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ANTONII [103]
3 years ago
13

Compare properties of squares and rhombi to properties of other quadrilaterals by answering each question. Write a brief explana

tion for each answer.
(a) Describe a property of squares that is also a property of rectangles.
(b) Describe a property of squares that is not a property of rectangles.
(c) Describe a property of rhombi that is also a property of parallelograms.
(d) Describe a property of rhombi that is not a property of parallelograms
Mathematics
1 answer:
GuDViN [60]3 years ago
3 0
Hey there! Hello!

Not sure if you still need these answers, but I'd love to help out if you do! 

Now, I want you to go ahead and think of some stuff that's true for squares. To name a few, the opposite sides are going to be parallel to one another, all the angles are 90°, all the sides are the same length, and both diagonals are going to be perpendicular and equal in length. I'm sure there's even more, but I'll leave that to you. (BTW, by diagonals, I mean the lines that go through the the opposite diagonal corners). 

What about rectangles? The opposite sides are going to be parallel to one another, the diagonals are going to be equal in length, and the angles are going to be 90°. 

Now, rhombi. All sides are going to be equal, opposite sides are going to be parallel, the diagonally opposite angles will be equal to each other, and the diagonals bisect each other at 90°. 

And lastly, parallelograms. Pretty similar to rhombi in that they have parallel opposite sides and that the opposite diagonal angles are equal to each other, but there's one thing that makes a parallelogram not a rhombus. 

If you differentiate the stuff I described, you'll be golden. There's a lot to choose from, and I personally like to have options. Hope this helped you out, feel free to ask me any additional questions you have! :-)
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lawyer [7]

Answer:

No

Edit:

Yes, based on original equation. (Credit to greenpumpkin for correction)

Step-by-step explanation:

For this problem, we simply need to find the values of x that can make the equation true.  So, let's begin by isolating the "x" variable.

sqrt(2x + 13) = x + 5

[sqrt(2x + 13)]^2 = (x + 5)^2

2x + 13 = x^2 + 10x + 25

0 = x^2 + 8x + 12

Note, we can remove the sqrt method by squaring both sides of the equation.  Doing this, we see we have a quadratic equation meaning we can apply the quadratic formula to find solutions for x.

[-b +/- sqrt( b^2 - 4(a)(c) ) ] / 2a

Let a = 1, b = 8, and c = 12

[-8 +/- sqrt( (8)^2 - 4(1)(12) ) ] / 2(1)

= [-8 +/- sqrt( 64 - 48 ) ] / 2

= [-8 +/- sqrt(16) ] / 2

= [ -8 +/- 4 ] / 2

So, x = [ -8 + 4 ] / 2  and x = [-8 - 4 ] / 2

x = [-4] / 2 = -2  and x = [-12] / 2 = -6

Hence, the two values of x that can solve this quadratic equation are x = -2 and x = -6.

Therefore, we know that x = -6 is not extraneous, meaning it is a solution to our equation.

Cheers.

----------------------------------------------------

Edit:

Plugging the value of -6 back into the original equation, we get the following:

sqrt(2x + 13) = x + 5

sqrt(2(-6) + 13) = (-6) + 5

sqrt (1) = -1

1 != -1

Given that 1 cannot equal negative 1, we can say that x = -6 is an extraneous solution. (Credit to greenpumpkin for correction)

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A class of 28 students has 13 boys and 15 girls. What is the ratio of girls to boys in the class? 15:28 13:15 O 13:28 15:13 Prev
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Answer:\frac{15}{13}

Step-by-step explanation:

Given

There are 28 students in a class out of which

13 are boys so the remaining 15 are girls

Ratio of girls to boys in the class=\frac{15}{13}

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