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WINSTONCH [101]
3 years ago
12

How many inches of lumber are remaining?​

Mathematics
1 answer:
belka [17]3 years ago
7 0

Answer:

I can't really answer this question because there is no picture of actual question but 1 foot = 12 inches so for instance we could do 6 inches + 3 inches = 9 inches.      

Step-by-step explanation: How we would solve this problem is simple.

6 + 3 = 9  if you have to turn this into a fraction it would be 9/12 = (3/4).

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Write the augmented matrix for each system of equations.
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Answer:

  see below

Step-by-step explanation:

List the coefficients and constant for an equation in one row of the matrix. The variables should be in the same order. Any missing terms are replaced by zero.

  \left[\begin{array}{ccc|c}9&-4&-5&9\\7&4&-4&-1\\6&-6&1&-5\end{array}\right]

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3 years ago
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zhuklara [117]
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15 liters of water for 40 flower pots
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Which set of ordered pairs represents y as a function of x? O {(-8,3), (0,7), (2,-4), (-3,7)} O {(-4,0), (-1,3), (-9,-3), (-4,-8
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Step-by-step explanation:

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8 0
3 years ago
Assume y≠60 which expression is equivalent to (7sqrtx2)/(5sqrty3)
Drupady [299]

Answer:

The equivalent will be:

\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)

Therefore, option 'a' is true.

Step-by-step explanation:

Given the expression

\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}

Let us solve the expression step by step to get the equivalent

\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}

as

\sqrt[7]{x^2}=\left(x^2\right)^{\frac{1}{7}}      ∵ \mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}

\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0

=x^{2\cdot \frac{1}{7}}

=x^{\frac{2}{7}}

also

\sqrt[5]{y^3}=\left(y^3\right)^{\frac{1}{5}}         ∵  \mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}

\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0

=y^{3\cdot \frac{1}{5}}

=y^{\frac{3}{5}}

so the expression becomes

\frac{x^{\frac{2}{7}}}{y^{\frac{3}{5}}}

\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}

=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)            ∵ \:\frac{1}{y^{\frac{3}{5}}}=y^{-\frac{3}{5}}

Thus, the equivalent will be:

\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)

Therefore, option 'a' is true.

5 0
3 years ago
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