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aleksandr82 [10.1K]
3 years ago
9

Prove algebraically that the recurring decimal 0.72 =8/11

Mathematics
2 answers:
miskamm [114]3 years ago
7 0

Answer:

see explanation

Step-by-step explanation:

We require 2 equations with the recurring decimal placed after the decimal point.

let x = 0.7272.... (1) ← multiply both sides by 100

100x = 72.7272... (2)

Subtract (1) from (2) thus eliminating the recurring decimal

99x = 72 ( divide both sides by 99 )

x = \frac{72}{99} = \frac{8}{11}

lbvjy [14]3 years ago
4 0

Answer:

true

Step-by-step explanation:

because 8÷11=0.72

72÷10

=8÷11

=0.72

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Y=x^2-6x+3 help write the vertex form of the equation
Artist 52 [7]

Answer:

y=(x-3)^2-6

Step-by-step explanation:

y=x^2-6x+3

This is written in the standard form of a quadratic function:

y=ax^2+bx+c

where:

  • ax² → quadratic term
  • bx → linear term
  • c → constant

You need to convert this to vertex form:

y=a(x-h)^2+k

where:

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To find the vertex form, you need to find the vertex. For this, use the equation for axis of symmetry, since this line passes through the vertex:

x=-\frac{b}{2a}

Using your original equation, identify the a, b, and c terms:

a=1\\\\b=-6\\\\c=3

Insert the known values into the equation:

x=-\frac{(-6)}{2(1)}

Simplify. Two negatives make a positive:

x=\frac{6}{2} =3

X is equal to 3 (3,y). Insert the value of x into the standard form equation and solve for y:

y=3^2-6(3)+3

Simplify using PEMDAS:

y=9-18+3\\\\y=-9+3\\\\y=-6

The value of y is -6 (3,-6). Insert these values into the vertex form:

(3_{h},-6_{k})\\\\y=a(x-3)^2+(-6)

Insert the value of a and simplify:

y=(x-3)^2-6

:Done

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Step-by-step explanation:

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