Geometry Proofs (Parallelograms)

1 answer:

5 0

Since measure REU equal measure SFT, RE=FT and SF=EU then the two triangles REU et SFT are similar.
Then we deduce that the two sides RU and ST are equal, RU=ST.
Also, since the two triangles above are similar, then the two angles FST and RUE are equal. We deduce that the two lines RU and ST are parallel (interior opposite angles principles.)
We have two facts now:
RU = ST and RU parallel to ST, we deduce that the quadrilateral is a parallelogram.

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3+(2+8)2ndpower ÷4×(1/2) 4th power

PIT_PIT [208]

Answer:

412

Step-by-step explanation:

You have to follow PEMDAS and solve the problem carefully.

$\frac{3+(2+8)^{2} }{4*(\frac{1}{2} )^{2} } =\frac{3+(10)^{2} }{4*(\frac{1}{2} )^{2} } = \frac{3+100}{4*(\frac{1}{2} )^{2} } = \frac{103 }{4*(\frac{1}{2} )^{2} }$

$\frac{103 }{4*(\frac{1}{2} )^{2} } = \frac{103 }{4*(\frac{1}{16})} } = \frac{103 }{\frac{1}{4} } =\\103*4\\=412$

8 0

3 years ago

Raul has 56 bouncy balls. He puts three times as many balls into red bags as he puts into green gift bags. If he puts the same n

Sati [7]

If we divide 56 by 4 we get 14. Why by4? So that one number is 3 times the other So he had 14 balls that went into the green bags and 42 (56-14) that went into the red bags. We could just answer the question and say 14 but I think they want to know how many in each green bag.
14 and 42 don't work because they are not the same number of balls. What number is a common factor? 7 is,
We could have 2 green bags and split the 14 balls into 2 groups of 7 and with the remaining 42 - put them into 6 red bags of 7 each.
And so the answer to your question is:
7 ball in each bag = 2 bags are green, and 6 bags are red
14 balls + 42 balls = 56 bouncy balls

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3 years ago

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Katyanochek1 [597]

The answer is 17/18.
You need to make the denominator the same on all the fractions.

Hope this helps.

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4 years ago

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