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Reika [66]
3 years ago
12

In ΔQRS, the measure of ∠S=90°, QR = 82 feet, and RS = 60 feet. Find the measure of ∠R to the nearest tenth of a degree.

Mathematics
2 answers:
dsp733 years ago
8 0

Answer:

43.0°

Step-by-step explanation:

cos∅=adj/hyp

cos∅=60/82

∅=cos-1 60/82

∅=cos-1  0.731707317073

∅=42.97028393174°

Round: 43.0°

MatroZZZ [7]3 years ago
6 0

Answer:

R=42.9703≈43  

Step-by-step explanation:

Deltamath

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\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ h=-16t^2+\stackrel{\stackrel{v_o}{\downarrow }}{65}t


now, take a look at the picture below, so for 2) and 3) is the vertex of this quadratic equation, 2) is the y-coordinate and 3) the x-coordinate.


\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+65}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left( -\cfrac{65}{2(-16)}~~,~~0-\cfrac{65^2}{4(-16)} \right) \implies \left( \cfrac{65}{32}~,~0- \cfrac{4225}{-64}\right)


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