$\dfrac{2m}{5}+3=7(4m-3)\ \ \ \ |use\ distributive\ property:a(b-c)=ab-ac\\\\\dfrac{2m}{5}+3=28m-21\ \ \ \ \ |multiply\ both\ sides\ by\ 5\\\\2m+15=140m-105\ \ \ \ |subtract\ 140m\ from\ both\ sides\\\\-138m+15=-105\ \ \ \ |subtract\ 15\ from\ both\ sides\\\\-138m=-120\ \ \ \ |divide\ both\ sides\ by\ (-138)\\\\m=\dfrac{120:6}{138:6}\\\\\boxed{m=\frac{20}{23}}$

7 0

3 years ago

A student wants to find point C on the directed line segment from A to B on a number line such that the segment is partitioned i

olganol [36]

Answer:

Point C at (-18/7)

Step-by-step explanation:

Point A is at -6 and point B is at 2.

So, the distance between A and B = 2 - (-6) = 8

point C on the directed line segment from A to B on a number line such that the segment is partitioned in a ratio of 3:4.

Let the distance AC = x

∴ BC = 8 - x

AC : CB = 3 : 4

∴ $\frac{AC}{CB} =\frac{3}{4} = \frac{x}{8-x}$

Using cross multiplication

3 (8-x) = 4x

24 - 3x = 4x

24 = 7x

x = 24/7

So, Point C = -6 + 24/7 = -18/7

Point C at (-18/7)

7 0

3 years ago

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