Here’s how you would work it out
Joe has eaten 2/5 of a pizza. Jane has eaten 1/3 of a pizza. How many times more pizza has joe eaten than jane?
1 answer:
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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
Answer:
(5,-5)
Step-by-step explanation:
<u>Eliminate</u><u> </u><u>y-term</u>
We can eliminate x-term but for this system of equations, eliminating y-term is faster.
<u>Substitute</u><u> </u><u>x</u><u> </u><u>=</u><u> </u><u>5</u><u> </u><u>in</u><u> </u><u>any</u><u> </u><u>given</u><u> </u><u>equations</u><u>.</u>
I will choose to substitute in the first equation.
<u>Answer</u><u> </u><u>Check</u>
Substitute both x-value and y-value in any given equations
The equation is true. (<em>It</em><em> </em><em>is</em><em> </em><em>better</em><em> </em><em>to</em><em> </em><em>check</em><em> </em><em>both</em><em> </em><em>the</em><em> </em><em>answer</em><em> </em><em>for</em><em> </em><em>both</em><em> </em>equation.)
Thus the answer is (5,-5)