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STALIN [3.7K]
3 years ago
5

A jewelry maker needs 5 inch pieces of wire from sections that are each 4 feet in length. What is the greatest number of 5 inch

pieces of wire the jewelry maker can cut from 6 sections?
Mathematics
1 answer:
Damm [24]3 years ago
3 0

For this case we have that by definition:

1 foot equals 12 inches

Then, the sections measure:

1ft ------------> 12in

4ft ------------> x

Where "x" represents the inches of each section.

x = \frac {4 * 12} {1}\\x = 48

Thus, each section measures 48 inches.

If the pieces of wire to be removed must be 5 inches each, then from a section they are removed:

\frac {48} {5} = 9.6

Now, from each section you can remove 6 pieces of wire.

Then, from 6 sections, it takes 6 * 6 = 36 pieces of wire.

Answer:

36

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v=1/x

Step-by-step explanation:

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Give two different methods you could use to put the fractions 1/2, 3/4, 3/8, 5/8, 5/16, 7/16, 9/16, and 11/16 in order
alina1380 [7]
You could give all of the fractions a common denominator. You could also solve each of them (1/2=0.5) to see their value and order them through the value.
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vladimir1956 [14]

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3 years ago
What are the roots of this equation? x2-4x+9=0
zysi [14]

Considering the definition of zeros of a function, the zeros of the quadratic function f(x) = x² + 4x +9 do not exist.

<h3>Zeros of a function</h3>

The points where a polynomial function crosses the axis of the independent term (x) represent the so-called zeros of the function.

In summary, the roots or zeros of the quadratic function are those values ​​of x for which the expression is equal to 0. Graphically, the roots correspond to the abscissa of the points where the parabola intersects the x-axis.

In a quadratic function that has the form:

f(x)= ax² + bx + c

the zeros or roots are calculated by:

x1,x2=\frac{-b+-\sqrt{b^{2}-4ac } }{2a}

<h3>This case</h3>

The quadratic function is f(x) = x² + 4x +9

Being:

  • a= 1
  • b= 4
  • c= 9

the zeros or roots are calculated as:

x1=\frac{-4+\sqrt{4^{2}-4x1x9 } }{2x1}

x1=\frac{-4+\sqrt{16-36 } }{2x1}

x1=\frac{-4+\sqrt{-20 } }{2x1}

and

x2=\frac{-4-\sqrt{4^{2}-4x1x9 } }{2x1}

x2=\frac{-4-\sqrt{16-36 } }{2x1}

x2=\frac{-4-\sqrt{-20} }{2x1}

If the content of the root is negative, the root will have no solution within the set of real numbers. Then \sqrt{-20} has no solution.

Finally, the zeros of the quadratic function f(x) = x² + 4x +9 do not exist.

Learn more about the zeros of a quadratic function:

brainly.com/question/842305

brainly.com/question/14477557

#SPJ1

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Answer:

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