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AleksandrR [38]
1 year ago
8

Which equation can be solved to find one of the missing side lengths in the triangle?aB60512 unitsАCOS(60°) =22 믕cos(60°) = 12Su

bmitNex응Save and Exitcos(609) =0=1or this and return
Mathematics
1 answer:
Mekhanik [1.2K]1 year ago
3 0

In a right rectangle, we have:

\sin \alpha=\frac{opposite}{hypotenuse}\cos \alpha=\frac{\text{adjacent}}{hypotenuse}

For your exercice, hypotenuse=12

The exercise also inform the angle 60°, then:

\sin \text{ 60}=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{b}{12}\cos 60=\frac{adjacent}{\text{hypotenuse}}=\frac{a}{12}

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An artist mixes yellow an blue paint to make the perfect shade of green. she uses 2 ounces of blue paint for every 7 ounces of y
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14 ounces of blue paint

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3 years ago
The annual snowfall in city A is 58.7 inches. This is 12.5 inches more than six times the snowfall in city B. Find the annual sn
Lelechka [254]
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3 0
3 years ago
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Write the equation of a parabola whose focus is at (-7, 3) and whose directrix is the line x = -3.
Sav [38]

Check the picture below.

now, let's keep in mind that, the vertex is half-way between the focus point and the directrix, it's a "p" distance from each other.

since this horizontal parabola is opening to the left-hand-side, "p" is negative, notice in the picture, "p" is 2 units, and since it's negative, p = -2.

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\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ 4p(y- k)=(x- h)^2 \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=-5\\ k=7\\ p=-2 \end{cases}\implies 4(-2)[x-(-5)]=[y-7]^2 \\\\\\ -8(x+5)=(y-7)^2\implies x+5=\cfrac{(y-7)^2}{-8}\implies \boxed{x=-\cfrac{1}{8}(y-7)^2-5}

7 0
4 years ago
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PtichkaEL [24]
Paul's rate before the pay rise = 7.0

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