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3 years ago

PLZ HELP!!!!! Which of these quadrilaterals can always be classified as a rhombus? A) kite

Mathematics

1 answer:

zaharov [31]3 years ago

3 0

Answer:

A) Kite

A rhombus is a simply a diamond. Think about the visual properties of a diamond. Isn't a kite a diamond?

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liq [111]

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Step-by-step explanation: Isolate the variable by dividing each side by factors that doesn't contain the variable.

Hope this helps!

7 0

3 years ago

Read 2 more answers

You are given the parametric equations x=2cos(θ),y=sin(2θ). (a) List all of the points (x,y) where the tangent line is horizonta

vladimir1956 [14]

Answer:

The solutions listed from the smallest to the greatest are:

x: $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$ $\sqrt{2}$

y: -1 1 -1 1

Step-by-step explanation:

The slope of the tangent line at a point of the curve is:

$m = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$m = -\frac{\cos 2\theta}{\sin \theta}$

The tangent line is horizontal when $m = 0$ . Then:

$\cos 2\theta = 0$

$2\theta = \cos^{-1}0$

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$\theta = \frac{1}{2}\cdot \left(\frac{\pi}{2}+i\cdot \pi \right)$ , for all $i \in \mathbb{N}_{O}$

$\theta = \frac{\pi}{4} + i\cdot \frac{\pi}{2}$ , for all $i \in \mathbb{N}_{O}$

The first four solutions are:

x: $\sqrt{2}$ $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$

y: 1 -1 1 -1

The solutions listed from the smallest to the greatest are:

x: $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$ $\sqrt{2}$

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2 years ago

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3 years ago

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RSB [31]

Answer:

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Step-by-step explanation:

20/4 = 5

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2 years ago