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Blizzard [7]
2 years ago
10

The size of a computer screen is measured along the diagonal. What is the approximate size, measured to the nearest inch, of a 1

2 inch by 10.5 inch computer screen?
Mathematics
1 answer:
SashulF [63]2 years ago
7 0

According to the Pythagoras theorem, the size of the diagonal of computer screen is 15.95 inches.

Pythagoras theorem.

Pythagoras theorem states that "sum of the squares on the legs of a right triangle is equal to the square on the hypotenuse"

And the general form of Pythagoras theorem is,

c² = a² + b²

where

a and b refers the sides of the triangle

c refers the hypotenuse

Given,

The size of a computer screen is measured along the diagonal.

Here we need to find the size of the diagonal of the computer screen.

Here we consider the computer screen with the right angle triangle.

When we compare this, we have the side of the triangle are 12 inch and 10.5 inch.

So, we have apply the value to the formula, then we get the hypotenuse as,

c² = 12² + (10.5)²

c² = 144 + 110.25

c² = 254.25

c = √254.25

c = 15.95

Therefore, the size of the diagonal of the computer screen is 15.95 inches.

To know more about Pythagoras theorem here.

brainly.com/question/343682

#SPJ1

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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:

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F = (4, -2)

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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:

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