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frutty [35]
2 years ago
6

obtain the equation of two points passing of lines 4x-3y-1=0 and 2x-3y+3=0 and equally inclined to the axes​

Mathematics
1 answer:
klasskru [66]2 years ago
8 0

Answer:By 8(4)P.5, the slope of the line equally inclined to axes is tan(±45  

o

)=±1. Hence by P+λQ=0,

5λ+3

2λ+4

​

=±1

⇒λ=  

3

1

​

,−1. Putting for λ, the two lines are

x−y=0,x+y−2=0.

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OleMash [197]

Answer:

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Step-by-step explanation:

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3 years ago
Given that the two triangles are similar, solve for x if AU = 20x + 108, UB = 273, BC = 703, UV = 444, AV = 372 and AC = 589.
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The answer:
<span>the two triangles are similar
in addition, the line UV is parallel to line BC, so we can use the theorem of Thales for proving the following ratios:

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</span>so we get    (372/589) AB - 80 =  20x, and x = ( (372/589) AB - 80  ) / 20

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Afina-wow [57]
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Read 2 more answers
-2+4(-x+4)=10 using distributive property and collecting like terms
stich3 [128]

Answer:

<h2>x = 1</h2>

Step-by-step explanation:

-2+4(-x+4)=10\qquad\text{use the distributive property:}\ a(b+c)=ab+ac\\\\-2+(4)(-x)+(4)(4)=10\\\\-2-4x+16=10\qquad\text{combine like terms}\\\\-4x+(-2+16)=10\\\\-4x+14=10\qquad\text{subtract 14 from both sides}\\\\-4x+14-14=10-14\\\\-4x=-4\qquad\text{divide both sides by (-4)}\\\\\dfrac{-4x}{-4}=\dfrac{-4}{-4}\\\\x=1

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