Find the values of X and Y that makes these triangles congruent by the HL theorem

Mathematics

1 answer:

muminat3 years ago

8 0

Answer:

C. x = 3, y = 2

Step-by-step explanation:

If both triangles are congruent by the HL Theorem, then their hypotenuse and a corresponding leg would be equal to each other.

Thus:

x + 3 = 3y (eqn. 1) => equal hypotenuse

Also,

x = y + 1 (eqn. 2) => equal legs

✔️Substitute x = y + 1 into eqn. 1 to find y.

x + 3 = 3y (eqn. 1)

(y + 1) + 3 = 3y

y + 1 + 3 = 3y

y + 4 = 3y

y + 4 - y = 3y - y

4 = 2y

Divide both sides by 2

4/2 = 2y/2

2 = y

y = 2

✔️ Substitute y = 2 into eqn. 2 to find x.

x = y + 1 (eqn. 2)

x = 2 + 1

x = 3

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What is the simplified form of the following expression? Assume x greater-than-or-equal-to 0 and y greater-than-or-equal-to 0 2

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<h2 /><h2>Expression </h2>

In mathematics, an expression is defined as a set of numbers, variables, and mathematical operations that are formed according to rules which are dependent on the context.

<h2 /><h2 /><h2>Given to us,</h2>

$2\sqrt[4]{16x}-2\sqrt[4]{2y}+3\sqrt[4]{81x}-4\sqrt[4]{32y}+5\sqrt[4]{x}-4\sqrt[4]{32y}+5\sqrt[4]{x}-16\sqrt[4]{2y}+13\sqrt[4]{4}-10\sqrt[4]{2y}+35\sqrt[4]{x}-18\sqrt[4]{2y}$

rearranging we get,

$=2\sqrt[4]{16x}+3\sqrt[4]{81x}+5\sqrt[4]{x}+5\sqrt[4]{x}+13\sqrt[4]{4}+35\sqrt[4]{x}-2\sqrt[4]{2y}-4\sqrt[4]{32y}-4\sqrt[4]{32y}-16\sqrt[4]{2y}-10\sqrt[4]{2y}-18\sqrt[4]{2y}$

we know that,

$\sqrt[4]{81}=3\\\sqrt[4]{16}=2\\\sqrt[4]{32}=2\sqrt[4]{2}\\$

therefore,

$=4\sqrt[4]{x}+9\sqrt[4]{x}+5\sqrt[4]{x}+5\sqrt[4]{x}+13\sqrt[4]{x}+35\sqrt[4]{x}-2\sqrt[4]{2y}-8\sqrt[4]{2y}-8\sqrt[4]{2y}-16\sqrt[4]{2y}-10\sqrt[4]{2y}-18\sqrt[4]{2y}$