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3 years ago

To the nearest tenth, what is the perimeter of the triangle with vertices at (−2, 3), (3, 6), and (2, −2)?

Mathematics

1 answer:

katovenus [111]3 years ago

7 0

9514 1404 393

Answer:

20.3

Step-by-step explanation:

The distance formula can be used to find the side lengths.

d = √((x2 -x1)^2 +(y2 -y1)^2)

For the first two points, ...

d = √((3 -(-2))^2 +(6 -3)^2) = √(5^2 +3^2) = √34 ≈ 5.83

For the next two points, ...

d = √((2 -3)^2 +(-2-6)^2) = √(1 +64) = √65 ≈ 8.06

For the last and first points, ...

d = √((-2-2)^2 +(3-(-2)^2) = √(16 +25) = √41 ≈ 6.40

Then the sum of the side lengths is ...

5.83 +8.06 +6.40 = 20.29 ≈ 20.3

The perimeter of the triangle is about 20.3 units.

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Karen bought 203 stickers. Each sheet had 13 stickers.She gave away 809 stickers to her friends. How many stickers does Karen ha

andrew11 [14]

Answer:

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Step-by-step explanation:

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3 years ago

Shirley spelled 12 of her 16 spelling words correctly on the test. What percent grade did shirley get on the test

Black_prince [1.1K]

Answer:

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Step-by-step explanation:

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3 years ago

The line l passes through the point P = (1,-1,1) and has direction vector d = [2,3,1]. Determine whether l and the plane 4x-y+5z

suter [353]

Hello :
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3 years ago

The roots of a quadratic equation are (1,0) and (-5,0). What is the axis of symmetry?

leva [86]

Answer:

Step-by-step explanation:

The axis of symmetry lies exactly halfway between the given x = 1 and x = -5.

The axis of symmetry is x = -2.

The 'halfway' value is also the 'average' of 1 and -5: -4/2, or -2.

The axis of symmetry is x = -2.

8 0

3 years ago

Given the set of vertices, determine whether parallelogram ABCD is a rhombus, rectangle or square. List all that apply. A(7,-4),

Sloan [31]

Given:

Vertices of a parallelogram ABCD are A(7,-4), B(-1,-4), C(-1,-12), D(7, -12).

To find:

Whether the parallelogram ABCD is a rhombus, rectangle or square.

Solution:

Distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula, we get

$AB=\sqrt{(-4-(-4))^2+(-1-7)^2}$

$AB=\sqrt{(-4+4)^2+(-8)^2}$

$AB=\sqrt{0+64}$

$AB=8$

Similarly,

$BC=\sqrt{(-1-(-1))^2+(12-(-4))^2}=8$

$CD=\sqrt{(7-(-1))^2+(-12-(-12))^2}=8$

$AD=\sqrt{(7-7)^2+(-12-(-4))^2}=8$

All sides of parallelogram are equal.

$AC=\sqrt{(-1-7)^2+(-12-(-4))^2}=8\sqrt{2}$

$BD=\sqrt{(7-(-1))^2+(-12-(-4))^2}=8\sqrt{2}$

Both diagonals are equal.

Since, all sides are equal and both diagonals are equal, therefore, the parallelogram ABCD is a square.

We know that, a square is special case of rectangles and rhombus.

So, parallelogram ABCD is a rhombus, rectangle or square. Therefore, the correct option is c.

7 0

3 years ago