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Pavel [41]
3 years ago
7

(0,0) is a solution to y> -3x + 4

Mathematics
1 answer:
solniwko [45]3 years ago
7 0

Answer:

(0,0) is a solution.

Step-by-step explanation:

To see if this is true you can put (0,0) into the y and x variables like so: 0>-3(0)+4. When you solve this you get 0>4 so this statement is true

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What is net worth why do people say he or she is worth that?
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Answer:

The net worth of a person is the difference between a person's assets (non-financial and financial) that he/she owns and the debt that person owes.

Value of Assets - Debt and liabilities = Net Worth

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what are two methods you can use to find the area of a square. I WILL GIVE BRAINLIEST I NEED HELP QUICKKKKK
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Answer:

To find the area of a rectangle multiply its height by its width. For a square you only need to find the length of one of the sides (as each side is the same length) and then multiply this by itself to find the area. This is the same as saying length2 or length squared.

Step-by-step explanation:

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3 years ago
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3 years ago
(x^2 - x^(1/2))/(1-x^(1/2))
Levart [38]
\frac { \left( { x }^{ 2 }-{ x }^{ \frac { 1 }{ 2 }  } \right)  }{ \left( 1-{ x }^{ \frac { 1 }{ 2 }  } \right)  }

\\ \\ =\frac { \left( { x }^{ 2 }-\sqrt { x }  \right)  }{ \left( 1-\sqrt { x }  \right)  } \cdot 1

\\ \\ =\frac { \left( { x }^{ 2 }-\sqrt { x }  \right)  }{ \left( 1-\sqrt { x }  \right)  } \cdot \frac { \left( 1+\sqrt { x }  \right)  }{ \left( 1+\sqrt { x }  \right)  }

\\ \\ =\frac { { x }^{ 2 }+{ x }^{ 2 }\sqrt { x } -\sqrt { x } -x }{ 1+\sqrt { x } -\sqrt { x } -x }

\\ \\ =\frac { -\sqrt { x } \left( 1-{ x }^{ 2 } \right) -x\left( 1-x \right)  }{ \left( 1-x \right)  }

\\ \\ =\frac { -\sqrt { x } \left( 1+x \right) \left( 1-x \right) -x\left( 1-x \right)  }{ \left( 1-x \right)  }

\\ \\ =\frac { \left( 1-x \right) \left\{ -\sqrt { x } \left( 1+x \right) -x \right\}  }{ \left( 1-x \right)  }

\\ \\ =-\sqrt { x } \left( 1+x \right) -x\\ \\ =-{ x }^{ \frac { 1 }{ 2 }  }\left( 1+{ x }^{ \frac { 2 }{ 2 }  } \right) -x

\\ \\ =-{ x }^{ \frac { 1 }{ 2 }  }-{ x }^{ \frac { 3 }{ 2 }  }-x\\ \\ =-\sqrt { x } -\sqrt { { x }^{ 3 } } -x
3 0
3 years ago
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densk [106]

Answer:

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4 0
2 years ago
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