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wolverine [178]
1 year ago
5

PLEASE HELP YOU WILL GET ALOT OF POINTS The triangle on the left is rotated to create the triangle on the right as its

Mathematics
2 answers:
Nana76 [90]1 year ago
7 0

Answer:

The third one

Step-by-step explanation:

It cannot be the first 2 as the corners wouldn't line up. It cannot be the last one as B and R are not the same angles

gregori [183]1 year ago
3 0
The first option is correct
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Give all solutions if the nonlinear system of equations, including those with no real complex components.
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\Delta=(-2)^2-4\cdot(-1)\cdot24=4+96=100\\\\x_1=\frac{-b-\sqrt\Delta}{2a};\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{100}=10\\\\x_1=\frac{2-10}{2\cdot(-1)}=\frac{-8}{-2}=4;\ x_2=\frac{2+10}{2\cdot(-1)}=\frac{12}{-2}=-6\\\\y_1=4^2+6\cdot4=16+24=40;\ y_2=(-6)^2+6\cdot(-6)=36-36=0\\\\Answer:\\x=4\ and\ y=40\ or\ x=-6\ and\ y=0
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Translate the sentence to an equation x minus 6 is 7
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The answer is x - 6 = 7
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3 years ago
Which classification describes AMNO with vertices M(2, -3), N(3, 1), and<br> 0(-3, 1)?
Sonja [21]

Answer:

Option D

Step-by-step explanation:

32). Given vertices of the triangle are M(2, -3), N(3, 1) and O(-3. 1).

Distance between two points (x_1,y_1) and (x_2,y_2) is given by the expression,

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

Distance between M(2, -3) and N(3, 1) will be,

MN = \sqrt{(3-2)^2+(1+3)^2}

      = \sqrt{1+16}

      = \sqrt{17}

Distance between M(2, -3) and O(-3, 1),

MO = \sqrt{(2+3)^2+(-3-1)^2}

      = \sqrt{25+16}

      = \sqrt{41}

Distance between N(3, 1) and O(-3, 1),

NO = \sqrt{(3+3)^2+(1-1)^2}

      = 6

Condition for right triangle,

c² = a² + b² [Here c is the longest side of the triangle]

By this property,

MO² = MN² + NO²

(\sqrt{41})^2=(\sqrt{17})^2+6^2

41 = 17 + 36

41 = 51

False.

Therefore, given triangle is not a right triangle.

Since, length of all sides are not equal, given triangle will be a scalene triangle.

Option D is the correct option.

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Answer:

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defon

Answer:

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