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3 years ago

2⁄3 − 1⁄2= Answer as a fraction

Mathematics

2 answers:

OverLord2011 [107]3 years ago

8 0

Answer: 1/6

Step-by-step explanation:

jeyben [28]3 years ago

7 0

Answer:

2/3 - 1/2 = 1/6

Hope this helps

-Amelia

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Solve the following system of equations graphically on the set of axes below.

professor190 [17]

The solution of the given equations is: (-1.5, 1.5)

Step-by-step explanation:

Linear equation usually represents a straight line. When a system of two liner equations is given, the point of intersection of lines represented by both equations is the solution of the system of equations.

We will plot the lines of both equations and find their point of intersection (Picture Attached)

We can see in the picture that two lines intersect at (-1.5, 1.5)

Hence,

The solution of the given equations is: (-1.5, 1.5)

Keywords: Linear equations, variables

Learn more about linear equations at:

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2 years ago

The cost of 2 cups of strawberries is $6. Write an equation for the relationship between the cost and cups os strawberries

Rasek [7]

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2 years ago

Read 2 more answers

Graph and label the image of the figure below after a dilation by a factor of 1/2.

neonofarm [45]

Answer:

M' (1.5, -1), F' (2, -1), L' (0.5 -2.5), W' (2.5, -2.5)

see graph below

Explanation:

Given:

The image of a quadrilateral on a coordinate plane

To find:

The coordinates of the new image after dilation of 1/2 have been applied to the original image.

Then graph the coordinates

First, we need to state the coordinates of the original image:

M = (3, -2)

F = (4, -2)

L = (1, -5)

W = (5, -5)

Next, we will apply a scale factor of 1/2:

$\begin{gathered} Dilation\text{ rule:} \\ (x,\text{ y\rparen}\rightarrow(kx,\text{ ky\rparen} \\ where\text{ k = scale factor} \\ \\ scale\text{ factor = 1/2} \\ M^{\prime}\text{ = \lparen}\frac{1}{2}(3),\text{ }\frac{1}{2}(-2)) \\ M^{\prime}\text{ = \lparen}\frac{3}{2},\text{ -1\rparen} \\ \\ F\text{ = \lparen}\frac{1}{2}(4),\text{ }\frac{1}{2}(-2)) \\ F^{\prime}\text{ = \lparen2, -1\rparen} \end{gathered}$ $\begin{gathered} L\text{ = \lparen}\frac{1}{2}(1),\text{ }\frac{1}{2}(-5)) \\ L^{\prime}\text{ = \lparen}\frac{1}{2},\text{ }\frac{-5}{2}) \\ \\ W\text{ = \lparen}\frac{1}{2}(5),\text{ }\frac{1}{2}(-5)) \\ W^{\prime}\text{ = \lparen}\frac{5}{2},\text{ }\frac{-5}{2}) \end{gathered}$

The new coordinates:

M' (3/2, -1), F' (2, -1), L' (1/2, -5/2), W' (5/2, -5/2)

M' (1.5, -1), F' (2, -1), L' (0.5 -2.5), W' (2.5, -2.5)

Plotting the coordinates: