George is 4 times as old as his daughter was 2 years ago. George is 36

Mathematics

2 answers:

NNADVOKAT [17]3 years ago

7 0

Answer:

I think 36×2×4 but am not sure

fredd [130]3 years ago

5 0

Where is the question?

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Marina has $145 to spend. She spends $68. 50 on jeans. T-shirts cost $12. 75. Write and solve an inequality to find the greatest

Kruka [31]

Answer:

Step-by-step explanation:

145-68.50 = 76.50

76.20/12.75 = no of t-shirts she can buy

ans= 6

6 0

2 years ago

When Arianna left her house,her cell phone was 45% charged started losing charge at a constant rate. 4 hours after leaving her h

4vir4ik [10]

Answer:

B = 3.75t

Step-by-step explanation:

Given;

initial charge of Arianna's phone = 45%

the charge remaining in Arianna battery = 15%

If after 4hours of leaving her house, the battery had 15% remaining battery life, then the battery is discharging at the rate of ( 15% / 4hours) 3.75% per hour.

Thus, the charge remaining in Arianna's battery after she left her house is given by;

B = 3.75t

where;

t is time in hours

B is the percentage of charge remaining

6 0

3 years ago

A spacecraft is traveling with a velocity of v0x = 5320 m/s along the +x direction. Two engines are turned on for a time of 739

OlgaM077 [116]

Answer:

The velocities after 739 s of firing of each engine would be 6642.81 m/s in the x direction and 5306.02 in the y direction

Step-by-step explanation:

For a constant acceleration: $v_{f}=v_{0}+at$ , where $v_{f}$ is the final velocity in a direction after the acceleration is applied, $v_{0}$ is the initial velocity in that direction before the acceleration is applied, a is the acceleration applied in such direction, and t is the amount of time during where that acceleration was applied.
<em>Then for the x direction</em> it is known that the initial velocity is $v_{0x} =$ 5320 m/s, the acceleration (the applied by the engine) in x direction is $a_{x}$ 1.79 m/s2 and, the time during the acceleration was applied (the time during the engines were fired) of the is 739 s. Then: $v_{fx}=v_{0x}+a_{x}t=5320\frac{m}{s} +1.79\frac{m}{s^{2} }*739s=6642.81\frac{m}{s}$
In the same fashion, <em>for the y direction</em>, the initial velocity is $v_{0y} =$ 0 m/s, the acceleration in y direction is $a_{y}$ 7.18 m/s2, and the time is the same that in the x direction, 739 s, then for the final velocity in the y direction: $v_{fy}=v_{0y}+a_{y}t=0\frac{m}{s} +7.18\frac{m}{s^{2} }*739s=5306.02\frac{m}{s}$