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Free_Kalibri [48]
3 years ago
10

Can you help me please?

Mathematics
1 answer:
Daniel [21]3 years ago
4 0

59pounds * 5days = 295pounds

6.95pounds * 3breakfasts = 20.85pounds

12.50pounds * 1eveningmeal = 12.50pounds

295pounds + 20.85pounds + 12.50pounds = 328.35pounds

Wrote out the units just for clarity! Liz payed 328.25 pounds altogether.

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ΔHAT is similar to ΔCAN.
LenKa [72]

In similar triangles, the ratios of corresponding sides are equal.

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y/6=5/(4+5)

Cross multiply

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3 years ago
There are 3 islands A,B,C. Island B is east of island A, 8 miles away. Island C is northeast of A, 5 miles away and northwest of
Nostrana [21]

Answer:

The bearing needed to navigate from island B to island C is approximately 38.213º.

Step-by-step explanation:

The geometrical diagram representing the statement is introduced below as attachment, and from Trigonometry we determine that bearing needed to navigate from island B to C by the Cosine Law:

AC^{2} = AB^{2}+BC^{2}-2\cdot AB\cdot BC\cdot \cos \theta (1)

Where:

AC - The distance from A to C, measured in miles.

AB - The distance from A to B, measured in miles.

BC - The distance from B to C, measured in miles.

\theta - Bearing from island B to island C, measured in sexagesimal degrees.

Then, we clear the bearing angle within the equation:

AC^{2}-AB^{2}-BC^{2}=-2\cdot AB\cdot BC\cdot \cos \theta

\cos \theta = \frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC}

\theta = \cos^{-1}\left(\frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC} \right) (2)

If we know that BC = 7\,mi, AB = 8\,mi, AC = 5\,mi, then the bearing from island B to island C:

\theta = \cos^{-1}\left[\frac{(7\mi)^{2}+(8\,mi)^{2}-(5\,mi)^{2}}{2\cdot (8\,mi)\cdot (7\,mi)} \right]

\theta \approx 38.213^{\circ}

The bearing needed to navigate from island B to island C is approximately 38.213º.

8 0
3 years ago
PLEASE HELP QUICK!!!!!! WILL GIVE BRAINIEST!!!!!! Don’t answer if you don’t know it please.
mojhsa [17]
I’m almost 100% sure it is: D. Graph D

Happy holidays and hope this helps!!
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