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bija089 [108]
3 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]3 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]3 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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If two angles are complementary. One angle measures 22 degrees. And the other angle measures 2x degrees, what is the value of x?
levacccp [35]

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Step-by-step explanation:

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5 0
3 years ago
Find the vectors T, N, and B at the given point. r(t) = < t^2, 2/3t^3, t >, (1, 2/3 ,1)
maxonik [38]

Answer with Step-by-step explanation:

We are given that

r(t)=< t^2,\frac{2}{3}t^3,t >

We have to find T,N and B at the given point t > (1,2/3,1)

r'(t)=

\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1

T(t)=\frac{r'(t)}{\mid r'(t)\mid}=\frac{}{2t^2+1}

Now, substitute t=1

T(1)=\frac{}{2+1}=\frac{1}{3}

T'(t)=\frac{-4t}{(2t^2+1)^2} +\frac{1}{2t^2+1}

T'(1)=-\frac{4}{9}+\frac{1}{3}

T'(1)=\frac{1}{9}=

\mid T'(1)\mid=\sqrt{(\frac{-2}{9})^2+(\frac{4}{9})^2+(\frac{-4}{9})^2}=\sqrt{\frac{36}{81}}=\frac{2}{3}

N(1)=\frac{T'(1)}{\mid T'(1)\mid}

N(1)=\frac{}{\frac{2}{3}}=

N(1)=

B(1)=T(1)\times N(1)

B(1)=\begin{vmatrix}i&j&k\\\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{-1}{3}&\frac{2}{3}&\frac{-2}{3}\end{vmatrix}

B(1)=i(\frac{-4}{9}-\frac{2}{9})-j(\frac{-4}{9}+\frac{1}{3})+k(\frac{4}{9}+\frac{2}{9})

B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k

B(1)=\frac{1}{3}

5 0
3 years ago
PLEASE HELP ASAP!!!!!!!!!!!
irina [24]

Answer:

132 in^2 but DOUBLE CHECK MY WORK!!!!

Step-by-step explanation:

Break it down into parts ok. So I see a 10x10 square, two 2x4 right triangles and a 4x6 rectangle.

10x10 is 100 in

1/2(2*4) = 4 x 2 = 8 in

6x4 = 24 in

add them together to get 132 in^2

7 0
3 years ago
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