Which two angles in the figure are obtuse angles?

Mathematics

1 answer:

ValentinkaMS [17]2 years ago

4 0

Answer: C. ∠SQP and ∠TQR

Step-by-step explanation:

Since, by the given diagram,

m∠TQR = m∠SQP (vertical angles)

And, m∠TQP < m∠TQR

By adding ∠TQR on both sides,

m∠TQP + m∠TQR < m∠TQR+m∠TQR

⇒m∠TQP + m∠TQR < 2 ∠TQR

⇒ 180° < 2 m∠TQR ( Because ∠TQP and ∠TQR are linear angles)

⇒ 90° < m∠TQR

Hence ∠TQR is a obtuse angle,

⇒ ∠SQP also is a obtuse angle.

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1. Simplify

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2 years ago

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yaroslaw [1]

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<h3>How to simplify a composite algebraic expression</h3>

In this question we must use properties of <em>real</em> algebra to simplify a given expression as most as possible. We proceed to show each step with respective reason.

$\frac{m^{2}-16\cdot n^{2}}{\frac{3\cdot m + 12\cdot n}{m\cdot n} }$ Given
$\frac{m\cdot n \cdot (m^{2}-16\cdot n^{2})}{3\cdot m + 12\cdot n}$ $\frac{\frac{a}{b} }{\frac{c}{d} } = \frac{a\cdot d}{c\cdot b}$
$\frac{m\cdot n \cdot (m-4\cdot n)\cdot (m+4\cdot n)}{3\cdot (m+4\cdot n)}$ Distributive property/ $a^{2}-b^{2} = (a +b)\cdot (a-b)$
$\frac{m\cdot n \cdot (m-4\cdot n)}{3}$ Existence of the multiplicative inverse/Modulative property
$\frac{m^{2}\cdot n}{3} -\frac{4\cdot m\cdot n^{2}}{3}$ Distributive property/ $x^a\cdot x^{b} = x^{a+b}$ /Result