Answer: The answer is <em><u>-2</u></em>
Step-by-step explanation:
that is where the line starts
hope this helped
Answer:
8 and 10
Step-by-step explanation:
the ratio of the two numbers is 4:5
So let's take 4 and 5 as our number, we find out that their LCM is 20.
The next pair would be 8 and 10. Their LCM is 40 , so they are the numbers.
Answer:
frequency= 165Hz
period= 
Step-by-step explanation:
1. To calculate the frequency using the formula:
f = v/λ
where,
f=frequency (what the question asked)
v=speed of wave (given in the question=330m/s)
λ=wavelength (given in the question=2m)
SO,
= 165Hz
2. To calculate the period use the formula:
since 
therefore, 
where,
f= frequency (what we solved above=165Hz)
T= period (what the question asked)
SO,
The answer to the problem is 13ab
(a) It looks like the ODE is
<em>y'</em> = 4<em>x</em> √(1 - <em>y </em>^2)
which is separable:
d<em>y</em>/d<em>x</em> = 4<em>x</em> √(1 - <em>y</em> ^2) => d<em>y</em>/√(1 - <em>y</em> ^2) = 4<em>x</em> d<em>x</em>
Integrate both sides. On the left, substitute <em>y</em> = sin(<em>t </em>) and d<em>y</em> = cos(<em>t</em> ) d<em>t</em> :
∫ d<em>y</em>/√(1 - <em>y</em> ^2) = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / √(1 - sin^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / √(cos^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / |cos(<em>t</em> )| d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
Since we want the substitutiong to be reversible, we implicitly assume that -<em>π</em>/2 ≤ <em>t</em> ≤ <em>π</em>/2, for which cos(<em>t</em> ) > 0, and in turn |cos(<em>t</em> )| = cos(<em>t</em> ). So the left side reduces completely and we get
∫ d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
<em>t</em> = 2<em>x</em> ^2 + <em>C</em>
arcsin(<em>y</em>) = 2<em>x</em> ^2 + <em>C</em>
<em>y</em> = sin(2<em>x</em> ^2 + <em>C </em>)
(b) There is no solution for the initial value <em>y</em> (0) = 4 because sin is bounded between -1 and 1.
