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1 year ago

KM and NP are parallel lines. which angles are supplementary angles?

Mathematics

1 answer:

Oksana_A [137]1 year ago

5 0

Supplementary angles = The angles add up to 180 degrees / form a straight line.

Angle KLO and angle KLJ form a straight line -> they are supplementary angles.

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Midpoint of R(-4,3) and S(5,6)

nirvana33 [79]

Answer:

(0,4.5)

Step-by-step explanation:

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2 years ago

There is no solution to the equation sec x = 0

kondaur [170]

The correct answer for the given statement above would be TRUE. It is true that there is no solution to the equation sec x = 0. Why?
<span>Sec(x) is actually 1/cos(x), which can't be absolute zero. Cos(x) ranges between -1 and 1; it would have to be unbounded for sec(x) to reach 0, or in short, it is undefined. Hope this answer helps. </span>

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2 years ago

The figure represents a traffic island that has angles measuring 60°, 20°, and 100°. Enter each angle next to its correct measur

suter [353]

Answer:

possblie awnser is 1:Possible answer for Triangle 1: m∠A = 70°; m∠B = ∠55°; m∠C = 55°.

The compass marks equal lengths on both sides of ∠A; therefore, AB

Step-by-step explanation:

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2 years ago

Read 2 more answers

Which of the following equations are linear or nonlinear?

gavmur [86]

Answer:

1. Linear

2. Nonlinear

3. Linear

4. Linear

5. Nonlinear

Step-by-step explanation:

Thies equation are in the form y=mx+b. Where m is the slope and b is the y-intercept. To solve them you neeed to see the power x is taken to. If the power on x is bigger than 1 than one, then it will curve at some point. This makes 2 and 5 nonlinear

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2 years ago

Which expression is equivalent to square root 55x^7y6/11x^11y^8

Novay_Z [31]

Answer:

$\large\boxed{\sqrt{\dfrac{55x^7y^6}{11x^{11}y^8}}=\dfrac{\sqrt5}{x^2y}}$

Step-by-step explanation:

$\sqrt{\dfrac{55x^7y^6}{11x^{11}y^8}}=\sqrt{\dfrac{55}{11}\cdot\dfrac{x^7}{x^{11}}\cdot\dfrac{y^6}{y^8}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\=\sqrt{5x^{7-11}y^{6-8}}=\sqrt{5x^{-4}y^{-2}}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\sqrt5\cdot\sqrt{x^{-4}}\cdot\sqrt{y^{-2}}=\sqrt5\cdot\sqrt{x^{(-2)(2)}}\cdot\sqrt{y^{(-1)(2)}}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=\sqrt5\cdot\sqrt{(x^{-2})^2}\cdot\sqrt{(y^{-1})^2}\qquad\text{use}\ \sqrt{a^2}=a$

$=\sqrt5\cdot x^{-2}\cdot y^{-1}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\to a^{-1}=\dfrac{1}{a}\\\\=\sqrt5\cdot\dfrac{1}{x^2}\cdot\dfrac{1}{y}=\dfrac{\sqrt5}{x^2y}$

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2 years ago