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seropon [69]
4 years ago
8

A parallelogram with all angles measuring 90 degrees is what type of quadrilateral

Mathematics
1 answer:
bixtya [17]4 years ago
7 0
I think the correct answer would be a rectangle
I hope this helps !
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Simplify (5x3y) (2xy4)
Anon25 [30]

Answer: 10x^4y^5

Step-by-step explanation:

8 0
3 years ago
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What is the mean of the following set?<br><br> 34, 31, 32, 38
larisa [96]

Answer:

33.75.

Step-by-step explanation:

The mean = sum of the numbers / the number count

= (34 + 31 + 32 +  38) / 4

= 135 / 4

= 33.75.

7 0
3 years ago
What is the difference between a repeating decimal and a terminating decimal​
Andrej [43]

Step-by-step explanation:

Terminating decimals: these have a finite number of digits after the decimal point.

Example: 9.5

Recurring decimals: these have one or more repeating numbers or sequences of numbers after the decimal point, which continue infinitely.

Example: 0.333333...

6 0
3 years ago
What else would need to be congruent to show that JKL= MNO by AAS?
taurus [48]
Answer: Choice B) Angle L = Angle O

---------------------------------------------------
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If we know that Angle L is congruent to Angle O, then we can use the AAS (angle angle side) congruence property. We have one pair of angles marked by the square marker (angle J and angle M). So they are congruent angles. We have a pair of congruent sides JK = MN = 3. So we're just missing a pair of angles. 

Note: The answer is NOT angle K = angle N because this would mean ASA would be used instead of AAS. The order of the letters is important as it establishes how the sides and angles relate. With ASA, the side is between the angles. With AAS, the side is not between the angles. 

3 0
3 years ago
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Cos(−θ)=√3/3, sinθ&lt;0
Delicious77 [7]

Answer:

\sin\theta=-\frac{\sqrt{6}}{3}


Step-by-step explanation:

The given trigonometric equation is \cos(-\theta)=\frac{\sqrt{3} }{3}.

We can either use the Pythagorean identity or the right angle triangle  to solve for \sin\theta.

According to the Pythagorean identity,

\cos^2\theta+\sin^2\theta=1


Recall that, the cosine function is an even function, therefore

\cos(-\theta)=\cos(\theta)


\Rightarrow \cos(\theta)=\frac{\sqrt{3} }{3}.

We substitute this value in to the above Pythagorean identity to get;


(\frac{\sqrt{3}}{3})^2+\sin^2\theta=1


\Rightarrow \frac{3}{9}+\sin^2\theta=1


\Rightarrow \sin^2\theta=1-\frac{3}{9}


\Rightarrow \sin^2\theta=\frac{6}{9}


\Rightarrow \sin\theta=\pm \sqrt{\frac{6}{9}}


\Rightarrow \sin\theta=\pm \frac{\sqrt{6}}{3}


But we were given that,

\sin\theta\:, so we choose the negative value.

\Rightarrow \sin\theta=-\frac{\sqrt{6}}{3}


The correct answer is B







7 0
4 years ago
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