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ivolga24 [154]
3 years ago
15

Those are the math questions I need the answers of ​

Mathematics
1 answer:
Elena L [17]3 years ago
7 0

Answer:

1)

12\sqrt{18}-3\sqrt{50}+\sqrt{32}

12\sqrt{9}\sqrt{2}-3\sqrt{25}   \sqrt{2} +\sqrt{16} \sqrt{2}

12\times3\sqrt{2} -3\times5\sqrt{2}+4\sqrt{2}

36\sqrt{2}-15\sqrt{2}+4\sqrt{2}

25\sqrt{2}

2)

\frac{1}{5+2\sqrt{6}}

Multiply 5-2\sqrt{6}:

\frac{1\times \left(5-2\sqrt{6}\right)}{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}

Apply Difference of Squares formula: (a+b)(a-b)=a^2-b^2

\frac{5-2\sqrt{6}}{5^2-\left(2\sqrt{6}\right)^2}

\frac{5-2\sqrt{6}}{25-2^2\left(\sqrt{6}\right)^2}

\frac{5-2\sqrt{6}}{25-4\times6}

\frac{5-2\sqrt{6}}{1}

{5-2\sqrt{6}}

a = 5, b = -2

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PLEASE HELP!! Ill mark the brainliest!!!!
ira [324]

Answer:

x = 5√2

y = 5√6

z = 5√3

ΔABC ~ ΔBDC ~ ΔADB

Step-by-step explanation:

ΔABC, ΔBDC, and ΔADB are all similar triangles to each other.

By definition of similar triangles, the corresponding sides have the same ratios.

CD from ΔBDC corresponds to BD from ΔADB, and BD from ΔBDC corresponds to AD from ΔADB. So:

CD / BD = BD / AD

10 / x = x / 5

x² = 50

x = 5√2

Since ΔBDC is right, we can use the Pythagorean Theorem to solve for y:

CD² + BD² = BC²

10² + (5√2)² = y²

y² = 100 + 50 = 150

y = 5√6

Again, since ΔΔABD is right, we can use the Pythagorean Theorem to solve for z:

AD² + BD² = AB²

5² + (5√2)² = z²

z² = 25 + 50 = 75

z = 5√3

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Answer:

Step-by-step explanation:

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k = 8

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,  Hope this helps :)

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