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prohojiy [21]
3 years ago
13

Help pleasee Qn is attached belowww PLEASE HELPP

Mathematics
1 answer:
Simora [160]3 years ago
7 0

Answer:

QR = 65.4 m

Step-by-step explanation:

a. Apply Law of Cosines to find QR:

p² = q² + r² - 2qr × Cos P

p = QR = ?

q = PR = 150 m

r = PQ = 120 m

P = 25°

Plug in the values

p² = 150² + 120² - (2)(150)(120) × Cos(25°)

p² = 22,500 + 14,400 - 36,000 × 0.9063

p² = 36,900 - 32,626.8

p² = 4,273.2

p = √4,273.2

p ≈ 65.4 m (nearest tenth)

QR = 65.4 m

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Write the given second order equation as its equivalent system of first order equations
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The second-order equation as its equivalent system of first-order equations is

\left[\begin{array}{ccc}u(1)\\\\v(1)\end{array}\right] = \left[\begin{array}{ccc}7.5\\\\9\end{array}\right]

An equivalent system that has the identical answer is known as an equivalent structure. Given a gadget of two equations, we can produce an equal system by way of replacing one equation by means of the sum of the 2 equations, or by way of changing an equation by means of a couple of of itself.

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In the structures sciences, a machine equivalent system is the conduct of a parameter or thing of a machine in a way just like a parameter or component of a distinctive system. Similarity means that mathematically the parameters and additives will be indistinguishable from each different.

Taking v = u, we have:

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\left[\begin{array}{ccc}u(1)\\\\v(1)\end{array}\right] = \left[\begin{array}{ccc}7.5\\\\9\end{array}\right]

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