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Komok [63]
2 years ago
5

Which part of this graph shows a nonlinear relationship? Distance from home 2. Time

Mathematics
2 answers:
MissTica2 years ago
3 0
Since 2 is not in a line
Answer=2
Alisiya [41]2 years ago
3 0

Answer:

thanks

Step-by-step explanation:

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Volgvan
Because he was wrong
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3 years ago
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Label each triangle
Ne4ueva [31]

Answer: Top right and left are ACUTE, Bottom right is RIGHT, and bottom left is OBTUSE.

4 0
3 years ago
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Triangle JKL has vertices J(2,5), K(1,1), and L(5,2). Triangle QNP has vertices Q(-4,4), N(-3,0), and P(-7,1). Is (triangle)JKL
Tems11 [23]

Answer:

Yes they are

Step-by-step explanation:

In the triangle JKL, the sides can be calculated as following:

  • J(2;5); K(1;1)

             => JK = \sqrt{(1-2)^{2} + (1-5)^{2}  } = \sqrt{(-1)^{2}+(-4)^{2}  } = \sqrt{1+16}=\sqrt{17}

  • J(2;5); L(5;2)

             => JL = \sqrt{(5-2)^{2} + (2-5)^{2}  } = \sqrt{3^{2}+(-3)^{2}  } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}

  • K(1;1); L(5;2)

             =>  KL = \sqrt{(5-1)^{2} + (2-1)^{2}  } = \sqrt{4^{2}+1^{2}  } = \sqrt{1+16}=\sqrt{17}

In the triangle QNP, the sides can be calculate as following:

  • Q(-4;4); N(-3;0)

             => QN = \sqrt{[-3-(-4)]^{2} + (0-4)^{2}  } = \sqrt{1^{2}+(-4)^{2}  } = \sqrt{1+16}=\sqrt{17}

  • Q (-4;4); P(-7;1)

   => QP = \sqrt{[-7-(-4)]^{2} + (1-4)^{2}  } = \sqrt{(-3)^{2}+(-3)^{2}  } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}

  • N(-3;0); P(-7;1)

             =>  NP = \sqrt{[-7-(-3)]^{2} + (1-0)^{2}  } = \sqrt{(-4)^{2}+1^{2}  } = \sqrt{16+1}=\sqrt{17}

It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP

=> They are congruent triangles

7 0
3 years ago
Read 2 more answers
You have 2 cats<br> and 3 dogs. Write<br> the ratio of cats to<br> dogs two different<br> ways.
fomenos

Answer:

2:3

3:2

Step-by-step explanation:

3 0
3 years ago
How do you solve (Arc)QPT if &lt;QZT = 120
yuradex [85]
By definition, the arc length is given by:
 arc = R * theta * ((2 * pi) / 360)
 Where,
 theta: angle in degrees
 R: radio
 We have then:
 (Arc) QPT if <QZT = 120:
 theta = 360-120 = 240 degrees
 R = 13.5 units
 Substituting values we have:
 (Arc) QPT = R * theta * ((2 * pi) / 360)
 (Arc) QPT = (13.5) * (240) * ((2 * pi) / 360)
 (Arc) QPT = 56.55 units
 Answer:
 
(Arc) QPT = 56.55 units
7 0
3 years ago
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