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Jlenok [28]
3 years ago
5

5 more than a number x is -4​

Mathematics
1 answer:
RSB [31]3 years ago
3 0

Answer:

x = -9

Step-by-step explanation:

x = unknown

x+5 = -4

reverse it

-5 + -4 = -9

this will work from the fact that if x is equal to 5 more than -4 you would reverse it so that it is negative 4 - 5 which would be -9

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Describe the process for solving a translation problem.
mojhsa [17]
Go to g o o g l e
translate is the answer
6 0
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
−125+|−136|+(−323) plzzzz need it asap
OLEGan [10]

Answer:

-312


Step-by-step explanation:


3 0
3 years ago
Read 2 more answers
E=6c^2−2c−1<br><br> F=−4c^2+7c+5<br> ​<br> E+F=
IRINA_888 [86]

The answer of given equation is 2c^{2} + 5c +4

Step-by-step explanation:

By putting the given values of E and F we get,

E + F

=( 6c^{2}-2c-1) + (-4c^{2}+7c+5)

Here we will take the coefficient of same variable together

=(6c^{2}-4c^{2}) +( -2c +7c) +( -1+5)

=2c^{2} + 5c +4

Hope this helps you.

4 0
3 years ago
PLEASE HELP I WILL GIVE BRAINLIEST IF TWO PPL ANSWER!! :C
olga55 [171]

Answer:

x+120 , if x < -120

x+12 , if x > -12

x+12 , if x < -12

x+12 , if x = -12

x÷3 , if x > 0

x÷3 , if x < 0

Step-by-step explanation:

Well whatever is in the absolute value sign is always positive.

5 0
3 years ago
Read 2 more answers
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