What's the square root of 68?

Mathematics

1 answer:

Aliun [14]3 years ago

3 0

$\sqrt{68}=\sqrt{4\cdot17}=\sqrt4\cdot\sqrt{17}=2\sqrt{17}\\\\\sqrt{68}\approx8.25$

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X + y = 30 2x + 4y = 84 Substitution or Elimination I

miv72 [106K]

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3 years ago

Solve 20x + 12 ≥ 14x + 30

Rashid [163]

Subtract 14x from both sides to get

$20x-14x+12 \geq 30$

Subtract 12 from both sides to get

$20x-14x \geq 30-12$

Now we have moved all terms involving x on one side, and all constant terms on the other. We can simplift both sides, i.e. sum like terms, to get

$6x \geq 18$

Now we have to divide both sides by 6. When dealing with inequality you have to be careful about dividing both sides by the same constant: if the constant is negative, the inequality side switches (i.e. $\geq \leftrightarrow \leq$ ). But this is not the case since 6 is positive, so we mantain the inequality sign: