Helppppp plsssss!!!Thanks!!
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Answer:
all work is shown and pictured
The markup percentage is 45.14%
Step-by-step explanation:
The given is:
- The selling price of a box of crackers is $1.75
- You mark the crackers up to $2.54
We need to find the markup percentage
The markup percentage = × 100%
∵ The selling price of a box of crackers is $1.75
∴ Old = 1.75
∵ You mark the crackers up to $2.54
∴ New = 2.54
- Substitute these values in the rule above
∵ The markup percentage = × 100%
∴ The markup percentage = × 100%
∴ The markup percentage = 0.4514 × 100%
∴ The markup percentage = 45.14%
The markup percentage is 45.14%
Learn more:
You can learn more about percentage in brainly.com/question/1834017
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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
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