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3 years ago

What is the difference of the polynomials?

Mathematics

2 answers:

alexandr402 [8]3 years ago

8 0

Answer:

C is the answer

Step-by-step explanation:

Vilka [71]3 years ago

3 0

Answer:

$8r^6s^3-5r^5s^4+r^4s^5+5r^3s^6$

Step-by-step explanation:

We want to simplify;

$(8r^6s^3-9r^5s^4+3r^4s^5)-(2r^4s^5-5r^3s^6-4r^5s^4)$

We expand the bracket to obtain;

$8r^6s^3-9r^5s^4+3r^4s^5-2r^4s^5+5r^3s^6+4r^5s^4$

We now group the like terms to obtain;

$8r^6s^3-9r^5s^4+4r^5s^4+3r^4s^5-2r^4s^5+5r^3s^6$

We now simplify to get;

$8r^6s^3-5r^5s^4+r^4s^5+5r^3s^6$

The correct answer is C

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The intensity of light with wavelength λ traveling through a diffraction grating with N slits at an angle θ is given by I(θ) = N

Ymorist [56]

Answer:

0.007502795

Step-by-step explanation:

We have

N = 10,000

$\bf d=10^{-4}$

$\bf \lambda = 632.8*10^{-9}$

Replacing these values in the expression for k:

$\bf k=\frac{\pi Ndsin\theta}{\lambda}=\frac{\pi10^4*10^{-4}sin\theta}{632.8*10^{-9}}=\frac{\pi 10^9sin\theta}{632.8}$

So, the intensity is given by the function

$\bf I(\theta)=\frac{N^2sin^2(k)}{k^2}=\frac{10^8sin^2(\frac{\pi 10^9sin\theta}{632.8})}{(\frac{\pi 10^9sin\theta}{632.8})^2}$

The total light intensity is then

$\bf \int_{-10^{-6}}^{10^{-6}} I(\theta)d\theta=\int_{-10^{-6}}^{10{-6}}\frac{10^8sin^2(\frac{\pi 10^9sin\theta}{632.8})}{(\frac{\pi 10^9sin\theta}{632.8})^2}d\theta$

Since $\bf I(\theta)$ is an even function

$\bf \int_{-10^{-6}}^{10^{-6}} I(\theta)d\theta=2\int_{0}^{10^{-6}}I(\theta)d\theta$

and we only have to divide the interval $\bf [0,10^{-6}]$ in five equal sub-intervals $\bf I_1,I_2,I_3,I_4,I_5$ with midpoints $\bf m_1,m_2,m_3,m_4,m_5$

The sub-intervals and their midpoints are

$\bf I_1=[0,\frac{10^{-6}}{5}]\;,m_1=10^{-5}\\I_2=[\frac{10^{-6}}{5},2\frac{10^{-6}}{5}]\;,m_2=3*10^{-5}\\I_3=[2\frac{10^{-6}}{5},3\frac{10^{-6}}{5}]\;,m_3=5*10^{-5}\\I_4=[3\frac{10^{-6}}{5},4\frac{10^{-6}}{5}]\;,m_4=7*10^{-5}\\I_5=[4\frac{10^{-6}}{5},10^{-6}]\;,m_5=9*10^{-5}$

By the midpoint rule

$\bf \int_{0}^{10^{-6}}I(\theta)d\theta\approx\frac{10^{-6}}{5}[I(m_1)+I(m_2)+...+I(m_5)]$

computing the values of I:

$\bf I(m_1)=I(10^{-5})=\frac{10^8sin^2(\frac{\pi 10^9sin(10^{-5})}{632.8})}{(\frac{\pi 10^9sin(10^{-5})}{632.8})^2}=13681.31478$

$\bf I(m_2)=I(3*10^{-5})=\frac{10^8sin^2(\frac{\pi 10^9sin(3*10^{-5})}{632.8})}{(\frac{\pi 10^9sin(3*10^{-5})}{632.8})^2}=4144.509447$

Similarly with the help of a calculator or spreadsheet we find

$\bf I(m_3)=3.09562973\\I(m_4)=716.7480066\\I(m_5)=211.3187228$

and we have

$\bf \int_{0}^{10^{-6}}I(\theta)d\theta\approx\frac{10^{-6}}{5}[I(m_1)+I(m_2)+...+I(m_5)]=\frac{10^{-6}}{5}(18756.98654)=0.003751395$

Finally the the total light intensity

would be 2*0.003751395 = 0.007502795

8 0

2 years ago

Multiply 2.4 x 108 by 5.2 x 103.

AfilCa [17]

Answer:

138827.52

Step-by-step explanation:

If by means divided then yea

4 0

2 years ago

Lnes made9 necklaces with 60 beads on each necklaces. How many beads did she use in all

Gala2k [10]

Made 9 necklaces.
60 beads per necklace.
How many beads for all the 9 necklaces?

60 × 9 = 540

9 necklaces = 540 beads

3 0

2 years ago

Factor of 42a5b3, 35a3b4, and 42ab4

Scrat [10]

Can u explain it a little more clearly...

4 0

3 years ago

What is the equation of the line that is parallel to 3x-5y=7 and passes through the point (-10,8)

Karo-lina-s [1.5K]

Answer: y= 3/5 + 18

Step-by-step explanation: Bruh i legit learned this a hour ago at my tuition, ok so basically simply to slope intercept from ye? so it would be y=-3/5x+7

and then you would use the distance formula and ye distance formula is square root of x sub 2 - x sub 1 squared+ y sub 2 - y sub 1 squared

4 0

3 years ago