Answer:
Not clear of the question
Step-by-step explanation:
The numbers are: 36 and 11 .
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Explanation:
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Let us represent the TWO (2) numbers with the variables;
"x" and "y" .
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x + y = 47 .
y − x = 25.
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Since: " y − x = 25 " ;
Solve for "y" in terms of "x" ;
y − x = 25 ;
Add "x" to each side of the equation:
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y − x + x = 25 + x ;
to get:
y = 25 + x .
Now, since:
x + y = 47 ;
Plug in "(25 + x)" as a substitution for "y"; to solve for "x" :
x + (25 + x) = 47 ;
x + 25 + x + 47 ;
2x + 25 = 47 ;
Subtract "25" from each side of the equation:
2x + 25 − 25 = 47 − 25 ;
2x = 22 ;
Divide EACH SIDE of the equation by "2" ;
to isolate "x" on one side of the equation; and to solve for "x" ;
2x / 2 = 22 / 2 ;
x = 11 ;
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x + y = 47<span> ;
</span>Plug in "11" for "x" into the equation ; to solve for "y" ;
11 + y = 47 ;
Subtract "11" from EACH SIDE of the equation;
to isolate "y" on one side of the equation; and to solve for "y" ;
11 + y − 11 = 47 − 11 ;
y = 36 .
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So: x = 11 , y = 36 ;
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Let us check our work:
y − x = 25 ;
36 − 11 =? 25 ? Yes!
x + y = 47 ;
36 + 11 =? 47 ? Yes!
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The numbers are: 36 and 11 .
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4b + 2 is your answer just to help you out if you use cymath you can easily get your answer like i did
Dont know this sorry.....
Domain is the numbers yo can use
we has a sqrt and a ln
we cant have any neagtive ln's or sqrts
x cannot be negative
it cannot be 0 either
domain is all numbers bigger than 0
deritivive
apply chain rule
dy/dx(f(g(x))=f'(g(x))g'(x)
and also
dy/dx (lnx)=1/x
and
dy/dx(f(x)g(x))=f'(x)g(x)+g'(x)f(x)
and from experience
dy/dx(sqrtx)=1/(2√x)
so
dy/dx(ln(2xsqrt(2+x)))=
(1/(2xsqrt(2+x))(2)(sqrt(2+x)+(x/(2sqrt(2+x))=
(3x+4)/(2x(x+2))
domain is all real numbers greater than 0
derivitive is
