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Vesnalui [34]
2 years ago
12

Please help me asap with my grade 11 trigonometry math! I'll give brainliest, please help, thank you. *URGENT*

Mathematics
2 answers:
stepladder [879]2 years ago
8 0

Answer:

Around 295m

Step-by-step explanation:

Vladimir79 [104]2 years ago
7 0

Answer:

295.44 m to the neaest hundredth

Step-by-step explanation:

A line drawn from the kite  perpendicular to the ground forms 2 right triangles. Let this  line be x m.

Also let y be the distance from Allison to the bottom of the line.

Consider 'Alison' triangle

cos 70 = y/150

y = 51.303 m

sin 70 = x/150

x = 150 sin 70 = 140.954 m

Consider 'Marcs' triangle

tan 30 = x / (AM - y)

tan 30 = 140.954 / (AM - 51.303)

AM - 51.303 =  140.954/tan 30

AM =  140.954/tan 30 + 51.303

AM = 295.442  m

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Solve equation (n-1)(n+6)(n+5)=0
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Answer:
1, -6 and -5 

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

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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

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