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
<span>(-yx^2)^2 • -2x^0y^3
= x^4y^2 (-2y^3)
= - 2x^4y^5</span>
Step-by-step explanation:
in ascending order;
4/15 ,1/5 ,2/3
Answer:
Step-by-step explanation:
<u>For old circular garden:</u>
take the radius as r.
then use the formula to find area of circle: πr² ......this is old garden area.
<u>For new enlarged garden:</u>
the radius is twice the old radius so, radius = 2 * r = 2r ......enlarged radius
now find area for this new garden: π(2r)² → 4πr²
In common fractions: (old garden)/(new garden)
: ( πr² ) / ( 4πr² )
: 1/4
Answer:
B. x<27
Step-by-step explanation:
300-165=135
135/5=27
