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zhuklara [117]
3 years ago
5

Can you work out G and H for me please. preferably telling me how you did it aswell

Mathematics
1 answer:
Veronika [31]3 years ago
8 0
Abd is a right angled triangle so by pythagoras we know
{g}^{2}  =  {11}^{2}  +  {13}^{2}  \\  {g}^{2}  = 290 \\ g =  \sqrt{290}  \: or \: g = 17.03cm
and to calculate h we need to use the sine rule
\frac{h}{ \sin(22) }  =  \frac{g}{ \sin(32) }  \\  \frac{h}{ \sin(22) }  =  \frac{17.03}{ \sin(32) }  \\ h =  \frac{17.03}{ \sin(32) } \times  \sin(22)   \\ h = 12.04cm
i hope this helps
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Answer:

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General Formulas and Concepts:

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Order of Operations: BPEMDAS

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  • Left to Right

<u>Algebra I</u>

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Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

f(x) = 2x - 6

g(x) = 3x + 9

(f + g)(x) is f(x) + g(x)

<u>Step 2: Find</u>

  1. Substitute in function values:                                                                          (f + g)(x) = 2x - 6 + 3x + 9
  2. Combine like terms:                                                                                         (f + g)(x) = 5x + 3
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Step-by-step explanation:

6 0
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Peter walked 10m from X to Y on bearing 020° and then he turned and walked 20m to Z with bearing 140° of Z from Y. Find the dist
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17.32m ; 110°

Step-by-step explanation:

Distance between X and Z

To calculate the distance between X and Z

y^2 = x^2 + z^2 - (2xz)*cosY

x = 20, Z = 10

y^2 = 20^2 + 10^2 - (2*20*10)* cos60°

y^2 = 400 + 100 - (400)* 0.5

y^2 = 500 - 200

y^2 = 300

y = sqrt(300)

y = 17.32m

Bearing of Z from X:

Using cosine rule :

Cos X = (y^2 + z^2 - x^2) / 2yz

Cos X = (300 + 100 - 400) / (2 * 20 '*10)

Cos X = 0 / 400

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