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2 years ago

The coordinates of the vertices of the triangle are

Mathematics

2 answers:

andrew-mc [135]2 years ago

8 0

Answer:

The height of the triangle PQR is given by:

h = 4 + 8

h = 12 units

Now, the area of the triangle PQR is:

= 108 square units.

Step-by-step explanation:

Alecsey [184]2 years ago

6 0

The area of the triangle PQR is 108 square units

<h3>How to determine the area of the triangle?</h3>

The given parameters are:

Height, h = 12 units
Base, b = 18 units

The area is then calculated using:

Area = 0.5 * base * height

So, we have:

Area = 0.5 * 18 * 12

Evaluate

Area = 108

Hence, the area of the triangle PQR is 108 square units

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Step-by-step explanation:

Due to WX being a Perpendicular Bisector WY = WZ

4x - 5 = 2x + 11

Add 5 to both sides

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2x = 16

Divide by 2

x = 8

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3 years ago

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3 years ago

PLEASE CHECK THIS WILL MARK BRAINLIEST

VikaD [51]

Answer:

This time all of them are right

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Step-by-step explanation:

7 0

3 years ago

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Simplify, state all restrictions.

Kipish [7]

The simplified expression is $\frac{1 - x - y}{x + y}$ and the restriction is $y \ne -x$

<h3>How to simplify the expression?</h3>

The expression is given as:

$\frac{x - y}{4x^2 - 8xy + 3y^2} \div \frac{2x + y}{2x - 3y} \times \frac{4x^2 - y^2}{x^2 - y^2} -1$

Express x^2 - y^2 as (x + y)(x - y) and factorize other expressions

$\frac{x - y}{(2x - y)(2x - 3y)} \div \frac{2x + y}{2x - 3y} \times \frac{4x^2 - y^2}{(x - y)(x + y)} -1$

Rewrite the expression as products

$\frac{x - y}{(2x - y)(2x - 3y)} \times \frac{2x - 3y}{2x + y} \times \frac{4x^2 - y^2}{(x - y)(x + y)} -1$

Cancel out the common factors

$\frac{1}{(2x - y)} \times \frac{1}{2x + y} \times \frac{4x^2 - y^2}{(x + y)} -1$

Express 4x^2 - y^2 as (2x - y)(2x + y)

$\frac{1}{(2x - y)} \times \frac{1}{2x + y} \times \frac{(2x - y)(2x + y)}{(x + y)} -1$

Cancel out the common factors

$\frac{1}{x + y} -1$

Take the LCM

$\frac{1 - x - y}{x + y}$

Hence, the simplified expression is $\frac{1 - x - y}{x + y}$ and the restriction is $y \ne -x$