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3 years ago

A school bought 32 new desks.Each desk cost $24.Which is the best estimate of how much the school spent on the new desks?

Mathematics

1 answer:

Nonamiya [84]3 years ago

3 0

32 new desks x 24 dollars per desk = 768$ spent on all 32 desks

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A line crosses the coordinates (-3,5) and (4,-2). What is the slope of the line?

ICE Princess25 [194]

Answer:

Slope = -1

Step-by-step explanation:

$\frac{-2-5}{4-(-3)}=\frac{-7}{7} =-1$

8 0

3 years ago

The cost of 4 notebooks and 3 pens is $14.50. The cost of 5 notebooks and 6 pens is $21.50. Part A: What is the cost of each ite

White raven [17]

Answer:

notebook: $2.5

pen: $1.5

part B, he can buy 3 pens

Step-by-step explanation:

let the cost of notebook be n, cost of pen be p

1. 4n + 3p = 14.5

2. 5n + 6p = 21.5

by 1, you can get 3p = 14.5 - 4n, put it into 2

5n + 2(14.5 - 4n) = 21.5

5n + 29 - 8n = 21.5

3n = 29 - 21.5 = 7.5

n = 2.5

put n = 2.5 into 1

4(2.5) + 3p = 14.5

3p = 4.5

p = 1.5

Waseem buy 3 notebooks for each pen, let number of pen be m, then the number of notebook will be 3m

m(2.5) + 3m(1.5) < 27

2.5m + 4.5m = 7m < 27

m < 27/7

maximum integer value for m = 3

6 0

2 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$