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densk [106]
2 years ago
5

There are 43 pine trees currently in the park. Park workers will plant more pine trees

Mathematics
1 answer:
Luden [163]2 years ago
8 0

Answer:

20 pine trees

Step-by-step explanation:

No of pine trees today = 63

No of pine trees there = 43

So no of pine trees planted today= 63-43= 20

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(Brainliest for fastest right answer!!) Which is correct?
Arisa [49]

Answer:

<em>(n - 5)(n + 5) </em>

Step-by-step explanation:

a² - b² = (a - b)(a + b)

n² - 25 = n² - 5² = <em>(n - 5)(n + 5)</em>

7 0
3 years ago
G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =
Svetach [21]
Presumably you should be doing this using calculus methods, namely computing the surface integral along \mathbf r(u,v).

But since \mathbf r(u,v) describes a sphere, we can simply recall that the surface area of a sphere of radius a is 4\pi a^2.

In calculus terms, we would first find an expression for the surface element, which is given by

\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv

\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k
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\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k
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So the area of the surface is

\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u
=-2\pi a^2(\cos\pi-\cos 0)
=-2\pi a^2(-1-1)
=4\pi a^2

as expected.
6 0
3 years ago
The function is y=6x.<br><br> Find the horizontal asymptote of the curve.
Lynna [10]

Answer:

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Step-by-step explanation:

hope this helps please mark me brainliest

6 0
2 years ago
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iragen [17]
It's C) -3/4



If you need the steps let me know. (:

8 0
3 years ago
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EastWind [94]

Answer:

Step-by-step explanation:

Bello,

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thanks

5 0
3 years ago
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