8.1g of sugar is needed for every cake made. How much sugar is needed for 6 cakes?

Mathematics

1 answer:

Lerok [7]2 years ago

8 0

Answer:

48.6

Step-by-step explanation:

If you use 8.1g of sugar for 1 cake then 6 cakes will be 48.6g of sugar

Just do 8.1*6 and you will get 48.6

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Ashley recently opened a store that uses only natural ingredients. She wants to advertise her products by distributing bags of s

jasenka [17]

Answer:

3 1/2 hours

Step-by-step explanation:

This is a problem in units conversion. We want to get from bags to hours by way of minutes per bag. One bag takes an effort of 2/3 person·minute, so we need to divide the total effort by the number of persons and convert minutes to hours.

(1575 bags)×(2/3 person·min/bag)/(5 person)/(60 min/h)

= (1575)(2/3)(1/5)(1/60) h = 3.5

It will take the 5 of them about 3 1/2 hours to prepare 1575 bags.

6 0

1 year ago

Read 2 more answers

Use Stokes' Theorem to evaluate S curl F · dS. F(x, y, z) = x2 sin(z)i + y2j + xyk, S is the part of the paraboloid z = 9 − x2 −

Korolek [52]

The vector field

$\vec F(x,y,z)=x^2\sin z\,\vec\imath+y^2\,\vec\jmath+xy\,\vec k$

has curl

$\nabla\times\vec F(x,y,z)=x\,\vec\imath+(x^2\cos z-y)\,\vec\jmath$

Parameterize $S$ by

$\vec s(u,v)=x(u,v)\,\vec\imath+y(u,v)\,\vec\jmath+z(u,v)\,\vec k$

where

$\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=(9-u^2)\end{cases}$

with $0\le u\le3$ and $0\le v\le2\pi$ .

Take the normal vector to $S$ to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k$

Then by Stokes' theorem we have

$\displaystyle\int_{\partial S}\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^3(\nabla\times\vec F)(\vec s(u,v))\cdot\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv$