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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?

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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Math question screen shot down below

ser-zykov [4K]

Answer:

11

Step-by-step explanation:

11,16,12,7,23,9,5,5

arrange to least to greatest

5,5,7,9,11 ,12,16,23

2 middle numbers

11+12= 23

23÷2= 11.5

Median is 11

3 0

3 years ago

Read 2 more answers

3<br> ..<br> 4) through: (-2,5), slope<br> 2.

ArbitrLikvidat [17]

Answer:

y - 5 = 2(x + 2)

General Formulas and Concepts:

Point-Slope Form: y - y₁ = m(x - x₁)

x₁ - x coordinate

y₁ - y coordinate

m - slope

Step-by-step explanation:

<u>Step 1: Define</u>

Slope <em>m</em> = 2

Point (-2, 5)

<u>Step 2: Create equation</u>

y - 5 = 2(x + 2)

8 0

3 years ago

Please answer<br> ~~~~~~~~~~~~<br> •••••••••••••••••

andrew-mc [135]

So to answer you just have to substitute all the given choices into the formula and see which come out true.
2(1) - 3(-3) >_ 12
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XNXOXPXEX

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VYVEVSV

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XNXOXPXEX

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4 0

4 years ago

At a college, 69% of the courses have final exams and 42% of courses require research papers. Suppose that 29% of courses have a

laila [671]

Answer:

a) 0.82

b) 0.18

Step-by-step explanation:

We are given that

P(F)=0.69

P(R)=0.42

P(F and R)=0.29.

a)

P(course has a final exam or a research paper)=P(F or R)=?

P(F or R)=P(F)+P(R)- P(F and R)

P(F or R)=0.69+0.42-0.29

P(F or R)=1.11-0.29

P(F or R)=0.82.

Thus, the the probability that a course has a final exam or a research paper is 0.82.

b)

P( NEITHER of two requirements)=P(F' and R')=?

According to De Morgan's law

P(A' and B')=[P(A or B)]'

P(A' and B')=1-P(A or B)

P(A' and B')=1-0.82

P(A' and B')=0.18

Thus, the probability that a course has NEITHER of these two requirements is 0.18.

3 0

3 years ago

Express the function as the sum of a power series by first using partial fractions. f(x) = 8 x2 − 4x − 12 f(x) = ∞ n = 0 find th

alexandr1967 [171]

I'm guessing the function is

$f(x)=\dfrac8{x^2-4x-12}=\dfrac8{(x-6)(x+2)}$

which, split into partial fractions, is equivalent to

$\dfrac1{x-6}-\dfrac1{x+2}$

Recall that for $|x|$ we have

$\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n$

With some rearranging, we find

$\dfrac1{x-6}=-\dfrac16\dfrac1{1-\frac x6}=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n$

valid for $\left|\dfrac x6\right|$ , or $|x|$ , and

$\dfrac1{x+2}=\dfrac12\dfrac1{1-\left(-\frac x2\right)}=\displaystyle\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n$

valid for $\left|-\dfrac x2\right|$ , or $|x|$ .

So we have

$f(x)=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n-\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n$

$f(x)=\displaystyle-\sum_{n=0}^\infty\left(\frac{x^n}{6^{n+1}}+\frac{(-x)^n}{2^{n+1}}\right)$

$f(x)=\displaystyle-\sum_{n=0}^\infty\frac{1+3(-3)^n}{6^{n+1}}x^n$

Taken together, the power series for $f(x)$ can only converge for $|x|$ , or $-2$ .

6 0

3 years ago