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Naddika [18.5K]
3 years ago
12

The long jump distances of four students are listed in the table below. How much farther

Mathematics
1 answer:
Papessa [141]3 years ago
7 0
I can’t answer this with out the distance they jumped.
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What is y in the equation 14-6y=44
Ostrovityanka [42]
14-6y=44
-14 Both sides
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Y=-5
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Aidan bought a jacket on sale for $38.40. If the original price of the jacket was $64, what was the percent of the discount?
UkoKoshka [18]

Answer:

60%

Step-by-step explanation:

$64*.60= $38.40

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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

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If f(x) = 3x^0 - 2x^-1 +4 then f(2)=
Assoli18 [71]

Answer:

try this link

Step-by-step explanation:

https://www3.nd.edu › WorkPDF

Web results

MATH 10550, EXAM 1 SOLUTIONS 1. If f(2) = 5, f(3) = 2, f(4) = 5, g(2 ...

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24% of what number 30
kenny6666 [7]

Answer:

125

Step-by-step explanation:

30*.24=125

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