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Rama09 [41]
2 years ago
13

Jimmy is making a rectangular sandbox in his backyard for his son. Jimmy will be using wood to frame the sides of the sandbox. H

e wants the length of the sandbox to be 4 feet longer than the width. Jimmy can use no more than 60 feet of wood. What inequality can show a possible width for the sandbox?
Mathematics
1 answer:
otez555 [7]2 years ago
3 0
10 x 6 = 60
6 + 4 = 10
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Twenty-four 4-inch wide square posts are evenly spaced with 5 feet between adjacent posts to enclose a square field, as shown. W
Nata [24]

Answer:

492 0/10

Step-by-step explanation:

We have that the perimeter is 4 times the length of the side, now we know that this side "l" is given by:

l = 4 in * 24 + 23 * 5 ft

l = 96 in + 115 ft

Now, we pass the inches to pes, knowing that 1 foot equals 12 inches, so

96 in * 1 ft / 12 in = 8 ft

therefore replacing it remains:

l = 8 ft + 115 ft

l = 123 ft

now the perimeter:

p = 123 * 4

p = 492

to change to a mixed number:

4920/10 = 492 0/10

4 0
2 years ago
Read 2 more answers
a model car is one tenth of the size of the real car the models measures 42cm longs what the length of length of the real car
trapecia [35]
Just multiply 42 by 10 and that gives you 420 , so the real car has a length of 420cm or 4.2 metres :)
8 0
2 years ago
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Amy has a box containing 6 white, 4 red, and 8 black marbles. She picks a marble randomly. It is red. The second time, she picks
Alika [10]
8/15 or 11/15 would b the probabiltity of those
6 0
1 year ago
An investment of $500 increases at a rate of 12.5% per year. What is the value of the investment after 30 years? Round to the ne
NemiM [27]
Here's the equation:
0.125(500) + 500
But because it is 30 years:
30(0.125(500)) + 500
3.75(500) + 500
We can make it simpler:
4.75(500)
2375
You will have $2375
7 0
2 years ago
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

=\displaystyle\int_0^{2\pi}\int_0^3u^3(2u\cos^3v\sin(u^2-9)+\cos^3v\sin v+2u\sin^3v+\cos v\sin^3v)\,\mathrm du\,\mathrm dv

which has a value of 0, since each component integral is 0:

\displaystyle\int_0^{2\pi}\cos^3v\,\mathrm dv=0

\displaystyle\int_0^{2\pi}\sin v\cos^3v\,\mathrm dv=0

\displaystyle\int_0^{2\pi}\sin^3v\,\mathrm dv=0

\displaystyle\int_0^{2\pi}\cos v\sin^3v\,\mathrm dv=0

4 0
2 years ago
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