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iVinArrow [24]
2 years ago
14

Which absolute value function has a graph that is wider than the parent function, f(x) = |x|, and is translated to the right 2 u

nits?
Which absolute value function has a graph that is wider than the parent function, f(x) = |x|, and is translated to the right 2 units? f(x) = 1.3|x| – 2f(x) = 3|x – 2|f(x) = 3/4 |x – 2|f(x) = 4/3 |x| + 2

Mathematics
2 answers:
Lesechka [4]2 years ago
8 0

Answer:

Required function - h(x)=|x-2|  

Step-by-step explanation:

Given : The parent function f(x)=|x|  and is translated to the right 2 units.

To find : Which absolute value function has a graph that is wider than the parent function?

Solution :

The parent function f(x)=|x|  

with the vertex (0,0)

The parent function is translated to the right 2 units.  

Transformation to the right,

f(x)→f(x-b) , the graph of f(x) is shifted towards right by b unit.

Same as the graph f(x) is shifted towards right by 2 unit and form graph of h(x).

h(x)=|x-2|  

If the graph is wider than the parent function then the function must be in the form of,  h(x)=k\times f(x)

Where the value of k must be less than of equal to 1. If k is more than 1 then the graph compressed.

So, let it be k=1

Therefore, The required absolute value function is  h(x)=|x-2|  

We plot the graph of both the equations in which translation is shown.

Refer the attached graph below.

Mariulka [41]2 years ago
4 0
Case 1:  If we multiply f(x) = |x| by a fraction greater than zero and less than 1, the width of the resulting graph will increase.  If the vertex of the original function is moved 2 units to the right, then we'd replace |x| with |x-2|  Only the coefficient (3/4) satisfies the "wider graph" requirement here.

Next time you list answr possibilities, please type them in only one per line, or separate them with commons, semicolons or the like.
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Step-by-step explanation:

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  • z = Number of standard deviation = 3
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  • Here, n = number of observations in each sample

  • UCL = p+zσ = 0.0336 + 3(0.0104) = 0.0336 + 0.0312 = 0.0648 = 0.065

  • LCL = p-zσ = 0.0336 - 0.0312 = 0.0024 = 0.002

b) Hence, Lower control limit cannot be a negative number as percent defective cannot be a negative number. As such, No. Percent of defective records cannot be a negative number.

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Leo is testing different types of greenhouse material to determine which type is most effective for growing strawberry bushes. H
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Jenny is taking a vacation to Florida. She travels 70 kilometers per hour for 2 hours, and 63 kilometers per hour for 5 hours. O
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Jenny traveled 70 km/h over 2 hours and 63 km/h over 5 hours. Her travel time was a total of 7 hours. 

We need to find out how far she traveled during this 7-hour period.

70 km/h * 2 h = 140 km 
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Then, we can add 140 km and 315 km to get a total distance traveled of <em>455 km/h.</em>

We were asked to find the average <em>speed.</em> We can find it now, since we have total distance traveled (455 km) and total time taken (7 hours).

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Jan accidentally ran 7 minutes longer than she was supposed to. Write an expression for the total amount of time Jan ran if she
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m+7

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Let's say m=60 minutes. She would've ran for 67 minutes, which is also m+7.

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2 years ago
Use Stokes' Theorem to evaluate S curl F · dS. F(x, y, z) = x2 sin(z)i + y2j + xyk, S is the part of the paraboloid z = 9 − x2 −
Korolek [52]

The vector field

\vec F(x,y,z)=x^2\sin z\,\vec\imath+y^2\,\vec\jmath+xy\,\vec k

has curl

\nabla\times\vec F(x,y,z)=x\,\vec\imath+(x^2\cos z-y)\,\vec\jmath

Parameterize S by

\vec s(u,v)=x(u,v)\,\vec\imath+y(u,v)\,\vec\jmath+z(u,v)\,\vec k

where

\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=(9-u^2)\end{cases}

with 0\le u\le3 and 0\le v\le2\pi.

Take the normal vector to S to be

\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k

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\displaystyle\int_{\partial S}\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S

=\displaystyle\int_0^{2\pi}\int_0^3(\nabla\times\vec F)(\vec s(u,v))\cdot\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^{2\pi}\int_0^3u^3(2u\cos^3v\sin(u^2-9)+\cos^3v\sin v+2u\sin^3v+\cos v\sin^3v)\,\mathrm du\,\mathrm dv

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\displaystyle\int_0^{2\pi}\cos^3v\,\mathrm dv=0

\displaystyle\int_0^{2\pi}\sin v\cos^3v\,\mathrm dv=0

\displaystyle\int_0^{2\pi}\sin^3v\,\mathrm dv=0

\displaystyle\int_0^{2\pi}\cos v\sin^3v\,\mathrm dv=0

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1 year ago
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