-1 > -2(x-4) - 5(4x - 7)
-1 > -2x + 8 - 20x + 35
-1 > -22x + 43
-1 - 43 > - 22x
-44 > - 22x
-44/-22 < x
2 < x....or x > 2
t + 23 > 1 subtract 23 from both sides
t + 23 - 23 > 1 - 23
t > -22
I: 12x-5y=0
II:(x+12)^2+(y-5)^2=169
with I:
12x=5y
x=(5/12)y
-> substitute x in II:
((5/12)y+12)^2+(y-5)^2=169
(25/144)y^2+10y+144+y^2-10y+25=169
(25/144)y^2+y^2+10y-10y+144+25=169
(25/144)y^2+y^2+144+25=169
(25/144)y^2+y^2+169=169
(25/144)y^2+y^2=0
y^2=0
y=0
insert into I:
12x=0
x=0
-> only intersection is at (0,0) = option B
Answer:
Vector u has u_x = (5 - 15) = -10, and u_y = -4 - 22 = -26, and its component form would be u = -10i - 26j.
If vector v is in the opposite direction: 10i + 26j
And if it is double in magnitude: v = 20i + 52j
Hope this helps you! Ask me anything if yu have any quistions!
Answer:
1. 570
2. 5,700
3. 57,000
Step-by-step explanation:
Hope it helped
