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balandron [24]
2 years ago
9

measure the lines with small paper clips and then with a centimeter ruler. Line 1, has 3 paper clips = 12 cm with ruler. Line 2,

has 2 paper clips = to 6 cm with a ruler. Line 3 has 2 paper clips = 9 cm ruler. explain why each measurements require more centimeters than paper clips
Mathematics
1 answer:
liberstina [14]2 years ago
3 0

Answer:

Its because the paper clips used here are of same length (4cm long).

So the measurements of lines two and three need to be multiples of 4 to me measured accurately with the paper.

i.e Line two needs to be 6cm + 2cm = 8cm (eight is a muliple of 4)

Line three needs to be 9cm + 3cm = 12cm (twelve is a multiple of 4)

Step-by-step explanation:

Since line 1 has 3 paper clips and its 12cm in length, the size of the paper clips is 4cm.

Line 2 is 6cm long and will require two more centimeters to measured accurately by the 4cm-paper clips.

Line 3 is 9cm and will require three more centimeters to be measured accurately by the 4cm-paper clips.

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Juanita will work for 4 7/8 hours at her job on Saturday. if she also volunteers for 2 1/3 hours at the hospital, how many hours
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Step-by-step explanation:

8 0
3 years ago
[10] In the following given system, determine a matrix A and vector b so that the system can be represented as a matrix equation
irina1246 [14]

Answer:

y=-\frac{158}{579}

Step-by-step explanation:

To find the matrix A, took all the numeric coefficient of the variables, the first column is for x, the second column for y, the third column for z and the last column for w:

A=\left[\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right]

And the vector B is formed with the solution of each equation of the system:b=\left[\begin{array}{c}3\\-3\\6\\1\end{array}\right]

To apply the Cramer's rule, take the matrix A and replace the column assigned to the variable that you need to solve with the vector b, in this case, that would be the second column. This new matrix is going to be called A_{2}.

A_{2}=\left[\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right]

The value of y using Cramer's rule is:

y=\frac{det(A_{2}) }{det(A)}

Find the value of the determinant of each matrix, and divide:

y==\frac{\left|\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right|}{\left|\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right|} =\frac{158}{-579}

y=-\frac{158}{579}

7 0
3 years ago
Which expression is equivalent to (16 x Superscript 8 Baseline y Superscript negative 12 Baseline) Superscript one-half?.
loris [4]

To solve the problem we must know the Basic Rules of Exponentiation.

<h2>Basic Rules of Exponentiation</h2>
  • x^ax^b = x^{(a+b)}
  • \dfrac{x^a}{x^b} = x^{(a-b)}
  • (a^a)^b =x^{(a\times b)}
  • (xy)^a = x^ay^a
  • x^{\frac{3}{4}} = \sqrt[4]{x^3}= (\sqrt[3]{x})^4

The solution of the expression is \dfrac{4x^4}{y^6}.

<h2>Explanation</h2>

Given to us

  • (16x^8y^{12})^{\frac{1}{2}}

Solution

We know that 16 can be reduced to 2^4,

=(2^4x^8y^{12})^{\frac{1}{2}}

Using identity (xy)^a = x^ay^a,

=(2^4)^{\frac{1}{2}}(x^8)^{\frac{1}{2}}(y^{12})^{\frac{1}{2}}

Using identity (a^a)^b =x^{(a\times b)},

=(2^{4\times \frac{1}{2}})\ (x^{8\times\frac{1}{2}})\ (y^{12\times{\frac{1}{2}}})

Solving further

=2^2x^4y^{-6}

Using identity \dfrac{x^a}{x^b} = x^{(a-b)},

=\dfrac{2^2x^4}{y^6}

=\dfrac{4x^4}{y^6}

Hence, the solution of the expression is \dfrac{4x^4}{y^6}.

Learn more about Exponentiation:

brainly.com/question/2193820

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lesya [120]

Answer:

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Step-by-step explanation:

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mr_godi [17]
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