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3 years ago

Pls help me !!! Thank you

Mathematics

1 answer:

madam [21]3 years ago

5 0

Answer:

First: Subtract each side to get y-b=mx

Secound:divide each side by x to get y-b/x=m

Third:we divide both sides b x to get m by itself.

(y-b)/x=(mx)/x

(y-b)/x= m

Therefore m is equal to(y-b)/x

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For every 5 chickens there were 10 eggs. What are the three ways to write this ratio?

mixas84 [53]

Answer:

The first 3

Step-by-step explanation:

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2 years ago

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Graph a triangle (STU) and reflect it over the y-axis to create triangle ST'U'.

lukranit [14]

The x-coordinates of $\triangle S'T'U'$ will be the negation of the x-coordinates of $\triangle STU$

The line segment from S to the y-axis equals the line segment from S' to the y-axis. Similarly, the line segment from T to the y-axis equals the line segment from T' to the y-axis

See attachment for $\triangle STU$ and $\triangle S'T'U'$

In order to solve this question, I will make the following assumptions.

Assume that the coordinates of $\triangle STU$ are

$S = (4,5)$

$T = (5,9)$

$U=(3,8)$

Refer to attachment for illustrations

(1) Reflect $\triangle STU$ over y-axis and describe the transformation

To reflect $\triangle STU$ across the y-axis, the following rule must be followed

$(x,y) \to (-x,y)$

This means that:

$S = (4,5) \to S' = (-4,5)$

$T = (5,9) \to T' = (-5,9)$

$U=(3,8) \to U'=(-3,8)$

The description of the transformation is as follows:

Notice that the signs of the x-coordinates $\triangle STU$ and $\triangle S'T'U'$ of both triangles are different.

In other words, if the x-coordinate of one is positive, then the other will have a negative x-coordinate; and vice versa.

(2) Compare the segments and the line of reflection

To reflect across the y-axis means that the reflecting line is the y-axis, itself.

The distance between a point to the y-axis is the absolute value of the x-coordinate.

So, the distance between S and the y-axis is:

$S = |4| = 4$

The distance between S' and the y-axis is:

$S' = |-4| = 4$

We can conclude that the two line segments are equal.

This is the same for other point T and T' because of the formula used above.

From T and T' to the y-axis is:

$T =|5| =5$

$T' =|-5| =5$