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4 years ago

Find the values of x and y that makes these triangles congruent by the HL therom.

Mathematics

1 answer:

Maurinko [17]4 years ago

4 0

(1) x = 2y
(2) x + 4 = 3y + 1

subtitute (1) to (2)

2y + 4 = 3y + 1 |subtract 4 from both sides

2y = 3y - 3 |subtract 3y from both sides

-y = -3 |change the signs

y = 3

subtitute the value of y to (1)

x = 2 · 3
x = 6

Answer: x = 6 and y = 3.

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Option 2. $x = \frac{4 \pm \sqrt{(-4)^{2}-4 (1)(-21)}}{2 (1)}$ shows the correct way to use the quadratic formula to solve the given equation.

Step-by-step explanation:

Step 1:

For an equation of the form $ax^{2} +bx+c=0$ the solution is $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$ .

Here a is the coefficient of $x^{2}$ , b is the coefficient of x and c is the constant term.

$x^{2} -4x=21$ can also be written as $x^{2} -4x-21=0.$

Comparing $x^{2} -4x-21=0$ with $ax^{2} +bx+c=0$ , we get that a is 1, b is -4 and c is -21.

To get the solution, we substitute the values of a, b, and c in $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$ .

Step 2:

Substituting the values, we get

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}= \frac{-(-4) \pm \sqrt{(-4)^{2}-4 (1)(-21)}}{2 (1)}.$

$\frac{-(-4) \pm \sqrt{(-4)^{2}-4 (1)(-21)}}{2 (1)} = \frac{4 \pm \sqrt{(-4)^{2}-4 (1)(-21)}}{2 (1)}.$

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3 years ago

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Answer:

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Step-by-step explanation:

For the first image, you have to do csc⁻¹ 2 [OR sin⁻¹ ½] because you are solving for an angle measure. When evaluated, you get 30°.

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Extended information on Trigonometric Ratios

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