The fractions One-fourth, One-half, Three-twelfthsand are shown below. Top: A fraction bar divided into 4 parts. 1 is shaded. Middle: A fraction bar divided into 2 parts. 1 is shaded. Bottom: A fraction bar divided into 12 parts. 3 are shaded. Which is a correct comparison of these fractions? One-fourth equals three-twelfths One-half is less than three-twelfths One-fourth is greater than one-half Three-twelfths is greater than one-fourth

One-fourth equals three-twelfths

Step-by-step explanation:

From the above question:

The fractions

One-fourth

: A fraction bar divided into 4 parts. 1 is shaded.

One-half

A fraction bar divided into 2 parts. 1 is shaded

Three-twelfths:

A fraction bar divided into 12 parts. 3 are shaded.

Which is a correct comparison of these fractions?

Comparing the options

a) One-fourth equals three-twelfths

= 1/4 = 3/12

Divide numerator and denominator by 3

= 3 ÷ 3 /12 ÷ 3

= 1/4

Option a is correct

b) One-half is less than three-twelfths

1/2 < 3/12

This statement above is wrong.

One half is greater than three twelfths

1/2 is > 3/12

1/2 > 1/4

c) One-fourth is greater than one-half

1/4> 1/2

This statement above is wrong.

1/2 is greater than 1/4

d) Three-twelfths is greater than one-fourth

This is statement is wrong. Three -twelfths is equal to one - fourths

The correct comparison of these fractions is One-fourth equals three-twelfths.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

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