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3 years ago

Precal on plato help please

1 answer:

8 0

We can use the Cosine Law:
c² = a² + b² - 2 ac cos C
180² = 50² + x² - 2 · 50 · x · cos 150°
32,400 = 2,500 + x ² - 100 x · ( - √3/2 )
32,400 - 2.500 - x² = 100 x · 0.866
29,900 - x² = 86.6 x
- x² - 86.6 x + 29,900 = 0 / · ( - 1 )
x² + 86.6 x - 29,900 = 0
x 1/2 = ( - 86.6 +/- √(86.6² + 4 · 1 · (-29,900) )) / 2
x = ( - 86.6 + 356.6 )/ 2 ( other solution is negative )
x = 135
Answer: C ) 135 miles/hour

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Answer:

Taylor series of sin(x) centered at x = 0.

Step-by-step explanation:

Taylor series expansion:

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Here, f(x) = sin (x) and a = 0

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Putting all the values and expanding, we get,

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Solving, we get

$\sin x = x - \displaystyle\frac{x^3}{3!} + \displaystyle\frac{x^5}{5!} - \displaystyle\frac{x^7}{7!} + ...\\\\\sin x = x - \displaystyle\frac{x^3}{6} + \displaystyle\frac{x^5}{120} - \displaystyle\frac{x^7}{5040} + ...$

3 0

3 years ago

Given: Circle k (O), OA and OC -radii, AP and CP - tangents, m∠AOC=150° Find: m∠APC

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Answer:

m∠APC = 30°

Step-by-step explanation:

To solve for the above question, we would be making use of circle theorems

Looking at the circle, we can see that

m∠APC is an Angle outside the circle

m∠AOC is a smaller arc in the circle.

Therefore, Circle Theorem states that for a circle with two tangents,

The Angle outside the circle + The smaller arc = 180°

Hence,

m∠AOC + m∠APC = 180°

m∠APC = 180° - m∠AOC

m∠APC = 180° - 150°

m∠APC = 30°

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