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satela [25.4K]
3 years ago
14

What’s the name of the shape below?

Mathematics
1 answer:
il63 [147K]3 years ago
6 0

Answer:

first shape is a trapezoid second shape is a triangle third shape is a rhombus

Step-by-step explanation:

hope this helped

You might be interested in
What is the taxidistance between (2, 11) and (17, 40)?
Novosadov [1.4K]

15 and 29 or (15,29)

2 + 15 = 17

11 + 29 = 40

Hope it Helped!


6 0
3 years ago
Which of the follow box-and-whisker plots correctly displays this data set?
sdas [7]

Answer:

d

Step-by-step explanation:

4 0
3 years ago
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 <em>total light intensity</em> 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 <em>even function</em>

\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}

<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)]

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
3 years ago
HELP ME PLEASE
Flauer [41]
Combine like terms by adding their FACES (Coefficients)

1) 4c² - 4               
2) 20m^4 + 6m
3) -6a³ - 19a² + 8a - 10
4) -10a² - 2b + 3ab³ + 4a²b² -5b^4 - 9a²b² - 12a³b + 15ab³ = -10a² - 2b + 18ab³ - 5a²b² - 12a³b - 5b^4
5) 5a³ - 5a²
8 0
3 years ago
Carl has 9 1/2 cups of juice for a party. If each party goer gets exactly 3/5 cup of juice each, how many party goers will get t
topjm [15]

Answer: 158

Step-by-step explanation:

Okay so the total is 9 5/10.

Each party goer needs 6/10

So 95/10 / 6/0=. 95/10x10/6

=950/6 = 158.

7 0
3 years ago
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