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3 years ago

The expression 3(x2 + 2x-3) -4(4x2 – 7x + 5) is equivalent to

Mathematics

1 answer:

tekilochka [14]3 years ago

7 0

The answer is B. Hope this helps!!

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Relationship B has a lesser rate than Relationship A. This graph represents Relationship A.Which equations could represent Relat

Black_prince [1.1K]

<h3>Answer:</h3>

y = x
y = (3/2)x

<h3>Explanation:</h3>

The rate shown in your graph is 3. (rise:run = 3:1) An equation with a lesser rate will have an x-coefficient that is less than 3. The x-coefficients in your answer choices appear to be ...

Of these values, only the first and last are less than 3.

3 0

3 years ago

What is the answer because I don't get it? The 1st one

ANTONII [103]

L = 10 mw = 11 mh = 5 md = 15.6844 mS = 430 m²V = 550 m³ x 2 = 1100m³
Agenda:l = lengthw = widthh = heightd = diagonalS = surface area V = volume

4 0

3 years ago

GIVING BRAINLIEST! Calculate the slope of the following line

Georgia [21]

The slope is 3 because if you find the rise and run of two points. Run=1 rise=3 (divide the rise by the run) which equals 3. The slope is 3

6 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

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