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devlian [24]
2 years ago
15

A number line goes from negative 5 to positive 5. Point D is at negative 4 and point E is at positive 5. A line is drawn from po

int D to point E.
What is the location of point F, which partitions the directed line segment from D to E into a 5:6 ratio?
Negative one-eleventh
One-eleventh
Two-fifteenths
Fifteen-halves
Mathematics
1 answer:
Ronch [10]2 years ago
7 0

<em>Directed numbers</em> are numbers that have either a <u>positive</u> or <u>negative </u>sign, which can be shown on a <em>number line</em>. Therefore, point F is Fifteen-halves of line <em>segment</em> DE.

A <u>number line</u> is a system that can show the positions of <em>positive</em> or <em>negative</em> numbers. It has its <em>ends</em> ranging from <em>negative infinity</em> to <em>positive infinity</em>. Thus any <em>directed</em> number can be located on the line.

Directed numbers are numbers with either a <u>negative</u> or <u>positive </u>sign, which shows their direction with respect to the <em>number line.</em>

In the given question, the <u>distance</u> between points D and E is <em>9 units</em>. So that <em>dividing</em> 9 units in the ratio of 5 to 6, we have;

\frac{5}{6} x 9   = \frac{45}{6}

          = \frac{15}{2}

Therefore, the <em>location</em> of point F, which <u>partitions</u> the directed line segment from d to E into a 5:6 ratio is    \frac{15}{2}. Thus the<em> answer</em> is <u>Fifteen-halves.</u>

For further clarifications on a number line, visit: brainly.com/question/23379455

#SPJ1

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Hat is the solution to the equation 3/3(4c + 16)=2c+9 ?<br><br> c =
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\frac{3}{3(4c+16)}=2c+9
\frac{1}{4c+16}=2c+9
times 4c+16 to both sides
1=(2c+9)(4c+16)
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(3/3)(4c+16)=2c+9
1(4c+16)=2c+9
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minus 2c both sides
2c+16=9
minus 16 both sides
2c=-7
divide both sides by 2
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c=-3.5





if it is \frac{3}{3(4c+16)}=2c+9,
c=\frac{-17- \sqrt{3} }{4} or \frac{-17+ \sqrt{3} }{4}


if it is (\frac{3}{3})(4c+16)=2c+9, c=-3.5



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