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Pavlova-9 [17]
1 year ago
6

On a coordinate plane, a line is drawn from point a to point b. point a is at (negative 8, negative 13) and point b is at (4, 11

). what are the x- and y- coordinates of point p on the directed line segment from a to b such that p is one-third the length of the line segment from a to b? x = (startfraction m over m n endfraction) (x 2 minus x 1) x 1 y = (startfraction m over m n endfraction) (y 2 minus y 1) y 1 (1, 5) (0, 3) (–4, –5) (–5, –7)
Mathematics
1 answer:
vladimir2022 [97]1 year ago
7 0

The coordinate for point P which is one-third the length of the line segment from a to b is (-4, -5)

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

Point a is at (-8, -13) and point b is at (4, 11). Hence the coordinates for point p(x, y) which is one-third the length of the line segment from a to b is:

x=\frac{1}{3}(4-(-8))+(-8)=-4\\ \\y=\frac{1}{3}(11-(-13))+(-13)=-5

The coordinate for point P which is one-third the length of the line segment from a to b is (-4, -5)

Find out more on equation at: brainly.com/question/2972832

#SPJ4

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A paper sheet is cut 28 inches to the nearest inches. Find the maximum and minimum length of the patient possible.
aliina [53]

Answer:

  • max: 28.5 inches
  • min: 27.5 inches

Step-by-step explanation:

If the actual dimension were different from 28 inches by more than 1/2 inch, it would be reported as a different dimension. So, the minimum that will be reported as 28 is 27.5. The maximum that will be reported as 28 will be 28.4999999.... ≈ 28.5

The maximum and minimum length of the sheet are 28.5 inches and 27.5 inches, respectively.

4 0
3 years ago
Given that x and y are positive integers, solve the equation x² - 4y² = 13​
zvonat [6]
<h3>Answer:  x = 7 and y = 3</h3>

=====================================================

Explanation:

Apply the difference of squares rule

x² - 4y² = 13

x² - (2y)² = 13

(x - 2y)(x + 2y) = 13

Since x and y are positive integers, this means x-2y and x+2y are both integers as well.

The value 13 is prime. Its only factors are 1 and 13

Since the above equation shows 13 factoring into x-2y and x+2y, then we have two cases:

  • A) x-2y = 1 and x+2y = 13
  • B) x-2y = 13 and x+2y = 1

----------------

Let's consider case A

We have this system of equations

\begin{cases}x-2y = 1\\x+2y = 13\end{cases}

Add the equations straight down

  • x+x becomes 2x
  • -2y+2y becomes 0y = 0 which goes away
  • 1+13 becomes 14

Therefore we have 2x = 14 solve to x = 7

From here, plug this into either equation to solve for y

x-2y = 1

7 - 2y = 1

-2y = 1-7

-2y = -6

y = -6/(-2)

y = 3

You should get the same result if you used x+2y = 13

----------------

Since we've found that x = 7 and y = 3, notice how case B is not possible

Example:  x-2y = 13 becomes 7-2(3) = 13 which is false.

Also, x+2y = 1 would turn into 7+2(3) = 1 which is also false.

-----------------

Let's check those x and y values in the original equation

x² - 4y² = 13

7² - 4*(3)² = 13

49 - 4(9) = 13

49 - 36 = 13

13 = 13

The answer is confirmed.

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Step-by-step explanation:

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