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3 years ago

Acellus... can you help please

Mathematics

1 answer:

Vedmedyk [2.9K]3 years ago

8 0

Based on the image we can see that angle B and angle E will be the same so we can write this into an equation.

x + 4 = 48°

Now, we can solve for x.

x + 4 - 4 = 48 - 4

x = 44 (answer)

Best of Luck!

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Alenkasestr [34]

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We can now apply the distributive property for the both the numerator and denominator. We can see also that the denominator is the expansion of the difference of squares:

$\begin{gathered} \frac{(2+\sqrt[]{3})\cdot(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})-(2-\sqrt[]{3})\cdot(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2})^2-(\sqrt[]{2-\sqrt[]{3}}))^2} \\ \frac{(2+\sqrt[]{3})\cdot(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})+(\sqrt[]{3}-2)\cdot(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{2^{}-(2-\sqrt[]{3})^{}} \\ \frac{\sqrt[]{2}\cdot(2+\sqrt[]{3})-\sqrt[]{2-\sqrt[]{3}}\cdot(2+\sqrt[]{3})+\sqrt[]{2}\cdot(\sqrt[]{3}-2)+\sqrt[]{2-\sqrt[]{3}}\cdot(\sqrt[]{3}-2)}{2-2+\sqrt[]{3}} \\ \frac{\sqrt[]{2}(2+\sqrt[]{3}+\sqrt[]{3}-2)+\sqrt[]{2-\sqrt[]{3}}(-2-\sqrt[]{3}+\sqrt[]{3}-2)}{\sqrt[]{3}} \\ \frac{\sqrt[]{2}(2\sqrt[]{3})+\sqrt[]{2-\sqrt[]{3}}(-4)}{\sqrt[]{3}} \\ 2\sqrt[]{2}-4\frac{\sqrt[]{2-\sqrt[]{3}}}{\sqrt[]{3}} \end{gathered}$

We then can continue rearranging this as:

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1 year ago

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3 years ago

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