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Temka [501]
3 years ago
13

Reynaldo rode his bike 2 miles north and 3 miles east. Which equation should he ise to find the distance, d, that takes him dire

ctly back home
Mathematics
2 answers:
yan [13]3 years ago
8 0

Answer:

√13

Step-by-step explanation:

Given :Reynaldo rode his bike 2 miles north and 3 miles east.

To Find :Which equation should he use to find the distance, d, that takes him directly back home

Solution:

Refer the attached figure.

Reynaldo rode his bike 2 miles north i.e. AB = 2 miles

Then he rode 3 miles east i.e. BC=3 miles

now to find the distance, d, that takes him directly back home i.e.AC

To find AC we use Pythagorean theorem:

Hypotenuse^{2} =Perpendicular^{2} +Base^{2}

d^{2} =AB^{2} +BC^{2}

d^{2} =2^{2} +3^{2}

hence this is the required equation .

d^{2} =4 +9

d^{2} =13

d=\sqrt{13}

Thus the distance, d, that takes him directly back home i.e.AC=√13

ValentinkaMS [17]3 years ago
6 0
Pythagorean theorem - a^2 + b^2 = c^2

2^2+3^2 = d^2
d^2 = 13
d= square root of 13
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4 years ago
How do I do 8b(ii) ? Please help me thank you!
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Step One
======
Find the length of FO (see below)

All of the triangles are equilateral triangles. Label the center as O
FO = FE = sqrt(5) + sqrt(2)

Step Two
======
Drop a perpendicular bisector from O to the midpoint of FE. Label the midpoint as J. Find OJ

Sure the Pythagorean Theorem. Remember that OJ is a perpendicular bisector.

FO^2 = FJ^2 + OJ^2
FO = sqrt(5) + sqrt(2)
FJ = 1/2 [(sqrt(5) + sqrt(2)]                                           \
OJ = ??

[Sqrt(5) + sqrt(2)]^2 = [1/2(sqrt(5) + sqrt(2) ] ^2 + OJ^2
5 + 2 + 2*sqrt(10) = [1/4 (5 + 2 + 2*sqrt(10) + OJ^2
7 + 2sqrt(10) = 1/4 (7 + 2sqrt(10)) + OJ^2      Multiply through by 4
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21 + 6 sqrt(10) = 4OJ^2   Divide both sides by 4
21/4 + 6/4* sqrt(10) = OJ^2
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sqrt OJ^2 = sqrt(21/4 + 3/2 sqrt(10) )
OJ = sqrt(21/4 + 3/2 sqrt(10) )

Step three
find h
h = 2 * OJ
h = 2* sqrt(21/4 + 3/2 sqrt(10) ) <<<<<< answer.


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