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Bri_Durbin

  • 2 years ago

Please help with a couple of questions!

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  1. Bri_Durbin
    • 2 years ago
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    A student scored 78 and 95 on his first two quizzes. Use a compound inequality to find the possible values for a third quiz score that would give him an average between 80 and 90, inclusive. (1 point) 67 ≤ x ≤ 97 78 ≤ x ≤ 95 80 ≤ x ≤ 90 86 ≤ x ≤ 87

  2. Bri_Durbin
    • 2 years ago
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    Would it be B???

  3. amorfide
    • 2 years ago
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    now it would not first of all you are given that he gets an average between 80 and 90 so you know it is between these two values you dont know his exact average so it could be 80 or higher but it must be 90 or belowe so it can be equal to 80 or above and it can be equal to 90 or below this gives two inequalities 80 ≤ x and x ≤ 90 this can be written as 1 inequality 80 ≤ x ≤ 90 ≤ this means less than or equal to the one that points the opposite way would be greater than or equal to think of it as a number line pointing right is positive meaning the numbers get bigger, getting greater pointing left is getting smaller so it gets less and less remember it is called greater than or equal to or less than or equal to because of the line underneath the sign

  4. Bri_Durbin
    • 2 years ago
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    can you help with another one???

  5. amorfide
    • 2 years ago
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    sure

  6. Bri_Durbin
    • 2 years ago
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    Solve. x – 1 over 5 = 3 x = –14/5 and x = 16/5 x = 14/5 and x = –16/5 x = –14/5 and x = –16/5 x = 14/5 and x = 16/5

  7. Bri_Durbin
    • 2 years ago
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    |dw:1354657718293:dw|

  8. amorfide
    • 2 years ago
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    |dw:1354657732769:dw|

  9. amorfide
    • 2 years ago
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    which one

  10. Bri_Durbin
    • 2 years ago
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    |dw:1354657798629:dw|

  11. amorfide
    • 2 years ago
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    okay absolute values okay so you should sketch a graph sketch |x-(1/5)|=3 if you do not know how first you must know |x|=x |-x|=x so you know you always get a positive number from whatever is in the | | so from this you know the graph will show only positive values so first you will draw the graph of y=x-(1/5) then you will reflect the negative part of the graph for example if i have |x-2|=4 i will draw the graphs for y=x-2 y=4 and y=|x-2| this would give |dw:1354657998510:dw| now i must draw y=|x-2| well since it is reflected i will have all of the negative part of the graph y=x-2 on the positive side of the graph so it will be like...

  12. amorfide
    • 2 years ago
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    |dw:1354658100161:dw|

  13. amorfide
    • 2 years ago
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    I labelled it -(x-2) because the graph y=|x-2| is essentially y=x-2 and the other part is that multiplied by -1 which gives y=-(x-2) now that we have drawn a graph you know it crosses with each other 2 times so you will make them equal to each other for example y=x-2 crosses y=4 so make them equal to each other x-2=4 re arrange to get x x=6 now i know that -(x-2) crosses with 4 so i make them equal -(x-2)=4 i re arrange to get x x=-2 sorry i made a mistake in the graph drawing it should be y=4 not y=3 okay now follow the same thing with your question

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