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3 years ago

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According to the Venn diagram what type of parallelogram is both a rectangle and a rhombus

2 answers:

julia-pushkina [17]3 years ago

7 0

According to the properties of a parallelogram.

Its opposite angles are equal and its opposite sides are also equal.

Now for a parallelogram to be a rectangle,

All its angles must be 90°.

But we also want the parallelogram to be a rhombus, which means all its sides must be equal.

Now any rectangle whose all sides are equal becomes a square.

And so parallelogram which is both a rectangle and a rhombus is a SQUARE.

Brut [27]3 years ago

4 0

A square can be both a rectangle and a rhombus

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Is this right? Algebra II

Doss [256]

I think it is. Dont take my word for it tho

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4 years ago

Find the exact value

Mnenie [13.5K]

By using properties for <em>trigonometric</em> functions and <em>trigonometric</em> expressions, we find that the <em>exact</em> value of the sine of the angle 5π/12 radians is $\frac{\sqrt{2+\sqrt{3}}}{2}$ .

<h3>How to find the exact value of a trigonometric expression</h3>

<em>Trigonometric</em> functions are <em>trascendent</em> functions, these are, that cannot be described algebraically. Herein we must utilize <em>trigonometric</em> formulae to calculate the <em>exact</em> value of a <em>trigonometric</em> function:

$\sin \frac{5\pi}{12} = \sqrt{\frac{1 - \cos \frac{5\pi}{6} }{2} }$

$\sin \frac{5\pi}{12} = \sqrt{\frac{1 + \cos \frac{\pi}{6} }{2} }$

$\sin \frac{5\pi}{12} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2} }{2} }$

$\sin \frac{5\pi}{12} = \sqrt{\frac{2+\sqrt{3}}{4} }$

$\sin \frac{5\pi}{12} = \frac{\sqrt{2+\sqrt{3}}}{2}$

By using properties for <em>trigonometric</em> functions and <em>trigonometric</em> expressions, we find that the <em>exact</em> value of the sine of the angle 5π/12 radians is $\frac{\sqrt{2+\sqrt{3}}}{2}$ .

To learn more on trigonometric functions: brainly.com/question/15706158

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7 0

2 years ago

(1/1+sintheta)=sec^2theta-secthetatantheta pls help me verify this

Xelga [282]

Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

$\displaystyle \frac{1}{1+\sin\theta} = \sec^2\theta - \sec\theta \tan\theta$

To start, we can multiply the fraction by (1 - sin(θ)). This yields:

$\displaystyle \frac{1}{1+\sin\theta}\left(\frac{1-\sin\theta}{1-\sin\theta}\right) = \sec^2\theta - \sec\theta \tan\theta$

Simplify. The denominator uses the difference of two squares pattern:

$\displaystyle \frac{1-\sin\theta}{\underbrace{1-\sin^2\theta}_{(a+b)(a-b)=a^2-b^2}} = \sec^2\theta - \sec\theta \tan\theta$

Recall that sin²(θ) + cos²(θ) = 1. Hence, cos²(θ) = 1 - sin²(θ). Substitute:

$\displaystyle \displaystyle \frac{1-\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta \tan\theta$

Split into two separate fractions:

$\displaystyle \frac{1}{\cos^2\theta} -\frac{\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta\tan\theta$

Rewrite the two fractions:

$\displaystyle \left(\frac{1}{\cos\theta}\right)^2-\frac{\sin\theta}{\cos\theta}\cdot \frac{1}{\cos\theta}=\sec^2\theta - \sec\theta \tan\theta$

By definition, 1 / cos(θ) = sec(θ) and sin(θ)/cos(θ) = tan(θ). Hence:

$\displaystyle \sec^2\theta - \sec\theta\tan\theta \stackrel{\checkmark}{=} \sec^2\theta - \sec\theta\tan\theta$

Hence verified.

8 0

3 years ago

Which line is parallel to line r?<br><br>line p<br><br>line q<br><br>line s<br><br>line t

Mila [183]

Line Q??? You don't really have an image of it

5 0

3 years ago

Read 2 more answers

Jack is cooking muffins the recipe calls for 5 cups of sugar he accidentally put in 9 cups. How many extra cups did he put in

Sedaia [141]

Answer:

4

Step-by-step explanation:

9-5=4

4 0

3 years ago