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3 years ago

Dpoiski2

1 answer:

5 0

For this question the answer is 2.5m (B)
You find this by: L=0.85m
W=0.4m

A rectangle had 4 sides (2 lengths and 2 widths)
So...
To find 1 length and 1 Width you do 0.85+0.4=1.25m
Now you want to find the other 2 sides (L and W)
You can do: the same method and add it to 1.25m OR...
You can do (I prefer this one): (0.85+0.4)x2=?
This is the same as 0.85+0.4=1.25
1.25x2=2.5m

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3 years ago

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3 years ago

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2 years ago

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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

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

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3 years ago

Given the system of equations, what is the solution? x + 3y = 5 x - 3y = -1

mr_godi [17]

$\begin{cases}x+3y=5\\x-3y=-1 \end{cases}\\---------(+)\\\\x+x+3y-3y=5+(-1)\\\\2x=4\qquad|:2\\\\\boxed{x=2}\\\\\\\\x+3y=5\\\\2+3y=5\qquad|-2\\\\2+3y-2=5-2\\\\3y=3\qquad|:3\\\\\boxed{y=1}$

The solution is (2; 1)

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3 years ago

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