em> = 7<em>i</em> - 7<em>j</em> - <em>k</em> (which is my closest guess to the given vectors' components), then the work done by <em>F</em> along <em>r</em> is
<em>W</em> = ((3<em>i</em> + <em>j</em> + <em>k</em>) • (7<em>i</em> - 7<em>j</em> - <em>k</em>))² / (√(3² + 1² + 1²) √(7² + (-7)² + (-1)²))
==> <em>W</em> ≈ 5.12 J
(assuming <em>F</em> and <em>r</em> are measured in Newtons (N) and meters (m), respectively).
Answer:
Angle Addition Postulate
Step-by-step explanation:
Given:
Prove:
Proof:
1. 
- Angle Addition Postulate
2. 
- given
3. 
- given
4. 
- Substitution Property of Equality
5. - Simplify
Answer:
1000, 2000, 3000, .... etc are the 4-digit numbers which when divided by 1000, give whole number answer.
Step-by-step explanation:
Scott divides some 4- digit numbers by 1000 and his answer is a decimal every time.
From this Scott has concluded that if you divide a 4-digit number by 1000, then the answer will never be a whole number.
I think Scott is wrong and the reason is that if we try to divide those 4-digit numbers which are multiples of 1000, i.e. 1000, 2000, 3000, 4000, ... etc. with 1000 then only you will get a whole number answer. (Answer)
