Destinyyyy
  • Destinyyyy
Help with a few problems..
Mathematics
chestercat
  • chestercat
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Destinyyyy
  • Destinyyyy
Solve by any method. (4x+3)(x+2)=3
dinamix
  • dinamix
4x^2+8x+3x+6 = 3 4x^2+11x+3=0 \[\Delta = 121-48 =73\]
Destinyyyy
  • Destinyyyy
Um your missing some numbers

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Destinyyyy
  • Destinyyyy
Final answer---> -11+- Square root 73 all over 8
Destinyyyy
  • Destinyyyy
Solve by any method. 1/x + 1/x+2 = 1/5
Destinyyyy
  • Destinyyyy
dinamix
  • dinamix
u complete it cuz i find u delta
Destinyyyy
  • Destinyyyy
What?
dinamix
  • dinamix
\[x(1) = \frac{ -b-\sqrt{\Delta} }{ 2a } , x(2) = \frac{ -b+\sqrt{\Delta} }{ 2a }\]
dinamix
  • dinamix
u have find solution
Destinyyyy
  • Destinyyyy
Um? Which problem are you talking about?
dinamix
  • dinamix
(4x+3)(x+2) = 3 @Destinyyyy
dinamix
  • dinamix
this i mean
Destinyyyy
  • Destinyyyy
Oh. I completed that question.
Destinyyyy
  • Destinyyyy
I now need help on this one- Solve by any method. 1/x + 1/x+2 = 1/5
dinamix
  • dinamix
\[\frac{ x+2+x }{ (x+2)(x) }=1/5\]
dinamix
  • dinamix
u understand that @Destinyyyy how i do it right ?
Destinyyyy
  • Destinyyyy
Thats no right... 1/x + 1/x+2 = 1/5
Destinyyyy
  • Destinyyyy
not *
dinamix
  • dinamix
hmm |dw:1440723213093:dw|
Destinyyyy
  • Destinyyyy
Umm I cant see the end part..
Destinyyyy
  • Destinyyyy
Um are you trying to say times everything by x+2 ?? Its hard to read
dinamix
  • dinamix
\[(\frac{ 1 }{ x } *\frac{ x+2 }{ x+2 }) +(\frac{ 1 }{ x+2}*\frac{ x }{ x })\]
dinamix
  • dinamix
\[\frac{ x+2 }{ x(x+2) }+\frac{ x }{ x(x+2) } = 1/5\]
dinamix
  • dinamix
\[\frac{ x+2+x }{ x(x+2) } =1/5\]
Destinyyyy
  • Destinyyyy
Sorry but im not understanding.. I thought im suppose to get rid of the fractions?
dinamix
  • dinamix
\[\frac{ 2x+2 }{ x^2+2x } =1/5\]
Destinyyyy
  • Destinyyyy
I dont get how your going from the original problem to the first image
Destinyyyy
  • Destinyyyy
Where are you getting x+2/x+2 from? And where did the parentheses come from?
dinamix
  • dinamix
u will get fuction now cuz we dont have way we must do that to solve problem
Destinyyyy
  • Destinyyyy
Huh?
Destinyyyy
  • Destinyyyy
dinamix
  • dinamix
tell to him if my mrthod is correct or no , cuz i am sure about my answer is true
Destinyyyy
  • Destinyyyy
Again huh? Can you please type in proper English so I can understand?
Destinyyyy
  • Destinyyyy
@.Gjallarhorn. @IrishBoy123
dinamix
  • dinamix
we sould have same denominator of this 1/x+(1/x+2)
dinamix
  • dinamix
should*
Destinyyyy
  • Destinyyyy
Where are you getting parenthesis from??
Destinyyyy
  • Destinyyyy
Can you show me step by step?
dinamix
  • dinamix
just parenthesis for u to understand .
Destinyyyy
  • Destinyyyy
Oh.. Well they seem to be confusing me instead
dinamix
  • dinamix
i give u example to understand okey
Destinyyyy
  • Destinyyyy
Um okay
dinamix
  • dinamix
|dw:1440724788988:dw| how collecting this numbers
dinamix
  • dinamix
Destinyyyy
  • Destinyyyy
Okay?
dinamix
  • dinamix
did u see my question or no
dinamix
  • dinamix
|dw:1440725128863:dw|
Destinyyyy
  • Destinyyyy
Yes I see it but I dont know what you want me to do or say
jim_thompson5910
  • jim_thompson5910
The original problem is this, right? \[\Large \frac{1}{x} + \frac{1}{x+2}= \frac{1}{5}\]
Destinyyyy
  • Destinyyyy
jim_thompson5910
  • jim_thompson5910
since we have an equation, we can multiply every fraction by the LCD 5x(x+2) to clear out the fractions \[\large \frac{1}{x} + \frac{1}{x+2}= \frac{1}{5}\] \[\large {\color{red}{5x(x+2)}}*\frac{1}{x} + {\color{red}{5x(x+2)}}*\frac{1}{x+2}= {\color{red}{5x(x+2)}}*\frac{1}{5}\] \[\large {\color{red}{5\cancel{x}(x+2)}}*\frac{1}{\cancel{x}} + {\color{red}{5x*\cancel{(x+2)}}}*\frac{1}{\cancel{x+2}}= {\color{red}{\cancel{5}*x(x+2)}}*\frac{1}{\cancel{5}}\] \[\large 5(x+2)*1 + 5x*1 = x(x+2)*1\] \[\large 5(x+2) + 5x = x(x+2)\] Hopefully you see how I was able to clear out the fractions?
Destinyyyy
  • Destinyyyy
Yes I thought thats what I was suppose to do but wasnt entirely sure. Thank you
jim_thompson5910
  • jim_thompson5910
are you able to solve `5(x+2)+5x=x(x+2)` from here?
Destinyyyy
  • Destinyyyy
Yes!
jim_thompson5910
  • jim_thompson5910
ok great
dinamix
  • dinamix
@jim_thompson5910 lol i think my method is easy look first
Destinyyyy
  • Destinyyyy
5x+10+5x= x^2+2x 10x+10=x^2+2x 8x+10=x^2 x^2-8x-10=0
jim_thompson5910
  • jim_thompson5910
now use the quadratic formula to solve `x^2-8x-10=0`
Destinyyyy
  • Destinyyyy
8+- square root 104 over 2
Destinyyyy
  • Destinyyyy
8+- 2 square root 26 over 2
jim_thompson5910
  • jim_thompson5910
There's one more step. You're doing great so far
Destinyyyy
  • Destinyyyy
Umm.. 8+2 square root 26/2 --> 10 square root 26 over 2 8-2 square root 26/2 --> 6 square root 26 over 2 That doesnt seem right though
jim_thompson5910
  • jim_thompson5910
|dw:1440726115004:dw|
jim_thompson5910
  • jim_thompson5910
divide each part (circled) by 2 |dw:1440726151837:dw|
Destinyyyy
  • Destinyyyy
Thats what im missing okay..
Destinyyyy
  • Destinyyyy
4+- square root 26
jim_thompson5910
  • jim_thompson5910
correct
Destinyyyy
  • Destinyyyy
Thank you.. I have two more. Do you mind helping? Both are word problems
jim_thompson5910
  • jim_thompson5910
sure, but post them one at a time (one per post) to avoid clutter or lag
Destinyyyy
  • Destinyyyy
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