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Nastasia [14]
2 years ago
9

What is the first step in solving the equation x2 – StartFraction 16 Over 25 EndFraction = 0? What is the second step in solving

the equation?
Mathematics
1 answer:
Furkat [3]2 years ago
6 0

Answer:

1. C

2. B

Step-by-step explanation:

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Answer:

Solution given:

South distance :base[b]=80milea

East distance :perpendicular [p]=35miles

Now

<S=?

we have

\tan( \alpha )  =  \frac{p}{b}  =  \frac{35 }{80}

\alpha  =  \tan {}^{ - 1} ( \frac{7}{16} )

\alpha  = 23.629°=24°

24° bearing should be taken from south airport to East airport.

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Add the equations in order to solve for the first variable. PLug this value into the other equations in order to solve for the remaining variables.
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Step-by-step explanation:

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3 years ago
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∆ABC is an equilateral triangle. Point P is on base BC such that <br>PC = 1 /3 BC. If AB = 12 cm, find AP
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3 years ago
An athletic field is a 60 yd60 yd​-by-120 yd120 yd ​rectangle, with a semicircle at each of the short sides. A running track 101
gregori [183]

Distance of each track are:

D₁ = 428.5 yd

D₂ = 436.35 yd

D₃ = 444.20 yd

D₄ = 452.05 yd

D₅ = 459.91 yd

D₆ = 467.76 yd

D₇ = 475.61 yd

D₈ = 483.47 yd

<u>Explanation:</u>

Given:

Track is divided into 8 lanes.

The length around each track is the two lengths of the rectangle plus the two lengths of the semi-circle with varying diameters.

Thus,

D = 2(120) + 2. \frac{1}{2} \pi d\\\\D = 240 + \pi d

Starting from the innermost edge with a diameter of 60yd.

Each lane is 10/8 = 1.25yd

So, the diameter increases by 2(1.25) = 2.5 yd each lane going outward.

So, the distances are:

D₁ = 240 + π (60) → 428.5yd

D₂ = 240 + π(60 + 2.5) → 436.35 yd

D₃ = 240 + π(60 + 5) → 444.20 yd

D₄ = 240 + π(60 + 7.5) → 452.05 yd

D₅ = 240 + π(60 + 10) → 459.91 yd

D₆ = 24 + π(60 + 12.5) → 467.76 yd

D₇ = 240 + π(60 + 15) → 475.61 yd

D₈ = 240 + π(60 + 17.5) → 483.47 yd

4 0
3 years ago
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