Please refer the pictures below-
Hence the area of the road is 9600m2
Answer:
n = 0
Step-by-step explanation:
1.2n + 1 = 1 - n
Add n and - 1 on both sides.
1.2n + n = 1 - 1
Combine like terms.
2.2n = 0
Divide both sides by 2.2.
n = 0
(28 : 100) *6 = 1.68 (tax)
28 + 1.68 = 29.68 (total cost)
The bearing of the individuals from each other are as indicated below;
- S45°E
- N25°E
- S65°E
- S15°E
- N45°E
- N75°E
<h3>What is bearing of Henley from Dinder?</h3>
It follows from the task content that the bearing of the individuals from each other are expected to be determined.
a) For Henley from Dinder; it follows from observation that Henley is; S45°E of Dinder's position.
b) For Dinder from Weare; it follows from observation that Dinder is; N25°E of Weare's position.
c) For Weare from Dinder; it follows from observation that Weare is; S65°E of Dinder's position.
d) For Weare from Henly; it follows from observation that Weare is; S15°E of Dinder's position.
e) For Dinder from Henly; it follows from observation that Dinder is; N45°E of Henly's position.
f) For Henly from Weare; it follows from observation that Henly is; N75°E of Weare's position.
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<span>(1 + cos² 3θ) / (sin² 3θ) = 2 csc² 3θ - 1
Starting with the left: Note that cos²θ + </span><span>sin²θ = 1.
In the same way: </span><span>cos²3θ + <span>sin²3θ = 1
</span></span>Therefore cos²3θ = 1 - <span>sin²3θ
</span> From the top: (1 + cos² 3θ) = 1 + 1 - sin²3θ = 2 - <span>sin²3θ
</span>
(1 + cos² 3θ) / (sin² 3θ) = (<span>2 - sin²3θ) / (sin² 3θ) = 2/</span><span>sin² 3θ - </span><span>sin²3θ/</span>sin²3θ
= 2/<span>sin² 3θ - 1; But 1/</span><span>sinθ = csc</span><span>θ, Similarly </span>1/sin3θ = csc3θ
= 2 *(1/sin<span>3θ)² - 1</span>
= 2csc²3θ - 1. Therefore LHS = RHS. QED.
