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2 years ago

Please help with this problem

Mathematics

2 answers:

HACTEHA [7]2 years ago

7 0

Answer:

x = 2

Step-by-step explanation:

Multiply both sides by (x-5) and expand:

(x+4)/(x-5)*(x-5) = -2(x-5)

x+4 = -2x+10

Subtract 4 from both sides:

x+4-4 = -2x+10-4

x = -2x+6

Add 2x to both sides:

x+2x = -2x+6+2x

3x = 6

Divide both sides by 3:

3x/3 = 6/3

x = 2

Fiesta28 [93]2 years ago

5 0

X=2

Your welcome.

(btw the comment i made was wrong)

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3 years ago

3x-2=2(x-5) find the value of x

kipiarov [429]

Now we have to,

find the required value of x.

Let's begin,

→ 3x-2 = 2(x-5)

→ 3x-2 = 2x-10

→ 3x-2x = -10+2

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3 years ago

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Romashka-Z-Leto [24]

Answer:

6/7

Step-by-step explanation:

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3 years ago

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AnnyKZ [126]

Answer:

$(\frac{(1+e^{2x}) ^{\frac{5}{2} } }{{5}} + \frac{(1+e^{2x} )^{\frac{3}{2} } }{{3}} )+C$

Step-by-step explanation:

Step(i):-

Given that the function

f(x) = $e^{4x} \sqrt{1+e^{2x} }$

Now integrating on both sides, we get

$\int\limits{f(x)} \, dx = \int\limits{e^{4x} \sqrt{1+e^{2x} } dx$

= $\int\limits{e^{2x} e^{2x} \sqrt{1+e^{2x} } dx$

Step(ii):-

Let $1 + e^{2x} = t$

$e^{2x} = t -1$

$2e^{2x}dx = d t$

$e^{2x}dx = \frac{1}{2} d t$

= $\int\limits{( \sqrt{1+e^{2x} }) e^{2x} e^{2x} dx$

= $\int\limits {\sqrt{t}(t-1)\frac{1}{2} dt }$

= $\frac{1}{2} \int\limits {\sqrt{t} (t) -\sqrt{t} ) dt }$

= $\frac{1}{2} \int\limits {(t^{\frac{1}{2} } t^{1} +t^{\frac{1}{2} } ) } \, dx$

$= \frac{1}{2} \int\limits {(t^{\frac{3}{2} } +t^{\frac{1}{2} } ) } \, dx$

= $\frac{1}{2} (\frac{t^{\frac{3}{2} +1} }{\frac{3}{2}+1 } + \frac{t^{\frac{1}{2} +1} }{\frac{1}{2}+1 } )+C$

= $\frac{1}{2} (\frac{t^{\frac{3}{2} +1} }{\frac{5}{2} } + \frac{t^{\frac{1}{2} +1} }{\frac{3}{2} } )+C$

= $\frac{1}{2} (\frac{t^{\frac{5}{2} } }{\frac{5}{2} } + \frac{t^{\frac{3}{2} } }{\frac{3}{2} } )+C$

= $(\frac{(1+e^{2x}) ^{\frac{5}{2} } }{{5}} + \frac{(1+e^{2x} )^{\frac{3}{2} } }{{3}} )+C$

Final answer:-

= $(\frac{(1+e^{2x}) ^{\frac{5}{2} } }{{5}} + \frac{(1+e^{2x} )^{\frac{3}{2} } }{{3}} )+C$

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3 years ago

Word phrase for y over 5

Gennadij [26K]

Y/5 im guessing thats the answer if you need it into number form

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3 years ago