When three or more lines intersect at one point, they are ____________.

1 answer:

Nookie1986 [14]3 years ago

5 0

The correct answer is Concurrent Lines. In geometry, when three or more lines in a plane or a higher dimensional space intersect at a single point, the lines are said to be concurrent.

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2d+7/6-d-5/3=0 Tell me the answer

Inessa [10]

Answer:

d=1／2

Step-by-step explanation:

$2d + \frac{7}{6} - d - \frac{5}{3} = 0 \\ 2d - d = \frac{5}{3} - \frac{7}{6} \\ d = \frac{10}{6} - \frac{7}{6} \\ d = \frac{3}{6} \\ d = \frac{1}{2}$

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3 years ago

What is equivalent to (3^6/y^-5)^2

Fantom [35]

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3 years ago

(4x^3 - x^2 +4x) + (x^3 -x^2 - 4x)

madreJ [45]

Answer:

5x³ - 2x²

Step-by-step explanation:

Step 1: Write out expression

4x³ - x² + 4x + x³ - x² - 4x

Step 2: Combine like terms (x³)

5x³ - x² + 4x - x² - 4x

Step 3: Combine like terms (x²)

5x³ - 2x² + 4x - 4x

Step 4: Combine like terms (x)

5x³ - 2x²

4 0

3 years ago

How do you find a vector that is orthogonal to 5i + 12j ?

Rashid [163]

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\
slope=\cfrac{a}{{{ b}}}\qquad negative\implies -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\
-------------------------------\\\\$

$\bf \boxed{5i+12j}\implies 
\begin{array}{rllll}
\ \textless \ 5&,&12\ \textgreater \ \\
x&&y
\end{array}\quad slope=\cfrac{y}{x}\implies \cfrac{12}{5}
\\\\\\
slope=\cfrac{12}{{{ 5}}}\qquad negative\implies -\cfrac{12}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{12}
\\\\\\
\ \textless \ 12, -5\ \textgreater \ \ or\ \ \textless \ -12,5\ \textgreater \ \implies \boxed{12i-5j\ or\ -12i+5j}$

if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.

so any other vector that has a negative reciprocal slope to it, will then be perpendicular or "orthogonal" to it.

so... for example a parallel to <-12, 5> is say hmmm < -144, 60>, if you simplify that fraction, you'd end up with <-12, 5>, since all we did was multiply both coordinates by 12.

or using a unit vector for those above, then

$\bf \textit{unit vector}\qquad \cfrac{\ \textless \ a,b\ \textgreater \ }{||\ \textless \ a,b\ \textgreater \ ||}\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{\sqrt{a^2+b^2}}\implies \cfrac{a}{\sqrt{a^2+b^2}},\cfrac{b}{\sqrt{a^2+b^2}}
\\\\\\
\cfrac{12,-5}{\sqrt{12^2+5^2}}\implies \cfrac{12,-5}{13}\implies \boxed{\cfrac{12}{13}\ ,\ \cfrac{-5}{13}}
\\\\\\
\cfrac{-12,5}{\sqrt{12^2+5^2}}\implies \cfrac{-12,5}{13}\implies \boxed{\cfrac{-12}{13}\ ,\ \cfrac{5}{13}}$

4 0

4 years ago