The answer you seek is 0.75, my friend.
Answer:
Step-by-step explanation:
Answer:
see the attachment
Step-by-step explanation:
The repetitive scaling is best handled by a spreadsheet.
<u>Part A</u>
We know the scale factor is 3/2, so we can multiply the number of servings and everything else by 3/2. The scaled recipe will make 9 servings.
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<u>Part B</u>
Since 15 = 6 + 9, we could arrive at this recipe by adding the Part A recipe to the original recipe. Instead, our spreadsheet uses the suggested 15/6 multiplier.
The formula used is shown in the spreadsheet attachment. It is filled to the right and down to cover all of the recipes and ingredients.
Answer:
remember the chain rule:
h(x) = f(g(x))
h'(x) = f'(g(x))*g'(x)
or:
dh/dx = (df/dg)*(dg/dx)
we know that:
z = 4*e^x*ln(y)
where:
y = u*sin(v)
x = ln(u*cos(v))
We want to find:
dz/du
because y and x are functions of u, we can write this as:
dz/du = (dz/dx)*(dx/du) + (dz/dy)*(dy/du)
where:
(dz/dx) = 4*e^x*ln(y)
(dz/dy) = 4*e^x*(1/y)
(dx/du) = 1/(u*cos(v))*cos(v) = 1/u
(dy/du) = sin(v)
Replacing all of these we get:
dz/du = (4*e^x*ln(y))*( 1/u) + 4*e^x*(1/y)*sin(v)
= 4*e^x*( ln(y)/u + sin(v)/y)
replacing x and y we get:
dz/du = 4*e^(ln (u cos v))*( ln(u sin v)/u + sin(v)/(u*sin(v))
dz/du = 4*(u*cos(v))*(ln(u*sin(v))/u + 1/u)
Now let's do the same for dz/dv
dz/dv = (dz/dx)*(dx/dv) + (dz/dy)*(dy/dv)
where:
(dz/dx) = 4*e^x*ln(y)
(dz/dy) = 4*e^x*(1/y)
(dx/dv) = 1/(cos(v))*-sin(v) = -tan(v)
(dy/dv) = u*cos(v)
then:
dz/dv = 4*e^x*[ -ln(y)*tan(v) + u*cos(v)/y]
replacing the values of x and y we get:
dz/dv = 4*e^(ln(u*cos(v)))*[ -ln(u*sin(v))*tan(v) + u*cos(v)/(u*sin(v))]
dz/dv = 4*(u*cos(v))*[ -ln(u*sin(v))*tan(v) + 1/tan(v)]
