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3 years ago

Solve x^2-3=0 using a quadratic equation

Mathematics

1 answer:

anzhelika [568]3 years ago

3 0

Answer:

I think the answers are 3 & 0 .

Step-by-step explanation:

Good luck .

You might be interested in

Which statement is true?

love history [14]

<h2>Hello!</h2>

The answer is:

The second option,

$(\sqrt[m]{x^{a} } )^{b}=\sqrt[m]{x^{ab} }$

Discarding each given option in order to find the correct one, we have:

<h2>First option,</h2>

$\sqrt[m]{x}\sqrt[m]{y}=\sqrt[2m]{xy}$

The statement is false, the correct form of the statement (according to the property of roots) is:

$\sqrt[m]{x}\sqrt[m]{y}=\sqrt[m]{xy}$

<h2>Second option,</h2>

$(\sqrt[m]{x^{a} } )^{b}=\sqrt[m]{x^{ab} }$

The statement is true, we can prove it by using the following properties of exponents:

$(a^{b})^{c}=a^{bc}$

$\sqrt[n]{x^{m} }=x^{\frac{m}{n} }$

We are given the expression:

$(\sqrt[m]{x^{a} } )^{b}$

So, applying the properties, we have:

$(\sqrt[m]{x^{a} } )^{b}=(x^{\frac{a}{m}})^{b}=x^{\frac{ab}{m}}\\\\x^{\frac{ab}{m}}=\sqrt[m]{x^{ab} }$

Hence,

$(\sqrt[m]{x^{a} } )^{b}=\sqrt[m]{x^{ab} }$

<h2>Third option,</h2>

$a\sqrt[n]{x}+b\sqrt[n]{x}=ab\sqrt[n]{x}$

The statement is false, the correct form of the statement (according to the property of roots) is:

$a\sqrt[n]{x}+b\sqrt[n]{x}=(a+b)\sqrt[n]{x}$

<h2>Fourth option,</h2>

$\frac{\sqrt[m]{x} }{\sqrt[m]{y}}=m\sqrt{xy}$

The statement is false, the correct form of the statement (according to the property of roots) is:

$\frac{\sqrt[m]{x} }{\sqrt[m]{y}}=\sqrt[m]{\frac{x}{y} }$

Hence, the answer is, the statement that is true is the second statement:

$(\sqrt[m]{x^{a} } )^{b}=\sqrt[m]{x^{ab} }$

Have a nice day!

6 0

3 years ago

Please help me find perimeter and area of this figure rounded to the nearest whole number! I will award brainliest!

Mice21 [21]

Brainly is erroneously preventing my response from being posted due to nonexistent profanity or links. So, I have converted my answer into an image file, which I have attached. I also attached an annotated diagram of the figure in the image you posted. Please let me know if you have trouble accessing either. Hope this helps.