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3 years ago

. In the adjoining fig-6 , name a) any 4 pairs of intersecting lines. b) 3 non – collinear points c) 4 collinear points

Mathematics

1 answer:

babymother [125]3 years ago

8 0

Answer:

a)pl,pm,pn,ql

b)B,C and F are non-collinear points

c) A,C and E are collinear points

B, D and F are collinear points

Step-by-step explanation:

a) any 4 pairs of intersecting lines.

Line p intersects line l , line m and line n

Line q intersects line l , line m and line n

So, 4 pairs of intersecting lines are : pl,pm,pn,ql

b)3 non - col-linear points

Three or more points that lie on a same straight line are called collinear points.

When points are not collinear , then they are said to be non-collinear

So, B,C and F do not lie on the same line

So, They are non-collinear points

c)collinear points

A,C and E lies on same line

So, They are collinear

B, D and F lies on same line

So, they are collinear

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eduard

Enrolement increased by 30% that means last year=100%
incease of 30%=30+100=130% now

now=182

percent means parts out of 100 so
130%=130/100=13/10

182=13/10 of x (x represents the number last year)
of means multipy so
182=13/10 times x
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3 years ago

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Answer:

Step-by-step explanation:

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2 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

Agatha is five times older than her son bob. three years ago she was nine times older. how old is agatha?

Naya [18.7K]

Write an equation system based on the problem
"Agatha is five times older than her son, Bob" could be written as:
a = 5b.......(first equation)
"Three years ago, she was nine times older" could be written as:
a - 3 = 9(b - 3)......(second equation)

Solve the equation system
To find the value of b, substitute 5b as a to the second equation
a - 3 = 9(b - 3)
5b - 3 = 9(b - 3)
5b - 3 = 9b - 27
5b - 9b = -27 + 3
-4b = -24
b = -24/-4
b = 6
Her son, Bob, is 6 years old

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3 years ago

Help me with question a please ! With full workings !

frosja888 [35]

A)

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
Q&({{ 0}}\quad ,&{{ 2}})\quad 
% (c,d)
P&({{ 0.5}}\quad ,&{{ 0}})
\end{array}\qquad 
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}$

$\bf QP=\sqrt{(0.5-0)^2+(0-2)^2}\implies QP=\sqrt{0.5^2+2^2}
\\\\\\
QP=\sqrt{\left( \frac{1}{2} \right)^2+4}\implies QP=\sqrt{ \frac{1^2}{2^2}+4}\implies QP=\sqrt{\frac{1}{4}+4}
\\\\\\
QP=\sqrt{\frac{17}{4}}\implies QP=\cfrac{\sqrt{17}}{\sqrt{4}}\implies QP=\cfrac{\sqrt{17}}{2}$

b)

since QR=QP, that means that QO is an angle bisector, and thus the segments it makes at the bottom of RO and OP, are also equal, thus RO=OP

thus, since the point P is 0.5 units away from the 0, point R is also 0.5 units away from 0 as well, however, is on the negative side, thus R (-0.5, 0)

c)

what's the equation of a line that passes through the points (-0.5, 0) and (0,2)?

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
Q&({{ 0}}\quad ,&{{ 2}})\quad 
% (c,d)
R&({{ -0.5}}\quad ,&{{ 0}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{0-2}{-0.5-0}\implies \cfrac{-2}{-0.5}$

$\bf m=\cfrac{\frac{-2}{1}}{-\frac{1}{2}}\implies \cfrac{-2}{1}\cdot \cfrac{2}{-1}\implies 4
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-2=4(x-0)\implies y=4x+2\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}$

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3 years ago