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TEA [102]
3 years ago
7

Anyone helppp meee please ....

Mathematics
1 answer:
borishaifa [10]3 years ago
4 0
Yep, this one seems sneaky and confusing.  But it's not so bad if you remember the things you learned about parallel lines.  (It can't be too tough ... I learned them
in 1954 and I still know how to use them.)

Look at the picture.  Line ' l ' is parallel to line ' m ', and the horizontal line on the bottom (which is not labeled) is a transversal that cuts the parallel lines.

Did you learn that interior angles on the same side of the transversal are equal ?
I'm sure you did, although it may have a new name nowadays.

Anyway, with the help of that 'tool', angle-'B' and angle-'D' are equal.  So . . .

(angle-A + angle-B) = 120

angle-B = 65

angle-A = 120 - 65 = <u>55 degrees</u>.
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hypotenuse = 102.69

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Simplify.<br><br> 1/4 (1 - 2/3)² + 1/3.<br><br> The answer must be a simplified fraction.
loris [4]

Answer:

\large\boxed{\dfrac{13}{36}}

Step-by-step explanation:

Use PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

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

\dfrac{1}{4}\left(1-\dfrac{2}{3}\right)^2+\dfrac{1}{3}=\dfrac{1}{4}\left(\dfrac{3}{3}-\dfrac{2}{3}\right)^2+\dfrac{1}{3}=\dfrac{1}{4}\left(\dfrac{3-2}{3}\right)^2+\dfrac{1}{3}\\\\=\dfrac{1}{4}\left(\dfrac{1}{3}\right)^2+\dfrac{1}{3}=\dfrac{1}{4}\cdot\dfrac{1^2}{3^2}+\dfrac{1}{3}=\dfrac{1}{4}\cdot\dfrac{1}{9}+\dfrac{1}{3}=\dfrac{1}{36}+\dfrac{1}{3}\qquad(*)\\\\\text{the common denominator is}\ 36.\\\\36=3\cdot12\to\dfrac{1}{3}=\dfrac{1\cdot12}{3\cdot12}=\dfrac{12}{36}\\\\(*)\qquad=\dfrac{1}{36}+\dfrac{12}{36}=\dfrac{1+12}{36}=\dfrac{13}{36}

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