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skad [1K]
2 years ago
13

F(g(4))=

ddle" class="latex-formula">
Mathematics
1 answer:
mash [69]2 years ago
5 0

Answer:

weird question kid

Step-by-step explanation:

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Which gives the correct values for points A, B, C, and D?
beks73 [17]

Answer:

The correct answer for this problem is the 2nd one because all you have to do is count how far each one is from -1 and -2

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
In quadrilateral ABCD what is m
sineoko [7]

Answer:

B.99°

Step-by-step explanation:

105°+ 66°+90°+ angle c=360°

angle c= 360° - 261

=99°

8 0
2 years ago
Nd the 52nd term and the term named in the problem <br>40, 140, 240, 340, ...<br>Find a 32​
Galina-37 [17]

Answer:

5140 and 3140

Step-by-step explanation:

Note there is a common difference d between consecutive terms in the sequence, that is

d = 140 - 40 = 240 - 140 = 340 - 240 = 100

This indicates the sequence is arithmetic with n th term

a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 40 and d = 100, thus

a_{52} = 40 + (51 × 100) = 40 + 5100 = 5140

a_{32} = 40 + (31 × 100) = 40 + 3100 = 3140

5 0
3 years ago
Continue each pattern with the next two numbers
kobusy [5.1K]
The 18th sequence is - 41
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3 years ago
Find the laplace transform of f(t) = cosh kt = (e kt + e −kt)/2
iren2701 [21]
Hello there, hope I can help!

I assume you mean L\left\{\frac{ekt+e-kt}{2}\right\}
With that, let's begin

\frac{ekt+e-kt}{2}=\frac{ekt}{2}+\frac{e}{2}-\frac{kt}{2} \ \textgreater \  L\left\{\frac{ekt}{2}-\frac{kt}{2}+\frac{e}{2}\right\}

\mathrm{Use\:the\:linearity\:property\:of\:Laplace\:Transform}
\mathrm{For\:functions\:}f\left(t\right),\:g\left(t\right)\mathrm{\:and\:constants\:}a,\:b
L\left\{a\cdot f\left(t\right)+b\cdot g\left(t\right)\right\}=a\cdot L\left\{f\left(t\right)\right\}+b\cdot L\left\{g\left(t\right)\right\}
\frac{ek}{2}L\left\{t\right\}+L\left\{\frac{e}{2}\right\}-\frac{k}{2}L\left\{t\right\}

L\left\{t\right\} \ \textgreater \  \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{t\right\}=\frac{1}{s^2} \ \textgreater \  L\left\{t\right\}=\frac{1}{s^2}

L\left\{\frac{e}{2}\right\} \ \textgreater \  \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{a\right\}=\frac{a}{s} \ \textgreater \  L\left\{\frac{e}{2}\right\}=\frac{\frac{e}{2}}{s} \ \textgreater \  \frac{e}{2s}

\frac{ek}{2}\cdot \frac{1}{s^2}+\frac{e}{2s}-\frac{k}{2}\cdot \frac{1}{s^2}

\frac{ek}{2}\cdot \frac{1}{s^2}  \ \textgreater \  \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \  \frac{ek\cdot \:1}{2s^2} \ \textgreater \  \mathrm{Apply\:rule}\:1\cdot \:a=a
\frac{ek}{2s^2}

\frac{k}{2}\cdot \frac{1}{s^2} \ \textgreater \  \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \  \frac{k\cdot \:1}{2s^2} \ \textgreater \  \mathrm{Apply\:rule}\:1\cdot \:a=a
\frac{k}{2s^2}

\frac{ek}{2s^2}+\frac{e}{2s}-\frac{k}{2s^2}

Hope this helps!
3 0
3 years ago
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