Pls i need this asap
1 answer:
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Answer:
Sipho catches up Jabu at 6:30 pm
Step-by-step explanation:
Jabu left town A to travel to town B at 6:00 a.m. he drove at an average speed of 80km/h.
At 8:30 am, when Sipho leaves, we will have traveled 2.5 hours at 80 km/h, so he will be 80*2.5 = 200 km ahead.
From then, his position will be given by:
Sipho left town A to travel to town B. He drove at an average speed of 100 km/h.
His position will be given by:
At what time will Sipho catch up jabu?
He will catch up in t hours, and t is found when:
So
He will catch up 10 hours after 8:30 am, that is, 6:30 pm.
Sipho catches up Jabu at 6:30 pm
Answer:
The correct options are;
1) ΔBCD is similar to ΔBSR
2) BR/RD = BS/SC
3) (BR)(SC) = (RD)(BS)
Step-by-step explanation:
1) Given that RS is parallel to DC, we have;
∠BDC = ∠BRS (Angles on the same side of transversal)
Similarly;
∠BCD = ∠BSR (Angles on the same side of transversal)
∠CBD = ∠CBD = (Reflexive property)
Therefore;
ΔBCD ~ ΔBSR Angle, Angle Angle (AAA) rule of congruency
2) Whereby ΔBCD ~ ΔBSR, we therefore have;
BC/BS = BD/BR → (BS + SC)/BS = (BR + RD)/BR = 1 + SC/BS = RD/BR + 1
1 + SC/BS = 1 + RD/BR = SC/BS = 1 + BR/RD - 1
SC/BS = RD/BR
Inverting both sides
BR/RD = BS/SC
3) From BR/RD = BS/SC the above we have by cross multiplication;
BR/RD = BS/SC gives;
BR × SC = RD × BR → (BR)(SC) = (RD)(BR).
