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Lelechka [254]
3 years ago
12

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:

A

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

B

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

=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv

=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0

\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}

8 0
3 years ago
In the diagram below segment DEF is shown(since all three letters are shown, they are collinear). If DE=3x+2, EF= x+5, and DF= 2
amm1812

The value of x is 4.

By Geometry, we know that line segment DF = DE + EF, which is equivalent to:

DF = 3\cdot x + 2 + x + 5

DF = 4\cdot x + 7 (1)

Now we solve algebraically the resulting expression.

If we know that DF = 23, then we solve the equation for x:

4\cdot x + 7 = 23

4\cdot x = 16

x = 4

The value of x is 4.

We kindly invite to see this question on line segments: brainly.com/question/23297288

6 0
2 years ago
Which of the following expression has the greatest value: -3+-4-(-2),-3-(-4)-(-2),-3+-4-2,-3-(-4)-2
AleksAgata [21]
The third one
U solve from left to right and when two negatives are next to each other it turns to a plus.
6 0
2 years ago
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