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bija089 [108]
2 years ago
6

An artist is creating a sculpture using bendable metal rods of equal length. One rod is formed into the shape of a square and an

other rod into the shape of an equilateral triangle. If each side of the triangle is 2 inches longer than each side of the square, how long, in inches, is each rod?

Mathematics
2 answers:
Alika [10]2 years ago
6 0
Check the picture below.

so notice, their perimeter is the same, because the perimeter is just one rod anyway, and all rods are the same length, thus

\bf P_s=P_t\implies 4s=3(s+2)\implies 4s=3s+6\implies s=6

Lapatulllka [165]2 years ago
5 0

Answer:

24 inches

Step-by-step explanation:

We are given that an artist creating a sculpture using bendable metal rods of equal length.

One rod is in square shape and second rod is in equilateral triangle shape.

We have to find the length of each rod

Let side of square= x inches

Length of each side of equilateral triangle =x+2 inches

We know that perimeter of square=4\times side

Perimeter of equilateral triangle=3\times side

Using the above formula

Then , Length of square shaped rod=perimeter of square =4x

Perimeter of equilateral triangle =Length of equilateral shaped rod=3(x+2)

Length of square shaped rod=Length of equilateral triangle  shaped rod

4x=3(x+2)

4x=3x+6

4x-3x=6

x=6

Length of each side of square =6 inches

Length of square shaped rod=4\times 6=24 inches

Length of each side of equilateral side=6+2=8 inches

Length of equilateral shaped rod=8\times 3=24 inches

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The final bill will cost $1,250.

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Interpret the graph to determine the statement the manager at a building supply store can use to help a customer determine how m
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Using the extended Euclidean algorithm, find the multiplicative inverse of a. 135 mod 61 b. 7465 mod 2464 c. 42828 mod 6407
rodikova [14]

Answer:

(a)1≡47 mod 61

(b)1≡2329 mod 2464

(c)Does not exist

Step-by-step explanation:

The operation a(mod b) has an inverse if the the two integers (a,b)

are co-prime. i.e. their g.c.d is 1.

(a)Given 135 mod 61

We first reduce it to its lowest form.

135 mod 61=13 mod 61

61=13(4)+9 ==> 9=61-13(4)

13=9(1)+4 ==> 4=13-9(1)

9=4(2)+1 ==> 1=9-4(2)

4=1(4)

Next we rewrite 1 as a linear combination of 13 and 61.

1=9-4(2)

=9-(13-9(1))2

=9(3)-13(2)

=(61-13(4))(3)-13(2)

=61(3)-13(12)-13(2)

1=61(3)-13(14)

1=61(3)+13(-14)

1≡-14 mod 61≡(-14+61)mod 61

1≡47 mod 61

(b)7465 mod 2464

Reducing it to its lowest form

7465 mod 2464=73 mod 2464

2464=73(33)+55 ==>55=2464-73(33)

73= 55(1)+18 ==> 18=73-55(1)

55=18(3)+1 ==>1=55-18(3)

18=1(18)

Rewriting 1 as a linear combination of 73 and 2464.

1=55-18(3)

=2464-73(33)-(73-55(1))(3)

=2464-73(33)-73(3)+55(3)

=2464-73(36)+55(3)

=2464-73(36)+(2464-73(33))(3)

=2464-73(36)+2464(3)-73(99)

=2464(4)-73(135)

1=2464(4)+73(-135)

Therefore:

1≡-135 mod 2464

1≡(-135+2464)mod 2464

1≡2329 mod 2464

(c)42828 mod 6407

The two numbers are not co-prime. In fact their g.c.d is 43.

Therefore their inverse does not exist.

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