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Kisachek [45]
2 years ago
7

Are all quadrilaterals parallelograms?

Mathematics
2 answers:
nika2105 [10]2 years ago
5 0

Answer:

no

Step-by-step explanation:

It is backwards.

Good luck Bud!

Stay safe from COVlD-19!

kondor19780726 [428]2 years ago
3 0

Answer:

No, but all parallelograms are quadrilaterals.

Step-by-step explanation:

Take a trapezoid, for example. A trapezoid is a quadrilateral because it has 4 sides, but it does not contain 2 sets of parallel sides thus making it not a parallelogram.

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What’s the appropriate volume of the cone 8in and 20in
Tasya [4]
I believe the appropriate volume is 160.
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3 years ago
A number cube labeled 1 through 6 is rolled.
lbvjy [14]

Answer: 3/6 or 1/2 or 50%

Step-by-step explanation: 1, 2, 3 4, 5, 6 are each face

2, 4, and 6 are even.

3 out of the 6 are even (3/6)

reduce (1/2)

1/2 = 50%

8 0
1 year ago
Hi help with these please and thank yu? :)
ryzh [129]

Answer:

the first one is A

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7 0
3 years ago
Find the area and perimeter of the shape below. Help plzzz.
kozerog [31]
Find the area of each shape and add.Let's start with the first semi-circle(half-circle)
If the area of a circle is πr² , then the area of a semi-circle is 1/2πr² where is 22/7 or 3.14, and r is the radius which is 1.8 meters
A=1/2πr²
A=1/2*22/7*1.8*1.8
A=5.09 m²(rounded to nearest hundredth)
Since the semi-circles are two and have the same radius, multiply the area of the first one by two or go through the same process again.
So 5.09 *2=10.18meters square is the area of the two semi-circles
Now, let's find the area of a rectangle  which is length times width, where length is 6 meters and width is (1.8+1.8, because 1.8 is half the width)=3.6 meters
A=l*w
A=6*3.6
A=21.6meters square
Therefore the area of the shape is 10.18m²+21.6m²=219.888m²
4 0
3 years ago
Simplify cotø(tanø+cotø)​
Sladkaya [172]

Answer:

\large\boxed{\cot\theta(\tan\theta+\cot\theta)=1+\cot^2\theta=\dfrac{1}{\sin^2\theta}=\csc^2\theta}

Step-by-step explanation:

\text{Use}\\\\\text{distributive property:}\ a(b+c)=ab+ac\\\cot\alpha\tan\alpha=1.\\\\======================\\\\\cot\theta(\tan\theta+\cot\theta)=(\cot\theta)(\tan\theta)+(\cot\theta)(\cot\theta)\\\\=1+\cot^2\theta\\\\\text{If you want next transformation, then use:}\\\\\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}\\\\\sin^2\alpha+\cos^2\alpha=1\\\\=======================

=1+\left(\dfrac{\cos\theta}{\sin\theta}\right)^2=1+\dfrac{\cos^2\theta}{\sin^2\theta}=\dfrac{\sin^2\theta}{\sin^2\theta}+\dfrac{\cos^2\theta}{\sin^2\theta}=\dfrac{\sin^2\theta+\cos^2\theta}{\sin^2\theta}\\\\=\dfrac{1}{\sin^2\theta}\\\\\text{If you want next transformation, then use:}\\\\\csc\alpha=\dfrac{1}{\sin\alpha}\\\\=\left(\dfrac{1}{\sin\theta}\right)^2=(\csc\theta)^2=\csc^2\theta

4 0
3 years ago
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