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3 years ago

I need help with algebra 2 pleasee

Mathematics

1 answer:

Aleonysh [2.5K]3 years ago

5 0

Answer:

$x=\pm \sqrt{21}/7$

Step-by-step explanation:

We can add 9 to both sides of the equation to get $21x^2=9$ .

We can divide both sides of the equation by 21 to get $x^2=9/21$ .

We take the square root of both sides of the equation to get $x=\pm \sqrt{9/21}$ .

However, that isn't in the form we need.

We know that $\sqrt{9/21}=\sqrt{9}/\sqrt{21}=3/\sqrt{21}=3\sqrt{21}/(\sqrt{21}*\sqrt{21})=3\sqrt{21}/21=\sqrt{21}/7$

so the answer is really $x=\pm \sqrt{21}/7$ and we're done!

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2 years ago

The sum of a 3 digit number and a 1 digit number is 217. The product of the numbers is 642. If one number is between 200 and 225

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Let, the numbers be "x" and "y". Consider "x" as the triple digit number.
Now, according to the question,
x + y = 217..........................equation (1)

x * y = 642............................equation (2)

Now,
Taking equation (2),
x * y = 642
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Substituting the value of y from equation (3), we get,

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Using zero product property,
EITHER,
x - 3 = 0
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x - 214 = 0
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Taking equation (2),
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