The geometrical relationships between the straight lines ab and cd
is the straight line ab is parallel to the straight line cd
Step-by-step explanation:
Let us revise some notes:
- If a line is drawn from the origin and passes through point A (a , b), then the equation of OA = ax + by
- If a line is drawn from the origin and passes through point B (c , d), then the equation of OB = cx + dy
- To find the equation of AB subtract OB from OA, then AB = (c - a)x + (d - b)y
- The slope of line AB =

∵ oa = 2 x + 9 y
∵ ob = 4 x + 8 y
∵ ab = OB - OA
∴ ab = (4 x + 8 y) - (2 x + 9 y)
∴ ab = 4 x + 8 y - 2 x - 9 y
- Add like terms
∴ ab = (4 x - 2 x) + (8 y - 9 y)
∴ ab = 2 x + -y
∴ ab = 2 x - y
∵ The slope of ab = 
∵ Coefficient of x = 2
∵ Coefficient of y = -1
∴ The slope of ab = 
∵ cd = 4 x - 2 y
∵ Coefficient of x = 4
∵ Coefficient of y = -2
∴ The slope of cd = 
∵ Parallel lines have same slopes
∵ Slope of ab = slope of cd
∴ ab // cd
The geometrical relationships between the straight lines ab and cd
is the straight line ab is parallel to the straight line cd
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Simplify 9^2/9^7 into
1/9^5 <——that could be the simple answer. If to fully simplify it solve 9^5 which would give you
1/59049
Answer:
The answer is C.
Step-by-step explanation:
Given formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height.
In this case, the initial postion is a platform 30ft above ground so h0=+30
The initial velocity is 38 ft/s straight up into the air so v0=+38
h(t)=-16t2+38t+30
When the object hits the ground, h=0.
h=-16t2+38t+30=0
Simplifying 8t2-19t-15=0
(8t+5)(t-3)=0
t=-5/8 or 3
As time cannot be -ve, t=3s. The answer is C.
14.6 lbs/4 weeks = 3.65 lbs /week
she lost 3.65 pounds each week
Answer:
4 batches
Step-by-step explanation:
4.8/1.2 = 4