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Andrei [34K]
2 years ago
8

ABCD is a rectangle. Find the mCED

Mathematics
1 answer:
mrs_skeptik [129]2 years ago
6 0

Answer:8.5...

Step-by-step explanation:

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Which number is the smallest? 5.6123 5.6245 5.6023 5.6985
alexandr402 [8]
5.6023, hope this helps!
4 0
2 years ago
A square is a quadrilateral always sometimes never
lbvjy [14]
A quadrilateral is a shape with 4 sides, regardless of their lengths, angles, or symmetry. 

Therefore, a square is always a quadrilateral.
6 0
3 years ago
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Algebra 2 question, please help!
yKpoI14uk [10]
Let the leading term of the polynomial, f(x), be axⁿ.

Examine the possibilities.
     n      a      x -> - ∞     x -> +∞
-------   ----     -----------  ------------
even   a>0    f -> +∞     f -> +∞   Not true
even   a<0    f-> - ∞      f-> -∞    True
odd     a>0    f-> -∞      f-> +∞    Not true
odd     a<0    f-> +∞     f-> -∞     Not true

Answer:
(a) the degree of the polynomial is even, and
(b) the coefficient of the leading term is negative.

8 0
2 years ago
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What type of model should be used to make a prediction from the given data? (0,3), (2,4), (5,5), (10,6)
mojhsa [17]
I would say a Coordinate Grid.

5 0
3 years ago
I need to use the Pythagorean Theorem to find X. I don’t know how to do that!!
frutty [35]

Answer:

17) x=8

18) x=\sqrt{189} or x= 3\sqrt{21}

Step-by-step explanation:

So the rule is a^{2} +b^{2} =c^{2}, "c" being the hypothenuse, or the long line that is opposite to the right angle.

17) We know that both values of x are equal to each other, which makes everything 10x easier!

                                               x^{2} +x^{2} =(8\sqrt{2})  ^{2}

(by the way we know the x values are our a and b values because they are legs! the way I like to remember the legs is that they are connected to the right angle box, and therefore support the hypothenuse)

<em>simplify</em> (╥︣﹏╥)    

                                                2x^2=8^2(\sqrt{2} )^{2}  

                                                  2x^2=64(2)

                                                   2x^2=128    

                                                  x^{2}= 128/2  

                                                     x^{2} =64

                                                        x=8      

18) Just pretend that the flipped triangle doesn't exist. It's parallel to the other triangle with values on it, and basically servers no purpose other than being parallel to the sister triangle :)

Anyways, since we know the hypothenuse (15) but we don't know one of our leg values (x), we're going to change our equation a bit!

                                                    c^{2} - b^{2} = a^{2}

It doesn't matter if you put the one leg value in a or b, just as long as you stick to that same equation you started with the entire time!

                                                   15^{2} -6^{2} =x^{2}

                                                   225-36=x^{2}  

                                                     x^{2} =189

                                               x=\sqrt{189}=3\sqrt{21}

        The more you do these, the easier they'll get, so don't worry!

                           

4 0
3 years ago
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