Answer:
z = 4
y = -1
x = 2
Step-by-step explanation:
x - y = 3
Therefore => x = 3 + y
y + 2z =7
Therefore => y= 7 - 2z
We have the equation :
2x + 3y + 4z = 17
2(3 + y) + 3(7 - 2z) + 4z = 17
6 + 2y + 21 - 6z +4z = 17
2(7 -2z) - 2z + 27 = 17
14 - 4z - 2z + 27 = 17
-6z + 41 = 17
-6z = -24
z = 4
Find the interquartile range of 17, 23, 8, 5, 9, 16, 22, 11, 13, 15, 17, 18
Yuri [45]
8.25
Hope I helped! ( Smiles )
Answer:
x1 =2-5i*sqrt(2)
x2 =2+5i*sqrt(2)
Step-by-step explanation:
-x^2 +4x-54=0 (quadratic equation)
a=-1, b=4, c=-54
x1=(-b+sqrt(b^2-4ac))/2a
x1=(-4+sqrt(4^2 - 4*(-1)(-54))/2*(-1)
x1=(-4+sqrt(16-216))/(-2)
x1 =(-4+sqrt(-200))/(-2)
x1 =(-4+sqrt(200i^2))/(-2) i^2=-1
x1 =(-4+sqrt(100*2*i^2))/(-2)
x1 =(-4+10i*sqrt(2))/(-2)
x1 =2-5i*sqrt(2)
x2 =(-b-sqrt(b^2-4ac))/2a
x2 =(-4-10i*sqrt(2))/(-2)
x2 =2+5i*sqrt(2)
3 and 4
Step-by-step explanation:
y=5x
y=5(0)
y=0
y=5x
y=5(10)
y=50
y=5x
y=5(51)
y=255
y=5x
y=5(400)
y=2000