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Lilit [14]
2 years ago
5

Identify the segment bisector of XY

Mathematics
1 answer:
zhuklara [117]2 years ago
3 0

The midpoint (W) of XY is also on segment PQ, so PQ is the bisector of XY.

Hope it helps you :)

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Find the measure of c. A. 40 B. 70 C. 90 D. 50
Juliette [100K]

Answer:

D. 50

Step-by-step explanation:

a = 90° (angle subtended in semicircle)

a + c + 40° = 180° (by angle sum postulate of a triangle)

90° + c + 40° = 180°

c + 130° = 180°

c = 180° - 130°

c = 50°

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What is z- 4/4 +8 simplified?​
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2 years ago
Write the equation for the line in slope-intercept form (2,-3) (-1/2) ( HELP PLEASE )
prisoha [69]

Answer:

A linear equation is a fancy term for a straight line, which can be created by joining 2 points.

A point can be defined by (x, y), where x and y are the horizontal distance and the vertical distance respectively from point (0,0) (called the origin).

There are 3 ways to define a line:

(1) The slope-intercept form

y = mx + b, in which m is the slope, b is the y-intercept, where the line crosses the vertical axis at (0, b). If values of m and b are given, you can substitute and write the equation right away.

The slope, m, is calculated by rise (vertical difference between 2 points on the line) divided by run (horizontal difference between the 2 points on the line)

Step-by-step explanation:

or example, if 2 points on the line are

(x1, y1) and (x2, y2)

Then slope = (y1– y2)/(x1 — x2)

If m > 0, then the line rises to the right.

If m = 0, then the line is horizontal.

If m < 0, then the line rises to the left.

If the line is in the form x = c, where c is a constant, then the line is vertical.

(2) Point slope form

m = (y — y1)/(x — x1), when (x1, y1) and m are given.

In other words,

m(x — x1) = y — y1

mx — m(x1) = y — y1

y = mx — m(x1) + (y1)

Note that b = -m(x1) + (y1)

(3) linear form

ax + by + c = 0, where a, b, and c are constants.

3 0
3 years ago
Please help me ))):??
Luden [163]

Y should be replaced in the equiation with 2

So it would be...

2y - 4

2·2 - 4

4 0
2 years ago
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