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lilavasa [31]
2 years ago
11

∠A and \angle B∠B are vertical angles. If m\angle A=(2x-10)^{\circ}∠A=(2x−10) ∘ and m\angle B=(x+8)^{\circ}∠B=(x+8) ∘ , then fin

d the measure of \angle A∠A.
Mathematics
1 answer:
AleksAgata [21]2 years ago
6 0

Answer:

m∠B=52

∘

Step-by-step explanation:

Two or more angles are said to be complementary if they sum up to 90 degrees.

Given that angles A and B are complementary, then:

\begin{gathered}\angle A+\angle B=90^\circ\\m\angle A=(2x+18)^{\circ}\\m\angle B=(6x-8)^{\circ}\\$Therefore:\\(2x+18)^{\circ}+(6x-8)^{\circ}=90^\circ\\2x+6x+18-8=90^\circ\\8x+10^\circ=90^\circ\\8x=90^\circ-10^\circ\\8x=80^\circ\\$Divide both sides by 8\\x=10^\circ\\$Therefore:\\m\angle B=(6x-8)^{\circ}\\m\angle B=(6(10)-8)^{\circ}\\=60-8\\m\angle B=52^{\circ}\end{gathered}

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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
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