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sp2606 [1]
3 years ago
14

a chef uses 4 3/4 cups of broth for 10 servings of soup.how much broth is is one serving . let x represent amount. 4 3/4/10 =x/1

Mathematics
1 answer:
enyata [817]3 years ago
6 0
Let's firstly convert the mixed fraction to "improper" and then proceed,

\bf \begin{array}{ccll}
\stackrel{broth}{cups}&servings\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
4\frac{3}{4}&10\\\\
x&1
\end{array}\implies \cfrac{\quad \stackrel{mixed}{4\frac{3}{4}}\quad }{x}=\cfrac{10}{1}\implies \cfrac{\quad \frac{4\cdot 4+3}{4}\quad }{x}=\cfrac{10}{1}

\bf \cfrac{\quad \stackrel{improper}{\frac{19}{4}}\quad }{x}=\cfrac{10}{1}\implies \cfrac{\quad \frac{19}{4}\quad }{\frac{x}{1}}=\cfrac{10}{1}\implies \cfrac{19}{4}\cdot \cfrac{1}{x}=\cfrac{10}{1}\implies \cfrac{19}{4x}=\cfrac{10}{1}
\\\\\\
19=40x\implies \cfrac{19}{40}=x
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The question is incomplete. The complete question is :

The volume of a right circular cone with radius r and height h is V = pir^2h/3. a. Approximate the change in the volume of the cone when the radius changes from r = 5.9 to r = 6.8 and the height changes from h = 4.00 to h = 3.96.

b. Approximate the change in the volume of the cone when the radius changes from r = 6.47 to r = 6.45 and the height changes from h = 10.0 to h = 9.92.

a. The approximate change in volume is dV = _______. (Type an integer or decimal rounded to two decimal places as needed.)

b. The approximate change in volume is dV = ___________ (Type an integer or decimal rounded to two decimal places as needed.)

Solution :

Given :

The volume of the right circular cone with a radius r and height h is

$V=\frac{1}{3} \pi r^2 h$

$dV = d\left(\frac{1}{3} \pi r^2 h\right)$

$dV = \frac{1}{3} \pi h \times d(r^2)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

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So, r = 5.9  and dr = 6.8 - 5.9 = 0.9

     h = 4  and dh = 3.96 - 4 = -0.04

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (5.9)(4)(0.9)+\frac{1}{3} \pi (5.9)^2 (-0.04)$

$dV=44.484951 - 1.458117$

$dV=43.03$

Therefore, the approximate change in volume is dV = 43.03 cubic units.

b).  The radius is changed from r = 6.47 to r = 6.45 and the height is changed from h = 10 to h = 9.92

So, r = 6.47  and dr = 6.45 - 6.47 = -0.02

     h = 10  and dh = 9.92 - 10 = -0.08

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (6.47)(10)(-0.02)+\frac{1}{3} \pi (6.47)^2 (-0.08)$

$dV=-2.710147-3.506930$

$dV= -6.22$

Hence, the approximate change in volume is dV = -6.22 cubic units

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