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marishachu [46]
3 years ago
9

An oil tank has to be drained for maintenance. The tank is shaped like a cylinder that is 4 ft long with a diameter of 1.6 ft. S

uppose oil is drained at a rate of 1.7 ft³ per minute. If the tank starts completely full, how many minutes will it take to empty the tank?
please answer if you know how to do this question. Thank you :)
Mathematics
1 answer:
oee [108]3 years ago
6 0
It will take 4.7 minutes to empty the tank. This is because you first would have to find the area of the tank. In order to do that you would have to find the area of just the flat circle, so to do that you do the radius squared times pi. So take the diameter divide it by 2 And get 0.8 (radius) and then square it (multiply it by itself) and get 0.64 than multiply it by 3.14. Than take the answer to that and multiply it by the height (4) and than divide by 1.7 .

Hope that helped :)
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Answer:

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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}

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

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.

(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 <em>v(t)</em> > 0 when -√11 < <em>t</em> < √11, so we split up the integral at <em>t</em> = √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 <em>v(t)</em> is the derivative of <em>s(t)</em>, 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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2 years ago
43x+ 67y = -24 and 67x + 43y = 24 solve
aleksandrvk [35]

Answer:

  (x, y) = (1, -1)

Step-by-step explanation:

We'll write these equations in general form, then solve using the cross-multiplication method.

  43x +67y +24 = 0

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∆1 = (43)(43) -(67)(67) = -2640

∆2 = (67)(-24) -(43)(24) = -2640

∆3 = (24)(67) -(-24)(43) = 2640

These go into the relations ...

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The solution is (x, y) = (1, -1).

_____

<em>Additional comment</em>

The cross multiplication method isn't taught everywhere. The attachment explains a bit about it. Our final relationship changes the order of the fractions to 1, x, y from x, y, 1. That way, we can use the equation coefficients in their original general-form order. (The fourth column in the 2×4 array of coefficients is a repeat of the first column.)

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Step-by-step explanation: We are given two equations and we need to find which is true. Since it is a simple algebraic question, so we just need to follow BODMAS rule to check the equations.

The equations are as follows -  

(I)~~L.H.S=(\dfrac{32}{8}-2)6=(4-2)6=2\times 6=12.

R.H.S=(32-1)2=31\times 2=62.

Since L.H.S ≠ R.H.S, so this equation is not correct.

(II)~~L.H.S=\dfrac{72}{9\times 4}=2.

R.H.S=\dfrac{62}{6+3}=\dfrac{62}{9}.

Since L.H.S ≠ R.H.S, so this equation is also not correct.

Thus, the correct option is (b) Neither I nor II.


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