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Mice21 [21]
3 years ago
11

If 2/3 of the distance from the y-axis to point A (−30, −45) is equal to 1/4 of the distance from the x-axis to point B(a, a), w

here a > 0, what is the value of a?
Mathematics
1 answer:
Yuliya22 [10]3 years ago
8 0

The value of a is 80

Step-by-step explanation:

The distance of a point (x_0,y_0) from the y-axis can be written as

d_y = |x_0|

because the x-coordinate of the y-axis is zero.

Similarly, the distance of a point (x_0,y_0) from the x-axis can be written as

d_x=|y_0|

Since the y-coordinate of the x-axis is zero.

In this problem:

- The distance of the point A (−30, −45) from the y-axis can be written as

d_A = |-30|=30

- The distance of point B (a,a) from the x-axis can be written as

d_B = |a| = a

Since a>0.

We are told that 2/3 of the distance from the y-axis to point A (−30, −45) is equal to 1/4 of the distance from the x-axis to point B(a, a), which means

\frac{2}{3}d_A = \frac{1}{4}d_B

Therefore,

\frac{2}{3}(30)=\frac{1}{4}a

And solving for a,

20=\frac{1}{4}a\\a=80

Learn more about distance:

brainly.com/question/3969582

#LearnwithBrainly

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neonofarm [45]
Answer:

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see graph below

Explanation:

Given:

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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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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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