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Anit [1.1K]
3 years ago
12

Clare is using little wooden cubes with edge length 1/2 inch to build a larger cube that has edge length 4 inches. How many litt

le cubes does she need?
Mathematics
2 answers:
Lorico [155]3 years ago
7 0

Answer: Clare would need 8 cubes.

               It's quite simple to get the answer for this, simply divide the edge lengths together

\frac{4}{1/2}

=\frac{4}{0.5}

= 8.

abruzzese [7]3 years ago
3 0

Answer:

Clare needs 8 little cubes.

Step-by-step explanation:

Edge length of larger cube = 4 inches

Edge length of little cubes = 1/2 inch

Number of little cubes needed = edge length of larger cube ÷ edge length of little cube = 4 ÷ 1/2 = 4 × 2/1 = 8

She needs 8 little cubes.

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1) Solve the following absolute value equation: 3x + 8) = 9
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Answer:

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Evaluate the integral below, where C is the curver(t) = ‹sin(t),cos(t), sin(2t)›, 0 ≤ t ≤2π. (Hint: Observe that C lies on the s
nordsb [41]

We can compute the integral directly: we have

\begin{cases}x(t)=\sin t\\y(t)=\cos t\\z(t)=\sin(2t)\end{cases}\implies\begin{cases}\mathrm dx=\cos t\,\mathrm dt\\\mathrm dy=-\sin t\,\mathrm dt\\\mathrm dz=2\cos(2t)\,\mathrm dt\end{cases}

Then the integral is

\displaystyle\int_C(y+9\sin x)\,\mathrm dx+(z^2+4\cos y)\,\mathrm dy+x^3\,\mathrm dz

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You could also take advantage of Stokes' theorem, which says the line integral of a vector field \vec F along a closed curve C is equal to the surface integral of the curl of \vec F over any surface S that has C as its boundary.

In this case, the underlying field is

\vec F(x,y,z)=\langle y+9\sin x,z^2+4\cos y,x^3\rangle

which has curl

\mathrm{curl}\vec F(x,y,z)=-\langle2z,3x^2,1\rangle

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\vec s(u,v)=\langle u\cos v,u\sin v,2u^2\cos v\sin v\rangle=\langle u\cos v,u\sin v,u^2\sin(2v)\rangle

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

Note that when viewed from above, C has negative orientation (a particle traveling on this path moves in a clockwise direction). Take the normal vector to S to be pointing downward, given by

\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=\langle2u^2\sin v,2u^2\cos v,-u\rangle

Then the integral is

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S\mathrm{curl}\vec F(x,y,z)\cdot\mathrm d\vec S

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Both integrals are kind of tedious to compute, but personally I prefer the latter method. Either way, you end up with a value of \boxed\pi.

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