Compute 1/(√3 -1) correct to two decimal places(√3≈1.73)
Answers
Answer:
3x+2y = 183x+2y = 183x+2y = 183x+2y = 183x+2y = 183x+2y = 18Given equations are
2x+3y=11−−−−(1)
2x−4y=−24−−−−(2)
Form (1)
2x+3y=11
⇒2x=11−3y
⇒x=
2
11−3y
−−−(3)
substituting x in(2)
2x−4y=−24
⇒2(
2
11−3y
)−4y=−24
⇒11−3y−4y=−24
⇒11−7y=−24
⇒7y=35
⇒y=35/7
⇒y=5.
putting y = 5 in (3)
x=
2
11−3(5)
⇒x=
2
11−15
⇒x=−4/2
∴x=−2.
Hence x = -2 and y = 5 is the solution of the
equation.
Now, we have to find m
y=mx+3 ∴m=−1
5=3(−2)+3
5−3=−2m⇒−2m=2
⇒m=−2/x=−1Given equations are
2x+3y=11−−−−(1)
2x−4y=−24−−−−(2)
Form (1)
2x+3y=11
⇒2x=11−3y
⇒x=
2
11−3y
−−−(3)
substituting x in(2)
2x−4y=−24
⇒2(
2
11−3y
)−4y=−24
⇒11−3y−4y=−24
⇒11−7y=−24
⇒7y=35
⇒y=35/7
⇒y=5.
putting y = 5 in (3)
x=
2
11−3(5)
⇒x=
2
11−15
⇒x=−4/2
∴x=−2.
Hence x = -2 and y = 5 is the solution of the
equation.
Now, we have to find m
y=mx+3 ∴m=−1
5=3(−2)+3
5−3=−2m⇒−2m=2
⇒m=−2/x=−1Given equations are
2x+3y=11−−−−(1)
2x−4y=−24−−−−(2)
Form (1)
2x+3y=11
⇒2x=11−3y
⇒x=
2
11−3y
−−−(3)
substituting x in(2)
2x−4y=−24
⇒2(
2
11−3y
)−4y=−24
⇒11−3y−4y=−24
⇒11−7y=−24
⇒7y=35
⇒y=35/7
⇒y=5.
putting y = 5 in (3)
x=
2
11−3(5)
⇒x=
2
11−15
⇒x=−4/2
∴x=−2.
Hence x = -2 and y = 5 is the solution of the
equation.
Now, we have to find m
y=mx+3 ∴m=−1
5=3(−2)+3
5−3=−2m⇒−2m=2
⇒m=−2/x=−1Given equations are
2x+3y=11−−−−(1)
2x−4y=−24−−−−(2)
Form (1)
2x+3y=11
⇒2x=11−3y
⇒x=
2
11−3y
−−−(3)
substituting x in(2)
2x−4y=−24
⇒2(
2
11−3y
)−4y=−24
⇒11−3y−4y=−24
⇒11−7y=−24
⇒7y=35
⇒y=35/7
⇒y=5.
putting y = 5 in (3)
x=
2
11−3(5)
⇒x=
2
11−15
⇒x=−4/2
∴x=−2.
Hence x = -2 and y = 5 is the solution of the
equation.
Now, we have to find m
y=mx+3 ∴m=−1
5=3(−2)+3
5−3=−2m⇒−2m=2
⇒m=−2/x=−1Given equations are
2x+3y=11−−−−(1)
2x−4y=−24−−−−(2)
Form (1)
2x+3y=11
⇒2x=11−3y
⇒x=
2
11−3y
−−−(3)
substituting x in(2)
2x−4y=−24
⇒2(
2
11−3y
)−4y=−24
⇒11−3y−4y=−24
⇒11−7y=−24
⇒7y=35
⇒y=35/7
⇒y=5.
putting y = 5 in (3)
x=
2
11−3(5)
⇒x=
2
11−15
⇒x=−4/2
∴x=−2.
Hence x = -2 and y = 5 is the solution of the
equation.
Now, we have to find m
y=mx+3 ∴m=−1
5=3(−2)+3
5−3=−2m⇒−2m=2
⇒m=−2/x=−1Given equations are
2x+3y=11−−−−(1)
2x−4y=−24−−−−(2)
Form (1)
2x+3y=11
⇒2x=11−3y
⇒x=
2
11−3y
−−−(3)
substituting x in(2)
2x−4y=−24
⇒2(
2
11−3y
)−4y=−24
⇒11−3y−4y=−24
⇒11−7y=−24
⇒7y=35
⇒y=35/7
⇒y=5.
putting y = 5 in (3)
x=
2
11−3(5)
⇒x=
2
11−15
⇒x=−4/2
∴x=−2.
Hence x = -2 and y = 5 is the solution of the
equation.
Now, we have to find m
y=mx+3 ∴m=−1
5=3(−2)+3
5−3=−2m⇒−2m=2
⇒m=−2/x=−1=
n
∑x
2
−(
n
∑x
)
2
=
10
1530
−(12)
2
=
153−144
=
9
=3
Coefficient of variation =
x
σ
×100
=
12
3
×100
=
4
1
×100
=25
Answer:
1•73 has place of decimal