Answer:
The slope is -1/3 and the y-intercept is (0, 2).
Step-by-step explanation:
2x+6y=12
6y=12-2x
6y=-2x+12
y=-2/6x+12/6
y=-1/3x+2
y=mx+b where m=slope and b=y-intercept
Hello :)
3x+y=27
-3x+4y= -42
+____________
(3x -3x) + (y+4y) = (27-42)
0 + 5y = -15
y = -3
Put "-3" for "y" in the system:
3x + y =27
3x -3=27
3x = 30
x = 10
Answer = x:10 y: -3
Have a nice day :)
Answer:
41/35
Step-by-step explanation:
Simplify the following:
4/7 + 3/5
Put 4/7 + 3/5 over the common denominator 35. 4/7 + 3/5 = (5×4)/35 + (7×3)/35:
(5×4)/35 + (7×3)/35
5×4 = 20:
20/35 + (7×3)/35
7×3 = 21:
20/35 + 21/35
20/35 + 21/35 = (20 + 21)/35:
(20 + 21)/35
| 2 | 1
+ | 2 | 0
| 4 | 1:
Answer: 41/35
The square (call it
) has one vertex at the origin (0, 0, 0) and one edge on the y-axis, which tells us another vertex is (0, 3, 0). The normal vector to the plane is
, which is enough information to figure out the equation of the plane containing
:

We can parameterize this surface by

for
and
. Then the flux of
, assumed to be
,
is


