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snow_tiger [21]
3 years ago
6

Answerrrrrrrrrrrrrrrrrrrrrr nowwwwwww

Mathematics
1 answer:
MrMuchimi3 years ago
7 0

\rm{\pink{\underline{\underline{\blue{ANSWER:-}}}}}

  • √64 = 8

  • \rm{8\dfrac{1}{7}\:=\:57/7} = 8.142

  • \rm{8.\overline{14}} = 8.1414..

  • 15/2 = 7.5

=> 8.142 > 8.1414 > 8 > 7.5 .

\rm\red{\therefore} 2) \rm{8\dfrac{1}{7}} , \rm{8.\overline{14}} , √64 , 15/2

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▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

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Solve the following quadratic equation 6x^2-5x-4=0
jeka94

Answer:

=

−

±

2

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4

√

2

x=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c}}}}{2{\color{#c92786}{a}}}

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Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.

6

2

−

5

−

4

=

0

6x^{2}-5x-4=0

6x2−5x−4=0

=

6

a={\color{#c92786}{6}}

a=6

=

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b={\color{#e8710a}{-5}}

b=−5

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c={\color{#129eaf}{-4}}

c=−4

=

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(

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5

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±

(

−

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)

2

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⋅

6

(

−

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)

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2 years ago
What are the endpoint coordinates for the midsegment of △BCD that is parallel to BC?
natulia [17]

Answer:

<em>The endpoint coordinates for the mid-segment are:  (-2,-1) and (1,0)</em>

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