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Natali5045456 [20]
3 years ago
11

D divided by 4 plus 12 = 20 show steps what is d

Mathematics
2 answers:
tester [92]3 years ago
4 0

Answer:

d=32!!!

Step-by-step explanation:

So first rewrite it as an equation: d/4+12=20

d/4+12=20 then subtract 12 from both sides to balance the equation

d/4=8 so multply both sides by 4  to isolate the variable, 4(d/4)=d =d and 8x4=32, so d=32

Ket [755]3 years ago
3 0

Answer:

d = 32

Step-by-step explanation:

Given

\frac{d}{4} + 12 = 20 ( subtract 12 from both sides )

\frac{d}{4} = 8 ( multiply both sides by 4 to clear the fraction )

d = 32

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

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

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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}}

(c) The particle is at rest when its velocity is zero:

\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 <em>t</em> in the interval (0, √11) or approximately (0, 3.317) s.

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

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

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\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 <em>v(t)</em> is the derivative of <em>s(t)</em>, this reduces to

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