What is the missing justification in the proof?

Mathematics

2 answers:

nataly862011 [7]3 years ago

7 0

The missing justification in the proof is
<span>B) Substitution property of equality

The expression for sin</span>² x and cos² x is substituted to the other side of the equation. Since sin x = a/c, then sin² x = a²/c². Similarly, since cos x = b/c, then cos² x = b²/c². Adding to two results to the third statement.

Elenna [48]3 years ago

5 0

The <em>correct answer</em> is:

B) Substitution property of equality

Explanation:

sin²x and cos²x have been added together, making

sin²x + cos²x.

Since we know that sin x = a/c, this means that sin²x = (a/c)² = a²/c².

Since we know that cos x = b/c, this means that cos²x = (b/c)² = b²/c².

These two values have been substituted for sin²x + cos²x, making

a²/c² + b²/c².

Since we substituted, this means it is the substitution property of equality.

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i got it correct:)

7 0

3 years ago

5 less than the quotient of a number y and 2

Marta_Voda [28]

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5 0

3 years ago

What are the zeros of the function f(x) = x2 + 5x + 5 written in simplest radical form?

Pavel [41]

$\boxed{x_{1}=\frac{-5 + \sqrt{5}}{2}} \\ \\ \\ \boxed{x_{2}=\frac{-5 - \sqrt{5}}{2}}$

<h2>Explanation:</h2>

Using the quadratic formula:

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ \\ Here: \\ \\ f(x) = x^2 + 5x + 5 \\ \\ \\ So: \\ \\ a=1 \\ \\ b=5 \\ \\ c=5 \\ \\ \\ x=\frac{-5 \pm \sqrt{5^2-4(1)(5)}}{2(1)} \\ \\ x=\frac{-5 \pm \sqrt{25-20}}{2} \\ \\ x=\frac{-5 \pm \sqrt{5}}{2} \\ \\ \\ Two \ solutions: \\ \\ \boxed{x_{1}=\frac{-5 + \sqrt{5}}{2}} \\ \\ \\ \boxed{x_{2}=\frac{-5 - \sqrt{5}}{2}}$