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Fudgin [204]
2 years ago
7

Solve the equation : 4x-(-2)=18

Mathematics
2 answers:
djyliett [7]2 years ago
7 0

\large\boxed{ \sf{Answer}}

4x - ( - 2) = 18 \\ 4x + 2 = 18 \\ 4x = 18 - 2 \\ 4x = 16 \\ x =  \frac{16}{4}  \\ x = 4

ʰᵒᵖᵉ ⁱᵗ ʰᵉˡᵖˢ

꧁❣ ʀᴀɪɴʙᴏᴡˢᵃˡᵗ2²2² ࿐

Usimov [2.4K]2 years ago
4 0

Step-by-step explanation:

4x-1-2=18

4x+2=18

4x=18-2

4x=16

x=16/4

x=4

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how to integrate <img src="https://tex.z-dn.net/?f=e%5E%7B2s%7D%20%2ACos%20%5Cfrac%7Bs%7D%7B4%7D" id="TexFormula1" title="e^{2s}
icang [17]

Answer:

\int\limits {e^{2s} cos\frac{s}{4} ds    =\frac{4 e^{2s} }{65 } ({8 cos (\frac{1}{4} ) s +  sin \frac{1}{4}  s} ))

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that  f(s) =  e^{2s} cos\frac{s}{4}

Now integrating

            \int\limits {f(s)} \, ds =  \int\limits {e^{2s} cos\frac{s}{4} ds

By using integration formula

   \int\limits { e^{ax} cos b x dx = \frac{e^{ax} }{a^{2}+b^{2}  } ( a cos b x + b sin b x )

<u><em>Step(ii):-</em></u>

 \int\limits {e^{2s} cos\frac{s}{4} ds    =   \frac{e^{2s} }{(2)^{2}+(\frac{1}{4}) ^{2}  } ( 2 cos (\frac{1}{4} ) s + \frac{1}{4}  sin \frac{1}{4}  s ))  

                    = \frac{e^{2s} }{(4+\frac{1}{16})} ( 2 cos (\frac{1}{4} ) s + \frac{1}{4}  sin \frac{1}{4}  s ))

                   = \frac{e^{2s} }{(\frac{65}{16} } ( \frac{8 cos (\frac{1}{4} ) s +  sin \frac{1}{4}  s}{4}  ))

                 = 16 X\frac{e^{2s} }{65 } ( \frac{8 cos (\frac{1}{4} ) s +  sin \frac{1}{4}  s}{4}  ))

                 =\frac{4 e^{2s} }{65 } ({8 cos (\frac{1}{4} ) s +  sin \frac{1}{4}  s} ))

<u><em>Final answer:-</em></u>

\int\limits {e^{2s} cos\frac{s}{4} ds    =\frac{4 e^{2s} }{65 } ({8 cos (\frac{1}{4} ) s +  sin \frac{1}{4}  s} ))

6 0
3 years ago
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