Find the tangent and normal plane equation of the intersection of surface x ^ 2 + y ^ 2 = 1 / 2 * Z ^ 2 and plane x + y + Z = 2 at point (1, - 1,2)

Find the tangent and normal plane equation of the intersection of surface x ^ 2 + y ^ 2 = 1 / 2 * Z ^ 2 and plane x + y + Z = 2 at point (1, - 1,2)

Intersection line: x ^ 2 + y ^ 2 = 1 / 2 * Z ^ 2
x+y+z=2
=>The equations are derived from Z respectively
2xx'+2yy' =z
x'+ y' + 1 =0
Point (1, - 1,2) substituting:
2x'-2y' =2 => x'-y'=1
x'+y'=-1
x'=0 y'=-1
Tangent vector s = (0, - 1,1)
Tangent equation x = 0, y + Z = - 1
Normal plane equation y + Z = - 1

0.75 + (negative 11 / 4) + 0.125 + (negative 5 / 7) + (negative 4 / 8) + 0.25 = how much, to process, thank you

0.75 + (11 / 4) + 0.125 + (5 / 7) + (1 / 8) + 0.25
=0.75 + 0.25-11 / 4-1 / 8-4 and 1 / 8
=1-11 / 4-4 and 1 / 4
=1-7
=-6

Calculation: (6a ^ 2-2a + 1) - 4 (2-5A + A ^ 2)

2a^2+18a-7

There are several groups of 33 (1 to 33) numbers, the sum of which is 104

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Binary linear equations 4x-2y = 4,2x + y = 2

4x-2y=4 （1）
2x+y=2 （2）
By adding (1) + (2) * 2, we can get: 1
8x=8
The solution is: x = 1 (3)
By substituting x = 1 into (1), we can get the following results
4-2y=4 （4）
The solution is y = 0
So:
x=1
y=0

5 plus 3 equals 28, 9 plus 1 equals 810, 8 plus 6 equals 214, 5 plus 4 equals 19, so how much is 7 plus 3?

410

Write - 1 - [- 3 - (- 5) - (+ 7)] - (+ 9) into algebraic formula and form

Write - 1 - [- 3 - (- 5) - (+ 7)] - (+ 9) into algebraic formula and form
=-1+3-5+7-9
=（-1）+3+（-5）+7+（-9）
=10-15
=-5；
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Two times one-third plus one-of-three times four to 49 times one-of-50

Two times one-third plus one-of-three times four to 49 times one-of-50
=(1 / 2-1 / 3) + (1 / 3-1 / 4) + +(1 / 49-1 / 50)
=1 / 2-1 / 3 + 1 / 3-1 / 4 + +1 / 49-1 / 50
=1 / 2-1 / 50
=12 out of 25

Known triangle area is 12 square centimeter, bottom is 8 centimeter, seek height?

12*2/8=3cm
h=2s/a

Solve equation 18-x = 12.4

18-x=12.4
x=18-12.4
x=5.6
Hope to help you