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seropon [69]
3 years ago
8

Which monomial is a perfect cube?

Mathematics
2 answers:
neonofarm [45]3 years ago
5 0

Answer:

Option (a) is correct.

1x^3 is a perfect cube.

Step-by-step explanation:

 Given : Some Monomial

We have to choose out of given monomial that is a perfect cube.

Perfect cubes are numbers that can be written as (ab)^3

So, consider 1x^3 , we know 1 can be written as 1^3

Thus,  1x^3 can be written as  1^3x^3=(1x)^3

Thus, Option (a) is correct.

fenix001 [56]3 years ago
4 0

Answer:

Option A is correct

1x^3

Step-by-step explanation:

Perfect cube is a number that is the cube of any integer number.

For example:

7^3 = 343

(ab)^3 = a^3b^3

From the given options we have to find which monomial is a perfect cube.

Consider the  monomial 1x^3.

We can write 1 as:

1 = 1 \cdot 1 \cdot 1 = 1^3

It can be written as:

1x^3 = 1^3 \cdot x^3 = (1x)^3

Therefore,  1x^3 monomial is a perfect cube

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omeli [17]

Answer:

1/4

Step-by-step explanation:

We can find the slope of a line given two points from the formula

m = (y2-y1)/(x2-x1)

   = (-2--1)/(-12--8)

   = (-2+1)/(-12+8)

    =-1/-4

   = 1/4

6 0
3 years ago
What is 5.316 - 1.942 (show ur work)
Fiesta28 [93]

Answer:

3.374

Step-by-step explanation:

\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}

\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}

\begin{matrix}\:\:&5&.&3&1&6\\ -&1&.&9&4&2\end{matrix}

\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}

\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:6-2=4

\frac{\begin{matrix}\:\:&5&.&3&1&\textbf{6}\\ -&1&.&9&4&\textbf{2}\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\textbf{4}\end{matrix}}

\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}

\frac{\begin{matrix}\:\:&5&.&3&\textbf{1}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}

\mathrm{The\:bottom\:number\:is\:larger\:than\:the\:upper\:number.\:\:Try\:to\:'borrow'\:a\:digit\:from\:the\:left.}

\mathrm{The\:top\:digit\:is\:not\:bigger\:than\:the\:bottom\:one.\:\:Try\:to\:'borrow'\:a\:digit\:from\:the\:left.}

\frac{\begin{matrix}\:\:&5&.&\textbf{3}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\mathrm{Borrow\:}1\mathrm{\:from\:}5\mathrm{.\:\:The\:remainder\:is\:}4

\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&10&\:\:&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&3&1&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{\:\:}&\:\:&\:\:&\:\:&4\end{matrix}}

\mathrm{Add\:}1\mathrm{\:ten\:to\:}3:\quad \:10+3=13

\frac{\begin{matrix}\:\:&4&\:\:&\textbf{13}&\:\:&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{3}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\mathrm{Borrow\:}1\mathrm{\:from\:}13\mathrm{.\:\:The\:remainder\:is\:}12

\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&10&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\mathrm{Add\:}1\mathrm{\:ten\:to\:}1:\quad \:10+1=11

\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}

\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:11-4=7

\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{7}&4\end{matrix}}

\mathrm{Place\:the\:decimal\:point\:in\:the\:answer\:directly\:below\:the\:decimal\:points\:in\:the\:terms}

\frac{\begin{matrix}\:\:&4&\textbf{\:\:}&12&11&\:\:\\ \:\:&\linethrough{5}&\textbf{.}&\linethrough{13}&\linethrough{1}&6\\ -&1&\textbf{.}&9&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\textbf{.}&3&7&4\end{matrix}}

\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:4-1=3

\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&12&11&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&\linethrough{13}&\linethrough{1}&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{3}&.&3&7&4\end{matrix}}

=3.374

Hence the correct answer is 3.374

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