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Bess [88]
2 years ago
12

Solve the inequality. 5/6t - 3≥3t + 6

Mathematics
1 answer:
Papessa [141]2 years ago
6 0

Answer:

t ≤ -\frac{54}{13}

Step-by-step explanation:

\frac{5}{6}t−3−3t≥3t+6−3t

-\frac{13}{6}t−3≥6

-\frac{13}{6}t−3+3≥6+3

\frac{13}{6}t≥9

-\frac{6}{13}(-\frac{13}{6}t) ≥ -\frac{6}{13}(9)

t ≤ -\frac{54}{13}

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Show work and Evaluate 3/4 + 2/5
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Answer:

23/20

Step-by-step explanation:

Find the common denominator of the 2 fractions (It's 20):

3/4 + 2/5            

Multiply both numerator and denominator by 5 for the first fraction and multiply by 4 for the second fraction:    

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Add the fractions together:

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5 0
2 years ago
If you roll a number cube 12 times, about how many times would you expect to roll a 5 or 6
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If Kevin wants a 90 average in math after 5 tests and his first 4 tests are 76, 92, 89 and 97 what does he need on the fifth tes
Keith_Richards [23]
You need to understand that you're solving for the average, which you already know: 90. Since you know the values of the first three exams, and you know what your final value needs to be, just set up the problem like you would any time you're averaging something.
Solving for the average is simple:
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Since you know you have that fourth exam, just substitute it into the total value as an unknown, X:
(87 + 88 + 92 + X) / 4 = 90
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Multiplying for four on each side cancels out the fraction.
So now you have:
87 + 88 + 92 + X = 360
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Negating the 267 on each side will isolate the X value, and give you your final answer:
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Now that you have an answer, ask yourself, "does it make sense?"
I say that it does, because there were two tests that were below average, and one that was just slightly above average. So, it makes sense that you'd want to have a higher-ish test score on the fourth exam.
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Step-by-step explanation:

x\:+\:x\:+\:y\:\cdot \:y

x+x+yy

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8 0
2 years ago
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