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3 years ago

A school ends the year with 651 students. This is a 5% increase from the start of the school year.

Mathematics

2 answers:

alisha [4.7K]3 years ago

6 0

651-n/n=0.05

the amount of change/original amount=percent change as a decimal

Jet001 [13]3 years ago

5 0

ANSWER:

n = 1.05n divided by 105 times 100

OR

n = (the number of students at the END of the school year) divided by 105 times 100

WORKING OUT:

1.05n = 651 students
1n = 651/105 times 100 = ANSWER

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3 years ago

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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!

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4 years ago