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Locust_Grub

  • 3 years ago

PLS HELP BEEN STUCK FOR HOURS ON THIS. Starting from 1.5 miles away, a car drives toward a speed checkpoint and then passes it. The car travels at a constant rate of 53 miles per hour. The distance of the car from the checkpoint is given by d = |1.5 – 53t|. At what times is the car 0.1 miles from the checkpoint? Calculate your answer in seconds. (1 point)95.1s and 108.7s 10.2s and 101.9s108.7s and 10.2s95.1s and 10.2s

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  1. Girly17
    • 3 years ago
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    AHhhh i have been too): i did steps and i still dont get it. Im from connexus

  2. 03225186213
    • 3 years ago
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    |dw:1351580898467:dw|

  3. Girly17
    • 3 years ago
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    huh

  4. Locust_Grub
    • 3 years ago
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    WRONG ALL OF YOU

  5. Girly17
    • 3 years ago
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    ok check this out bruh. this is what i first did. I subtracted 1.5 from both sides of this equation. 1.5-53t=0.1 and i got 1.4 then i divided 1.4/5.3 and i got 0.026415(etc) NEED HELP

  6. Locust_Grub
    • 3 years ago
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    :P ANSWER THE TOPIC QUESTION MY BRAINS ABOUT TO BLOW

  7. Girly17
    • 3 years ago
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    LMAO I AGREE IVE BEEN SITTIN ON MY retriceFOR 4 HOURS MAN

  8. Fellowroot
    • 3 years ago
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    I have t=0.026 and t=0.030 Does anyone else agree?

  9. Hero
    • 3 years ago
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    d = |1.5 – 53t| 0.1 = -(1.5 - 53t) 0.1 = 1.5 - 53t 0.1 =53t - 1.5 1.5 + 0.1 = 53t 1.6/53 = t t = 108.67 secs 53t = 1.5 - 0.1 53t = 1.4 t = 1.4/53 t = 95.09 secs

  10. Girly17
    • 3 years ago
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    fellow i got the same on these two equations i just did

  11. Hero
    • 3 years ago
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    Good on you

  12. Girly17
    • 3 years ago
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    so 0.026 and 0.030 is correct? hero

  13. Hero
    • 3 years ago
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    I don't know where you get that from. It wants in seconds

  14. Hero
    • 3 years ago
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    In hours, yes, that is correct

  15. Girly17
    • 3 years ago
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    wait nvm. gotcha. thanks!;)

  16. Hero
    • 3 years ago
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    I don't know why it took you four hours

  17. Hero
    • 3 years ago
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    @Girly17, I help you with a question and you block me and call me a weirdo? Thanks

  18. Fellowroot
    • 3 years ago
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    Yeah, double checked I got it. You can solve this problem in 2 ways. 1st way by pure math abs( anything ) = something means anything =+something and/or anything =-something and solve for t also you can think of this as a physics problem with zero acceleration, deduce your position function and find the times at the desired distances.

  19. Hero
    • 3 years ago
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    I solved it the "easiest" way

  20. Girly17
    • 3 years ago
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    calm down calm down you were rude and said post it like everyone else. well look here yo i woulda done that but i couldnt find how to do that so i seen send a message thing and it was easier for me.

  21. Girly17
    • 3 years ago
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    took me 4 hours cuz im awesome like that

  22. Hero
    • 3 years ago
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    I wasn't being rude. I was trying to explain to you that it is inconvenient to help you in the messages.

  23. Hero
    • 3 years ago
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    Messages are not for posting questions.

  24. Hero
    • 3 years ago
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    Anyone else would tell you the same thing

  25. Girly17
    • 3 years ago
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    which means you took it the wrong way, how are the inconvenient? lets see.

  26. Girly17
    • 3 years ago
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    and how was i suppose to know?!?!?!?!?!?!?!?!

  27. Hero
    • 3 years ago
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    You didn't know, but I was explaining to you. That's all I can try to do is communicate with you and explain to you. All you have to do is not overreact.

  28. Girly17
    • 3 years ago
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    i am not overreacting. you're just giving me bs. but wtvr goodnight and thanks

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