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3 years ago

The coordinates of the vertices of ∆LMN are L(3, 0), M(2, -3), and N(0, 3).

Mathematics

1 answer:

fgiga [73]3 years ago

7 0

Answer:

The first triangle has been rotated 180 degrees reflected on point r (as in x)

Step-by-step explanation:

Draw it

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What is the surface area of the rectangular prism, if its length is 17 meters?

wel

182 m^ (Third answer) because it is not the volume

2 rectangles measuring 17x2 2(17x2)= 2x34 = 68 m^

2rectangles measuring 17x3 2(17x3)=2x51+102 m^

2rectangles measuring 3x2 2(3x2)=2x6=12 m^

TOTAL---------------------------------------------------------182 m^

8 0

3 years ago

I’ll mark brainliest !! Please help me with these answers

Galina-37 [17]

Answer:

Problem1= a Problem 2=a

Step-by-step explanation:

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3 years ago

What is the slope intercept form of the equation -7x +14y=-35

Mademuasel [1]

Answer:

-7x+14y-35

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

T is a point inside parallelogram ABCD. The area of ΔTAB = 11, the area of ΔTBC = 4, and the area of ΔTCD = 6. Find the area of

antoniya [11.8K]

wrong question

the ∆TAC is not the triangle

5 0

3 years ago

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.

3 0

2 years ago