What is the answer to this problem.

A car travels for 4(t+3) hours at a constant speed of 10(t+3) km/h. If the total distance travelled by the car is 810 km, find t

IceJOKER [234]

Step-by-step explanation:

$\because \: time \times speed = distance \\ \therefore \: 4(t + 3) \times 10(t + 3) = 810 \\ \therefore \: 4(t + 3)^{2} = 81 \\ \therefore \: \{2(t + 3) \}^{2} = {9}^{2} \\ \therefore \: \{2(t + 3) \} = {9} \\ \therefore \: t + 3 = \frac{9}{2} \\ \therefore \: t = \frac{9}{2} - 3 \\ \therefore \: t = \frac{9 - 6}{2} \\ \therefore \: t = \frac{3}{2} \\ \therefore \: t = 1.5 \: hrs \\$

$\because \: time \times speed = distance \\ \therefore \: 4(t + 3) \times 10(t + 3) = 810 \\ \therefore \: 4(t + 3)^{2} = 81 \\ \therefore \: \{2(t + 3) \}^{2} = {9}^{2} \\ \therefore \: \{2(t + 3) \} = {9} \\ \therefore \: t + 3 = \frac{9}{2} \\ \therefore \: t = \frac{9}{2} - 3 \\ \therefore \: t = \frac{9 - 6}{2} \\ \therefore \: t = \frac{3}{2} \\ \therefore \: t = 1.5 \: hrs \\ speed \: of \: car = 10(t + 3) \\ \hspace{60 pt} = 10(1.5 + 3) \\ \hspace{60 pt}= 10 \times 4.5 \\ \red{ \boxed{ \bold{ \therefore \: speed \: of \: car = 45 \:km/ h}}} \\$

3 0

3 years ago

7b+11>9b-13 need to find b>?

STALIN [3.7K]

$7b+11 > 9b-13\ \ \ \ /-11\\\\7b+11-11 > 9b-13-11\\\\7b > 9b-24\ \ \ \ /-9b\\\\7b-9b > 9b-9b-24\\\\-2b > -24\ \ \ \ /:(-2) < 0\ then\ " > "\ change\ to\ "\ < "\\\\b < 12\\\\b\in(-\infty;\ 12)$