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4 years ago

What is f(x)=|1/4x| translated 2 units to the left

Mathematics

2 answers:

yarga [219]4 years ago

6 0

Answer:

y=l(1/4)x + (1/2)l

Step-by-step explanation:

When shifting non-linear functions to the left or right, you must put the opposite operation inside the parenthesis. This means when you move to the left, your number must be positive. If you move to the right, your number must be negative. In this scenario, we want to shift the graph to the left 2 units. However, because our m value is a fraction, the shifting vale must also be a fraction. So instead of 2, the value becomes (1/2). Putting this all together, your answer is y=l(1/4)x + (1/2)l

Viktor [21]4 years ago

3 0

Answer:

f(x)=|1/2x|

Step-by-step explanation:

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Archy [21]

(a) If the particle's position (measured with some unit) at time t is given by s(t), where

$s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}$

then the velocity at time t, v(t), is given by the derivative of s(t),

$v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}$

(b) The velocity after 3 seconds is

$v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}$

$\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}$

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

$\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|$

In interval notation, this happens for t in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt$

By definition of absolute value, we have

$|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)$

In part (d), we've shown that v(t) > 0 when -√11 < t < √11, so we split up the integral at t = √11 as

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt$

and by the fundamental theorem of calculus, since we know v(t) is the derivative of s(t), this reduces to

$s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}$

7 0

3 years ago

Is the relation { (3, 2), (-1, 2), (7, 2), (0, 2)} a function? Why or why not? List the domain and range.

Makovka662 [10]

Answer:

its a function because each input (X) only has one output (y)

domain (3,-1,7,0)

range (2)

Step-by-step explanation:

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3 years ago

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Their distance would be 50 feet in 2.5 minutes, if the speeds remain constant.

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Webassign find the area of the region bounded by the parabola y = x2, the tangent line to this parabola at the point (6, 36), an

Dvinal [7]

First find the tangent line

dy/dx=2x
at x=6, the slope is 2(6)=12
so
use point slope form
y-y1=m(x-x1)
point is (6,36)
so
y-36=12(x-6)
y-36=12x-72
y=12x-36

alright, so we know they intersect at x=6
and y=12x-36 is below y=x^2

so we do $\int\limits^6_0 {x^2} \, dx - \int\limits^6_0 {12x-36} \, dx = \int\limits^6_0 {x^2-(12x-36)} \, dx = \int\limits^6_0 {x^2-12x+36} \, dx =$
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the area under the curve bounded by the lines and the x axis is 72 square units

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3 years ago