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

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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ROOTS AND CUBES Can someone please help me answer all 4 following?

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cubed (³) means multiply that number by itself 3 times
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Step-by-step explanation:

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