A. Evaluate the following expressions:
a. 202*3)(3**2)(-1**3)+6/2-3*2
b. (2*3) + 2(3**2) + (-1)
(xy + 2y2+z)/(x+y+z) for x=3, y=2, and z=1
C.(xy + 2y2+z)/(x+y+z) for x=3,y=2, and z=1
Answers
Answered by
1
Answer:
ANSWER
x=3+2
2
x
1
=
3+2
2
1
×
3−2
2
3−2
2
=3−2
2
(
x
−
2
1
)
2
=x+
x
1
−2
=3+2
2
+3−2
2
−2=4
So,
x
−
x
1
=
4
=±12 Ans
solution
Answered by
0
Explanation:
a)
(202)(3)(3)(2)((−1)(3))+6/2−(3)(2)
=606(3)(2)((−1)(3))+6/2−(3)(2)
=(606)(6)((−1)(3))+6/2−(3)(2)
=3636((−1)(3))+6/2−(3)(2)
=(3636)(−3)+6/2−(3)(2)
=−10908+6/2−(3)(2)
=−10908+3−(3)(2)
=−10905−(3)(2)
=−10905−6
=−10911
b)
(2)(3)+2(3)(2)−1
=6+2(3)(2)−1
=6+(2)(6)−1
=6+12−1
=18−1
=17
c)
xy+2y2+z/x+y+z
Evaluate for x=3,y=2,z=1
(3)(2)+2(22)+1/3+2+1
=5/2(Decimal: 2.5)
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