In <span>equilateral triangle all sides are equal =x,
from right triangle that is formed by </span>apothem , height 7*3= 21
hypotenuse /opposite leg = sin angle,
hypotenuse is a side of triangle =x, opposite leg is apotheme =21,
angle in equilateral triangle =60⁰
21/x= sin 60, x*sin60=21, x=21/sin60, x=21/(√3/2), x=42/√3
Area of triangle =1/2 x*x*sin angle
Area of triangle =1/2 *42/√3*42/√3*sin 60=1/2*(42²/3)*(√3/2) ≈ 255 cm²
this is just second way to do this problem, that I did at first (either way is correct)
<span> apothem divides one side by half,
so we get small right triangle with sides x hypotenuse, x/2 is one leg , and 21 is another leg
</span>by Pythagorean theorem
x²=(x/2)² +21²
x²-x²/4=441
3x²/4=441
3x²=441*4=1764
x²=1764/3
x=42/√3
Area of triangle =1/2*base *height =1/2*42/√3 *21≈255 cm²
Answer:
≈ 23 years
Step-by-step explanation:
Initial price= $995
Value increase= 3.1%
a) $995 +3.1%= $995*1.031= $1025.85
b) $995*1.031^x= $2000
1.031^x= 2000/995
1.031^x= 2.01
x= log 2/log 1.031= 22.7 ≈ 23 years
Answer:
37.65
Step-by-step explanation:
Take 2447.25, divide by 52 (there are 52 weeks in a year) and mutiply by 0.8 (80%), and then you have the answer!
Denote the cylindrical surface by
, and its interior by
. By the divergence theorem, the integral of
across
(the outward flow of the fluid) is equal to the integral of the divergence of
over the space it contains,
:
![\displaystyle\iint_S\vec v\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec v)\,\mathrm dV](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_S%5Cvec%20v%5Ccdot%5Cmathrm%20d%5Cvec%20S%3D%5Ciiint_R%28%5Cnabla%5Ccdot%5Cvec%20v%29%5C%2C%5Cmathrm%20dV)
The given velocity vector has divergence
![\vec v=z\,\vec\imath+y^2\,\vec\jmath+x^2\,\vec k\implies\nabla\cdot\vec v=\dfrac{\partial(z)}{\partial x}+\dfrac{\partial(y^2)}{\partial y}+\dfrac{\partial(x^2)}{\partial z}=2y](https://tex.z-dn.net/?f=%5Cvec%20v%3Dz%5C%2C%5Cvec%5Cimath%2By%5E2%5C%2C%5Cvec%5Cjmath%2Bx%5E2%5C%2C%5Cvec%20k%5Cimplies%5Cnabla%5Ccdot%5Cvec%20v%3D%5Cdfrac%7B%5Cpartial%28z%29%7D%7B%5Cpartial%20x%7D%2B%5Cdfrac%7B%5Cpartial%28y%5E2%29%7D%7B%5Cpartial%20y%7D%2B%5Cdfrac%7B%5Cpartial%28x%5E2%29%7D%7B%5Cpartial%20z%7D%3D2y)
Then the total outward flow is
![\displaystyle\iiint_R2y\,\mathrm dV](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciiint_R2y%5C%2C%5Cmathrm%20dV)
Converting to cylindrical coordinates gives the integral
![\displaystyle2\int_0^{2\pi}\int_0^3\int_0^4r^2\sin\theta\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\boxed{0}](https://tex.z-dn.net/?f=%5Cdisplaystyle2%5Cint_0%5E%7B2%5Cpi%7D%5Cint_0%5E3%5Cint_0%5E4r%5E2%5Csin%5Ctheta%5C%2C%5Cmathrm%20dz%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta%3D%5Cboxed%7B0%7D)