Y + 4x < 8
y < -4x + 8
2 points that satisfy this are (0,8) and (2,0)....and those happen to be ur x and y intercepts (where the line crosses the x and y axis)
graph...so go ahead and plot ur x and y intercepts (0,8) and (2,0).....ur slope is - 4.....so start at ur y int (0,8) and go down 4 spaces, and to the right 1...plot that point, then go down 4 spaces and to the right 1, then plot that point...keep doing this and u will have ur line...u should have crossed the x axis at (0,2)......ur line will be a dashed line since the problem has no equal sign.... the shading will go below the line because it is less then.
y - 3 > = 1/2x
y > = 1/2x + 3
2 points that satisfy this are : (0,3) and (-6,0)...ur x and y intercepts
graph : plot ur intercepts (0,3) and (-6,0)....u have a slope of 1/2...so start at ur x intercept (-6,0) and go up 1 space, and to the right 2 spaces, plot that point...then go up 1 and to the right 2, plot that point...keep doing this and u will cross the y axis at (0,3)....this line will be a solid line....the shading will go above the line.
Basically, you just move every individual point up, down, left, or right by the amount indicated.
For example, point G is graphed on the point (-3, -1). Moving it right 5 and up 1 will give you G’, which is (2,0)
T is graphed on the point (-1, -1). Moving it right 5 and up 1 will give you T’, which is (4,0)
B is graphed on the point (-3, -5). Moving it right 5 and up 1 will give you B’, which is
(2,4)
Answer:
c. 12
Step-by-step explanation:
PA² = (BC + PB) × PB => tangent-secant theorem
PA = 8
PB = 4
BC = ?
Substitute
8² = (BC + 4) × 4
64 = 4BC + 16
64 - 16 = 4BC + 16 - 16
48 = 4BC
48/4 = 4BC/4
12 = BC
BC = 12
12)
9/10 / 3/4
= 9/10 * 4/3
= 12/10
= 6/5
