During a school event, Noah is selling pizza at the food stand for $1.25 per slice.

Mathematics

1 answer:

Eddi Din [679]3 years ago

7 0

Answer:

1. When he sold 20 slices of pizza at $1.25 each, he made $20

When he sold 50 slices of pizza at $1.25 each, he made $62.5

2. 96 slices of pizza. You would get that by dividing 120 by 1.25, as he needs to make $120, and each slice costs $1.25

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Evaluate c (y + 7 sin(x)) dx + (z2 + 9 cos(y)) dy + x3 dz where c is the curve r(t) = sin(t), cos(t), sin(2t) , 0 ≤ t ≤ 2π. (hin

saw5 [17]

Treat $\mathcal C$ as the boundary of the region $\mathcal S$ , where $\mathcal S$ is the part of the surface $z=2xy$ bounded by $\mathcal C$ . We write

$\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r$

with $\mathbf f=(y+7\sin x,z^2+9\cos y,x^3)$ .

By Stoke's theorem, the line integral is equivalent to the surface integral over $\mathcal S$ of the curl of $\mathbf f$ . We have

$\nabla\times\mathbf f=(-2z,-3x^2,-1)$

so the line integral is equivalent to

$\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$
$=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

where $\mathbf s(u,v)$ is a vector-valued function that parameterizes $\mathcal S$ . In this case, we can take

$\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)$

with $0\le u\le1$ and $0\le v\le2\pi$ . Then

$\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$

and the integral becomes

$\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$
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3 years ago

PLEASE HELP ME WITH THIS QUESTION ITS URGENT!!!

Alex Ar [27]

Start by distributing the exponent to each of the terms in $(2u)^{3}$ . This will become $2^{3} u^{3}$ , and $2^{3} =8$ . Now, the expression is:

$\frac{2u^{3} v^{2} }{8u^{3} * u^{2} }$ .

We can now simplify the bottom to read: $8u^{5}$ because when multiplying variables raised to an exponent, we add the exponents. The expression now looks like:

$\frac{2u^{3} v^{2} }{8u^{5} }$

The 2/8 simplifies to 1/4:

$\frac{u^{3} v^{2} }{4u^{5} }$

Now, we have two u terms on the top and bottom. When dividing variables raised to an exponent, we subtract the exponents. However, since 3-5=-2, the term will be on the bottom to avoid the negative exponent. The final answer is:

$\frac{v^{2} }{4u^{2} }$