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White raven [17]
3 years ago
13

if a taxi charges $3.50 for the first on-fifth of a mile, then $0.55 cents for each additional one-fifth mile, how far can one t

ravel for $13.95?
Mathematics
1 answer:
34kurt3 years ago
7 0
Let's forget for the tie being the cost ($3.5)of the 1st fifth of a mile:

Taxi charge excluding $3.5, is $13.95 - $3.5 = $10.45.

If each of the 5th (1/5) of a mile costs $0.55, then with $10.45 we can drive:
10.45/0.55 = 19 fifth of a mile. Add to that the 1st fifth = 20 fifth of a mile
 20 fifth = 20 x 1/5 = 4 miles (answer)

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2 years ago
5. Can someone explain how to solve this?
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Answer:

AC = 5 cm

BD = 12.5 cm (3 sf) [or 2 × root 39]

BE = 6.93 cm (3 sf) [or 4 × root 3]

Step-by-step explanation:

CE = 8cm [CE is radius of circle]

AC + 3 = 8

<u>A</u><u>C</u><u> </u><u>=</u><u> </u><u>5</u><u> </u><u>c</u><u>m</u>

BC = 8cm [BC is a radius of circle]

(AC)^2 + (AB)^2 = (BC)^2 [Pythagoras theorem]

25 + (AB)^2 = 64

AB = 6.2450 cm (5 sf) [or root 39]

BD = 2(BA)

= 2(6.2450)

<u>B</u><u>D</u><u> </u><u>=</u><u> </u><u>1</u><u>2</u><u>.</u><u>5</u><u> </u><u>c</u><u>m</u><u> </u><u>(</u><u>3</u><u> </u><u>s</u><u>f</u><u>)</u><u> </u><u>[</u><u>o</u><u>r</u><u> </u><u>2</u><u> </u><u>×</u><u> </u><u>r</u><u>o</u><u>o</u><u>t</u><u> </u><u>3</u><u>9</u><u>]</u>

(BA)^2 + (AE)^2 = (BE)^2 [Pythagoras theorem]

39 + 9 = (BE)^2

<u>B</u><u>E</u><u> </u><u>=</u><u> </u><u>6</u><u>.</u><u>9</u><u>3</u><u> </u><u>c</u><u>m</u><u> </u><u>(</u><u>3</u><u> </u><u>s</u><u>f</u><u>)</u><u> </u><u>[</u><u>o</u><u>r</u><u> </u><u>4</u><u> </u><u>×</u><u> </u><u>r</u><u>o</u><u>o</u><u>t</u><u> </u><u>3</u><u>]</u>

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