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abruzzese [7]
3 years ago
9

A parallelogram has four angles that are the same measure. Is it a square?

Mathematics
2 answers:
Simora [160]3 years ago
3 0

Answer:

Yes

Step-by-step explanation:

Yes. It could be a square if all sides are also the same measure.

Over [174]3 years ago
3 0

Answer:

yes yes yes yes

Step-by-step explanation:

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-3.5 , 2 , -3 , -1 , -1.5<br> Least to greatest
gulaghasi [49]

Answer:

-3.5, -3, -1.5, -1, 2

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Beth earns $54 per day and $10 for each extra hour she works. Ray earns $60 per day and $8 for each extra hour he puts in. They
inysia [295]
For Beth, we know that she works 5 days a week and that she earns $54 dollars per day. Next, add the $10 dollars for each extra hour of work. p stands for the amount she gets paid.
(54 * 5) + (10 * x) = p
 
 For Ray, do exactly the same things you did for Beth but put in his earnings.

(60 * 5) + (8 * x) = p 
6 0
3 years ago
Simplify the expression. (2q5)4 2q625 16q9 16q20 2q20
Katen [24]

Answer:

2^4= 16

(q^5)^4= q^20

16q^20. answer is c

3 0
3 years ago
David is catching fish. If he would have caught three times more fish, then the total number of fish would be 18 more. How many
AleksandrR [38]

Answer:

6 i think

Step-by-step explanation:

It woud be 6 because 18 divided by 3 is 6.

Sorry if I am wrong but if it is right i hope it helped:)

3 0
3 years ago
Find the volume V obtained by rotating the region bounded by the curves about the given axis.
BartSMP [9]

Using the disk method, the volume is given by the integral

\displaystyle \pi \int_{\pi/2}^\pi (9\sin(x))^2\,\mathrm dx = 81\pi \int_{\pi/2}^\pi \sin^2(x)\,\mathrm dx

That is, each disk has a radius of <em>y</em> = 9 sin(<em>x</em>) and hence area = <em>π</em> (9 sin(<em>x</em>))². Add up infinitely many such disks by integrating. Then the volume is

\displaystyle 81\pi \int_{\pi/2}^\pi \sin^2(x)\,\mathrm dx = \frac{81\pi}2 \int_{\pi/2}^\pi (1-\cos(2x))\,\mathrm dx \\\\ =\frac{81\pi}2 \left(x-\frac{\sin(2x)}2\right)\bigg|_{\pi/2}^\pi \\\\ = \frac{81\pi}2 \left( \left(\pi-\frac{\sin(2\pi)}2\right) - \left(\frac\pi2 - \frac{\sin(\pi)}2\right) \right) \\\\ = \boxed{\frac{81\pi^2}4}

8 0
3 years ago
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