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2 years ago

Tried to think didnt work it out help

Mathematics

2 answers:

Anna [14]2 years ago

6 0

Answer:

Givn : OK = 3, OJ = 30, KN = 1, JM = 3

ON=OK+KN=3+1=4

OM=OJ+JM=30+10=40

In triangles OKJ & ONM,

OKOJ=330=110

ONOM=440=110

Angle O is common in both the triangles.

Two sides are in same proportion and the included angle is common (SAS) . Hence both the triangles are similar.

That means KJNM=110 or the two sides are parallel.

Hence ˆK=ˆN,ˆJ=ˆM corresponding angles.

AlladinOne [14]2 years ago

3 0

Answer:

Givn : OK = 3, OJ = 30, KN = 1, JM = 3

ON=OK+KN=3+1=4

OM=OJ+JM=30+10=40

In triangles OKJ & ONM,

OKOJ=330=110

ONOM=440=110

Angle O is common in both the triangles.

Two sides are in same proportion and the included angle is common (SAS) . Hence both the triangles are similar.

That means KJNM=110 or the two sides are parallel.

Hence ˆK=ˆN,ˆJ=ˆM corresponding angles.

Step-by-step explanation:

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Step-by-step explanation:

Cramer's Rule

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

$\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.$

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$\Delta_x=\begin{vmatrix}p &b \\q &d \end{vmatrix}$

And

$\Delta_y=\begin{vmatrix}a &p \\c &q \end{vmatrix}$

The solution for x and y is

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$\displaystyle y=\frac{\Delta_y}{\Delta}$

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$\Delta=\begin{vmatrix}1 &1 \\1 &-2 \end{vmatrix}=-2-1=-3$

$\Delta_x=\begin{vmatrix}1 &1 \\4 &-2 \end{vmatrix}=-2-4=-6$

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