Answer:
See explanation
Step-by-step explanation:
If PK and PF are tangent to the circle, then the radii OK and OF are perpendicular to the lines PK and PF respectively.
Consider quadrilateral PFOK. The sum of the measures of all interior angles in quadrilateral PFOK is equal to 360°. Since OK and OF are perpendicular to the lines PK and PF, you have that
∠FOK+90°+∠KPF+90°=360°,
∠FOK=360°-90°-90°-∠KPF=180°-∠KPF.
Angles FOK and KOE are supplementary angles, then
∠KOE=180°-∠FOK=180°-(180°-∠KPF)=∠KPF.
Since angle KOE is central angle, the arc EK has the same measure as angle KOE and as angle KPF.
Find the slope of this line
slope=(y2-y1)/(x2-x1)
slope=(-9-(-4))/(0-(-3))=(-9+4)/(3)=-5/3
yint is y=-9
y=mx+b
y=-5/3x-9
parllell means has same slope
y=-5/3x+b
find b by inputing point
(3,-7)
(x,y)
x=3
y=-7
-7=-5/3(3)+b
-7=-5+b
add 5
-2=b
the equation of the line is y=-5/3x-2
A) 24
24 x 10
b)
103
103 x 10
C) 35
35 x 10
D) 73
73 x 10
E) 28
28 x 10
F) 41
41 x 10
Answer:
The graph of the inequality is attached below.
Step-by-step explanation:
Given the inequality
The graph of the inequality is attached below.
As i said before put the center instead of (h,k) in the general formula and put r=1
so (x-(-2))^2 + (y-(-5)^2 = 1
(x+2)^2 + (y+5)^2 = 1
