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Karolina [17]
3 years ago
11

FIND THE THIRD,FIFTH AND TENTH TERM SEQUENCE DESCRIBED BY EACH RULE

Mathematics
1 answer:
Karolina [17]3 years ago
8 0
In order to solve for a nth term in an arithmetic sequence, we use the formula written as:<span>

an = a1 + (n-1)d

where an is the nth term, a1 is the first value in the sequence, n is the term position and d is the common difference.

</span><span>THIRD
</span><span>A3=4+(3-1)(-5)
A3 = -6

A(3)=-2(3-1)(-5)
A3 = 20
 </span><span>
FIFTH
</span>A5=4+(5-1)(-5)
A5 = -16

A(5)=-2(5-1)(-5)
A5 = 40<span>

TENTH
</span>A10=4+(10-1)(-5)
A10 = -41

A(10)=-2(10-1)(-5)
A10 = 90
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How do you simplify this?<br>x²y+xy² / y²+2/5 × xy​​​
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\huge \boxed{\mathbb{QUESTION} \downarrow}

  • How do you simplify this?
  • x²y+xy² / y²+2/5 × xy

\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}

\sf\frac{  { x  }^{ 2  }  y+x { y  }^{ 2  }    }{  { y  }^{ 2  }  + \frac{ 2  }{ 5  }   \times  xy  } \\

Factor the expressions that are not already factored.

_____

<u>How </u><u>to</u><u> factorise</u><u> </u><u>:</u><u>-</u>

<u>NUMERATOR</u> \downarrow

\sf \: x ^ { 2 } y + x y ^ { 2 }

Factor out xy.

\sf \: xy\left(x+y\right)

<u>DENOMINATOR</u> \downarrow

\sf{ y  }^{ 2  }  + \frac{ 2  }{ 5  }   \times  xy \\

Factor out 1/5.

\sf \: {\frac{1}{5}y\left(2x+5y\right)}  \\

_____

Continuing...

\sf\frac{xy\left(x+y\right)}{\frac{1}{5}y\left(2x+5y\right)}  \\

Cancel out y in both the numerator and denominator.

\sf\frac{x\left(x+y\right)}{\frac{1}{5}\left(2x+5y\right)}  \\

Expand the expression.

\sf\frac{x^{2}+xy}{\frac{2}{5}x+y}  \\

This can further simplified to as \downarrow

=    \boxed{\boxed{\bf\frac{5x\left(x  +y\right)}{2x+5y}}}

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