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strojnjashka [21]
3 years ago
7

Calculate the area of each figure.

Mathematics
1 answer:
guapka [62]3 years ago
6 0

Answer:

230cm^{2}

Step-by-step explanation:

17*10=170

12*10*1/2=60

170+60=230

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Nice. We're not really told which is the right angle. We can tell from the squared distances:


AB^2=(1 - -5)^2 + (1-5)^2=6^2+4^2=52


BC^2=(3-1)^2+(4-1)^2=2^2+3^2=13


AC^2=(3- -5)^2+(4 -5)^2=8^2+1^2=65


We see AC^2=AB^2+BC^2 so B is the right angle.


That was preliminary and probably indicated in an associated figure.


Now we want the line through the midpoint of the legs, AB and BC, in standard form.


The midpoint of AB is the average of the coordinates of A and B:


\left( \dfrac{-5 +1}{2}, \dfrac{5+1}{2} \right) = (-2,3)


The midpoint of BC is


\left( \dfrac{1 + 3}{2}, \dfrac{1 + 4}{2} \right) = ( 2, \frac 5 2)


The line through those points is


(y - 3)(2 - -2)=(x - -2)(\frac 5 2 - 3)


4y - 12 =  (x+2)(- \frac 1 2)


8y - 24 = -x -2


x + 8y = 22


That's the answer. Let's check it.


Midpoint of AB is (-2,3), -2 + 8(3)=22 good.


Midpoint of BC is (2,5/2), 2+8(5/2)=22 good.


\textrm{Answer: } x + 8y = 22


4 0
3 years ago
Which statement cannot be used to determine whether quadrilateral XYZW must be a parallelogram?
Semenov [28]
The Answer is B) <span>XY =WZ and XW = YZ
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aniked [119]
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Those triples are
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The On-Line Encyclopedia of Integer Sequences (OEIS) lists the hypotenuses of primitive triples as sequence number A020882.
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4 years ago
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3 years ago
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atroni [7]

Answer:

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2/3 * 5 = 10/3, which is also an improper fraction.

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