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ipn [44]
3 years ago
6

find three consecutive integers such that the sum of twice he smallest and 3 times the largest is 126.

Mathematics
1 answer:
vladimir1956 [14]3 years ago
8 0
a,a+1,a+2\ \ -\ three\ consecutive\ integers\\\\
2a+3(a+2)=126\\\\
2a+3a+6=126\\\\
5a+6=126\ \ \ |Subtract\ 6\\\\
5a=120\ \ \ |Divide\ by\ 5\\\\
a=24\\\\Numbers\ are:\ 24,25,26.
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3 years ago
5=4b-12 hurry and make sure you put how you got the answer
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Read 2 more answers
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Alex Ar [27]

NOTES:

  • squared (²) means multiply that number by itself 2 times
  • cubed (³) means multiply that number by itself 3 times
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Answer: (C) 41

<u>Step-by-step explanation:</u>

\quad 6^2+\sqrt[3]{125} \\= 6 \cdot 6+\sqrt[3]{5\cdot 5 \cdot 5}\\= 36 + 5\\= 41

***************************************************************************************

Answer: (C) 10

<u>Step-by-step explanation:</u>

   \bigg(\dfrac{7}{3}\times \sqrt[3]{27}-2\bigg)\times \dfrac{1}{5} + \sqrt{81}

=\bigg(\dfrac{7}{3}\times \sqrt[3]{3\cdot 3 \cdot 3}-2\bigg)\times \dfrac{1}{5} + \sqrt{9\cdot 9}

=\bigg(\dfrac{7}{3}\times3-2\bigg)\times \dfrac{1}{5}+9

=(7 - 2)\times \dfrac{1}{5}+9

=5 \times \dfrac{1}{5}+ 9

= 1 + 9

= 10

***************************************************************************************

8² = 8 · 8 = 64                                     11² = 11 · 11 = 121

5³ = 5 · 5 · 5 = 125                               3³ = 3 · 3 · 3 = 27

\sqrt{1600}=\sqrt{40\cdot 40}=40                       \sqrt[3]{64}=\sqrt[3]{4\cdot 4\cdot 4}=4

\sqrt{144}=\sqrt{12\cdot 12}=12                            \sqrt[3]{8000}=\sqrt[3]{20\cdot 20\cdot 20}=20

3 0
3 years ago
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