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____ [38]
3 years ago
10

Limit as x approaches 0 of (sin^2x)/x

Mathematics
1 answer:
melisa1 [442]3 years ago
5 0

Answer:

<h3>0</h3>

Step-by-step explanation:

Given the expression

\lim_{x \to \ 0} \frac{sin^2x}{x}

Substitute the value of x in the function

= \frac{sin ^2(0)}{0}\\= 0/0 (indeterminate) \\

Apply l'hospital rule

\lim_{x \to \ 0} \frac{d/dx(sin^2x)}{d/dx(x)}  \\=  \lim_{x \to \ 0} \frac{(2sinxcosx)}{1}  \\

Substitute the value of x

= 2 sin(0)cos(0)

= 2 * 0 * 1

= 0

Hence the limit of the function is 0

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Use Laplace transform methods to solve the differential equation: The initial conditions are f(0) = 0 and <img src="https://tex.
Llana [10]

Answer:

Step-by-step explanation:

\frac{d^2 y}{dt^2} +12\frac{dy}{dt} +32y = 10e^{-2t} where f(x) is replaced by y

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s^2 Y -s(0)-0 +12s Y-0+32Y = \frac{10}{s+2} \\Y(s^2+12s+32) =  \frac{10}{s+2}\\Y =  \frac{10}{(s+2)(s+4)(s+8)}

We can resolve into partial fraction to get Y = L(y)

Let this equals

\frac{A}{s+2} +\frac{B}{s+4} +\frac{C}{s+8} \\10 = A(s+4)(s+8) + B(s+2)(s+8) +Cs+4)(s+2) \\

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s=-9

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zhuklara [117]
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Can someone please help me with this?
dedylja [7]

Answer:

e = 29.0 will be the answer.

Step-by-step explanation:

Use the sine rule in the given triangle,

\frac{EF}{\text{sin}(\angle D)}=\frac{DE}{\text{sin}(\angle F)}=\frac{DF}{\text{sin}(\angle E)}

\frac{d}{\text{sin}(35)}=\frac{16}{\text{sin}(30)}=\frac{e}{\text{sin}(115)}

d=\frac{16\times \text{sin}(35)}{\text{sin}(30)}

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e=\frac{16\times \text{sin}(115)}{\text{sin}(30)}

e = 29.002

e ≈ 29.0

Therefore, e = 29.0 will be the answer.

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3 years ago
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