Find x. and rounddddd
1 answer:
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Hello,
In details:
s= 1 +3 +5+7+9+.... +2(k-2)+1 +2(k-1)+1
s= 2(k-1)+1 +2(k-2)+1 +.... +9+7+5 +3 +1
2s=(2k-2+2) +(2k-4+4)+...........................................+(2k-2+2)
=2k+2k+2k+....+2k (there is k terms)
2s=2k*k
2s=2k²
s=k²
Answer J
In details:
s= 1 +3 +5+7+9+.... +2(k-2)+1 +2(k-1)+1
s= 2(k-1)+1 +2(k-2)+1 +.... +9+7+5 +3 +1
2s=(2k-2+2) +(2k-4+4)+...........................................+(2k-2+2)
=2k+2k+2k+....+2k (there is k terms)
2s=2k*k
2s=2k²
s=k²
Answer J
The correct answer is: [A]: " 6 to the power of (1 over 6) " .
Explanation:
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√ [(∛6)] = √ [6⁽¹/³⁾ ] = [6⁽¹/²⁾ ] ⁽¹/³⁾ = 6^ ( ⁽¹/²⁾ * ⁽¹/³⁾ ) = ???? ;
<u>Note</u>: ;
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→ 6^ ( ⁽¹/²⁾ * ⁽¹/³⁾ ) ;
= 6^ ( ⁽¹/⁶⁾ ) ;
→ which is: Answer choice: [A]: " 6 to the power of (1 over 6) " .
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