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3 years ago

8

Simplify. 1/4 (1 - 2/3)² + 1/3. The answer must be a simplified fraction.

1 answer:

loris [4]3 years ago

4 0

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)

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

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2 years ago

Carter draws one side of equilateral △PQR on the coordinate plane at points P(-3,2) and Q(5,2). Which ordered pair is a possible

Law Incorporation [45]

Step-by-step explanation:

Hey, there!!!

Let me simply explain you about it.

We generally use the distance formula to get the points.

let the point R be (x,y)

As it an equilateral triangle it must have equal distance.

now,

let's find the distance of PQ,

we have, distance formulae is;

$pq = \sqrt{( {x2 - x1)}^{2} + ( {y2 - y1)}^{2} }$

$or \: \sqrt{( {5 + 3)}^{2} + ( {2 - 2)}^{2} }$

By simplifying it we get,

$8$

Now,

again finding the distance between PR,

$pr = \sqrt{( {x2 - x1}^{2} + ( {y2 - y1)}^{2} }$

or,

$\sqrt{( {x + 3)}^{2} + ( {y - 2)}^{2} }$

By simplifying it we get,

$= \sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13 }$

now, finding the distance of QR,

$qr = \sqrt{( {x - 5)}^{2} + ( {y - 2)}^{2} }$

or, by simplification we get,

$\sqrt{ {x}^{2} + {y}^{2} - 10x - 4y + 29 }$

now, equating PR and QR,

$\sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13} = \sqrt{ {x}^{2} + {y}^{2} - 10x - 4y + 29 }$

we cancelled the root ,

${x}^{2} + {y}^{2} + 6x - 4y + 13 = {x}^{2} + {y}^{2} -10x - 4y + 29$

or, cancelling all like terms, we get,

6x+13= -10x+29

16x=16

x=16/16

Therefore, x= 1.

now,

equating, PR and PQ,

$\sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13 } = 8}$

cancel the roots,

${x}^{2} + {y}^{2} + 6x - 4y + 13 = 8$

now,

(1)^2+ y^2+6×1-4y+13=8

or, 1+y^2+6-4y+13=8

y^2-4y+13+6+1=8

or, y(y-4)+20=8

or, y(y-4)= -12

either, or,

y= -12 y=8

Therefore, y= (8,-12)

by rounding off both values, we get,

x= 1

y=(8,-12)

So, i think it's (1,8) is your answer..

<em>Hope it helps</em><em>.</em><em>.</em><em>.</em>

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Step-by-step explanation:

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