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alexira [117]
4 years ago
14

Solve for x. If the equation is an identity write identity or if it has no solution write no solution.

Mathematics
2 answers:
AlladinOne [14]4 years ago
8 0

3(5x−2)−6x=3(3x+2)

distribute

15x-6 -6x = 9x+6

combine like terms

9x-6 = 9x+6

subtract 9x from each side

-6 =6

no solution

Zina [86]4 years ago
7 0

3(5x−2)−6x=3(3x+2)


distribute


15x-6 -6x = 9x+6


combine like terms


9x-6 = 9x+6


subtract 9x from each side


-6 =6


no solution

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Write the equation of the line perpendicular to y=-1/4x-7 that passes through the point (3,-2)
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the equation of line is given by;

y-y1=m(x-x1)

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In the diagram below, we have $ST = 2TR$ and $PQ = SR = 20$. Find the length $UV$.
Maslowich

Answer:

12

Step-by-step explanation:

1. ST=2TR and SR=20, then by segment addition postulate

ST+TR=SR\\ \\2TR+TR=SR\\ \\3TR=20\\ \\TR=\dfrac{20}{3}\\ \\ST=2\cdot \dfrac{20}{3}=\dfrac{40}{3}

2. Consider triangles PQU and TSU. These triangles are similar by AA similarity theorem (triangles have congruent vertical angles PUQ and TUS and congruent alternate interior angles PQU and TSU). Similar triangles have proportional corresponding sides, so

\dfrac{QU}{US}=\dfrac{PQ}{ST}=\dfrac{20}{\frac{40}{3}}=\dfrac{60}{40}=1.5

QU=1.5US

3. Consider triangles QUV and QSR. These triangles are similar by AA similarity theorem. Similar triangles have proportional corresponding sides, so

\dfrac{QU}{QS}=\dfrac{UV}{SR}\\ \\\dfrac{QU}{QU+US}=\dfrac{1.5US}{1.5US+US}=\dfrac{1.5}{2.5}=0.6

so

\dfrac{UV}{SR}=0.6\Rightarrow UV=0.6SR=0.6\cdot 20=12

7 0
3 years ago
Read 2 more answers
PLEASE help me with this question! This is really urgent! No nonsense answers please.
s344n2d4d5 [400]

Answer:

140°

Step-by-step explanation:

\because m\widehat{BG} = 360\degree - m\widehat{GCB} \\\therefore m\widehat{BG} = 360\degree - 300\degree \\\therefore m\widehat{BG} = 60\degree \\\because m\widehat{BGD} = m\widehat{BG} +m\widehat{GD}\\\therefore m\widehat{BGD} = 80\degree+60\degree\\\therefore m\widehat{BGD} = 140\degree\\\because m\angle BAD = m\widehat{BGD} \\\huge\purple {\boxed {\therefore m\angle BAD =140\degree}}

7 0
3 years ago
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