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Elena-2011 [213]
3 years ago
12

At noon the outside temperature was –5°F. By 4 p.m. the same day, the temperature had dropped 13°F. What was the outside tempera

ture at 4 p.m.? At 4 p.m. the outside temperature was °F.
Mathematics
2 answers:
Radda [10]3 years ago
7 0
The temperature turned to -18°F
-5-13=-18

MAXImum [283]3 years ago
5 0
<span>-5 + (-13) = -18ºF
</span><span>Good luck!</span>
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Step-by-step explanation:

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Archy [21]

(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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If 15 litres of paint covers an area of 250 m2, what area (in m2) will 20 litres of paint cover?
Korvikt [17]

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Step-by-step explanation:

Since we are informed that 15 litres of paint covers an area of 250 m2, the the area that 1 liter of paint will cover will be:

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= 16.666667m²

Therefore, the area that 20 litres of paint will cover will be:

= 16.666667m² × 20

= 333.33m²

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kkurt [141]

Answer:

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Step-by-step explanation:

It can be helpful to draw a diagram of a parallelogram ABCD and identify the angles in the problem statement. You will find they are adjacent angles.

Adjacent angles in a parallelogram are supplementary, so the sum of the given angle measures will be 180°.

  102 + (3x +2) = 180 . . . equation used to solve for x

  3x = 76 . . . . . . . . . . . . . subtract 104

  x = 76/3 . . . . . . . . . . . . divide by 3

  x = 25 1/3 . . . the value of x

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