Given:
Equation of a circle

Required:
Find the center and radius of the circle. Then graph of circle.
Explanation:
We have a standard equation of circle

Whose radius is (h, k) and radius r.
We have equation of a circle

Now, we will compare with standard equation of a circle

Graph of circle is
Answer:
Hence, this is the answer.
Answer:
0.007502795
Step-by-step explanation:
We have
N = 10,000


Replacing these values in the expression for k:

So, the intensity is given by the function

The <em>total light intensity</em> is then

Since
is an <em>even function</em>

and we only have to divide the interval
in five equal sub-intervals
with midpoints 
The sub-intervals and their midpoints are
![\bf I_1=[0,\frac{10^{-6}}{5}]\;,m_1=10^{-5}\\I_2=[\frac{10^{-6}}{5},2\frac{10^{-6}}{5}]\;,m_2=3*10^{-5}\\I_3=[2\frac{10^{-6}}{5},3\frac{10^{-6}}{5}]\;,m_3=5*10^{-5}\\I_4=[3\frac{10^{-6}}{5},4\frac{10^{-6}}{5}]\;,m_4=7*10^{-5}\\I_5=[4\frac{10^{-6}}{5},10^{-6}]\;,m_5=9*10^{-5}](https://tex.z-dn.net/?f=%20%5Cbf%20I_1%3D%5B0%2C%5Cfrac%7B10%5E%7B-6%7D%7D%7B5%7D%5D%5C%3B%2Cm_1%3D10%5E%7B-5%7D%5C%5CI_2%3D%5B%5Cfrac%7B10%5E%7B-6%7D%7D%7B5%7D%2C2%5Cfrac%7B10%5E%7B-6%7D%7D%7B5%7D%5D%5C%3B%2Cm_2%3D3%2A10%5E%7B-5%7D%5C%5CI_3%3D%5B2%5Cfrac%7B10%5E%7B-6%7D%7D%7B5%7D%2C3%5Cfrac%7B10%5E%7B-6%7D%7D%7B5%7D%5D%5C%3B%2Cm_3%3D5%2A10%5E%7B-5%7D%5C%5CI_4%3D%5B3%5Cfrac%7B10%5E%7B-6%7D%7D%7B5%7D%2C4%5Cfrac%7B10%5E%7B-6%7D%7D%7B5%7D%5D%5C%3B%2Cm_4%3D7%2A10%5E%7B-5%7D%5C%5CI_5%3D%5B4%5Cfrac%7B10%5E%7B-6%7D%7D%7B5%7D%2C10%5E%7B-6%7D%5D%5C%3B%2Cm_5%3D9%2A10%5E%7B-5%7D)
<em>By the midpoint rule</em>
![\bf \int_{0}^{10^{-6}}I(\theta)d\theta\approx\frac{10^{-6}}{5}[I(m_1)+I(m_2)+...+I(m_5)]](https://tex.z-dn.net/?f=%20%5Cbf%20%5Cint_%7B0%7D%5E%7B10%5E%7B-6%7D%7DI%28%5Ctheta%29d%5Ctheta%5Capprox%5Cfrac%7B10%5E%7B-6%7D%7D%7B5%7D%5BI%28m_1%29%2BI%28m_2%29%2B...%2BI%28m_5%29%5D)
computing the values of I:


Similarly with the help of a calculator or spreadsheet we find

and we have
![\bf \int_{0}^{10^{-6}}I(\theta)d\theta\approx\frac{10^{-6}}{5}[I(m_1)+I(m_2)+...+I(m_5)]=\frac{10^{-6}}{5}(18756.98654)=0.003751395](https://tex.z-dn.net/?f=%20%5Cbf%20%5Cint_%7B0%7D%5E%7B10%5E%7B-6%7D%7DI%28%5Ctheta%29d%5Ctheta%5Capprox%5Cfrac%7B10%5E%7B-6%7D%7D%7B5%7D%5BI%28m_1%29%2BI%28m_2%29%2B...%2BI%28m_5%29%5D%3D%5Cfrac%7B10%5E%7B-6%7D%7D%7B5%7D%2818756.98654%29%3D0.003751395)
Finally the the total light intensity
would be 2*0.003751395 = 0.007502795
The dimensions to the poster is 3’x5’
Answer: negative, positive, zero, and undefined.
Step-by-step explanation: as x increases. The slope of a line can also be interpreted as the “average rate of change”.
Answer:
≈ 50.7 cm²
Step-by-step explanation:
The area (A) of Δ XYZ is calculated as
A =
bh ( b is the base and h the perpendicular height ), that is
A =
× XY × YZ
Calculate YZ using the tangent ratio in the right triangle
tan40° =
=
=
( multiply both sides by 11 )
11 × tan40° = x , that is
YZ ≈ 9.23 cm
Thus
A =
× 11 × 9.23 = 5.5 × 9.23 ≈ 50.7 cm² ( to 3 significant figures )