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aleksandr82 [10.1K]
3 years ago
13

Helpppp plsssss!!thanks

Mathematics
1 answer:
DedPeter [7]3 years ago
5 0
A
2x - 6 - (3x + 9)
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The answer is V=384.
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Help me please:’)
Lelechka [254]

Using the condition given to build an inequality, it is found that the maximum number of junior high school student he can still recruit is of 17.

<h3>Inequality:</h3>

Considering s the number of senior students and j the number of junior students, and that he cannot recruit more than 50 people, the inequality that models the number of students he can still recruit is:

s + j \leq 50

In this problem:

  • Already recruited 28 senior high students, hence s = 28.
  • Already recruited 5 junior high students, want to recruit more, hence j = j + 5.

Then:

28 + j + 5 \leq 50

33 + j \leq 50

j \leq 17

The maximum number of junior high school student he can still recruit is of 17.

You can learn more about inequalities at brainly.com/question/25953350

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2 years ago
Please help on number 12 i don’t know how to do it
Ilia_Sergeevich [38]

Answer:

try using the point (1,13)

Step-by-step explanation: I pulled up and online graph to help and the distance from A to B is 7/2. So then you would do B, plus 7/2. If that's wrong, I'm sorry, but i tried.

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4 years ago
Describe in words the parameter of interest. The parameter is the average number of hours students at the school watch TV per da
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Answer:

The parameter is the average number of hours students at the school watch TV per day.

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The correct option is : The parameter is the average number of hours students at the school watch TV per day.

8 0
3 years ago
Help Q-Q 7/10c =4 1/5
shepuryov [24]

\huge\mathfrak\green{answer}

= q -   \frac{q7}{10} c =  \frac{41}{5}

\red{multiply}

= q -  \frac{cq7}{10} =  \frac{41}{5}

\red{subtract \: q \: from \: both \: sides}

= q -  \frac{cq7}{10}  - q =  \frac{41}{5} - q

\red{now \: simplify}

=  -  \frac{cq7}{10}  =  \frac{41}{5}  - q

\red{multiply \: 10 \: both \: sides}

= 10( -  \frac{cq7}{10} ) = 10. \frac{41}{5}  - 10q

\red{simplify}

=  - cq7 = 82 - 10q

\red{now \: divide}

=  \frac{ - cq7}{ - q7}  -  \frac{82}{ - q7}  -  \frac{10q}{ - q7}

\red{simplify}

= c =   - \frac{82 - 10q}{q7} (q≠0)

Brainliest? :) (I'd really appreciate if you mark me as brainliest)

\huge\mathfrak\green{thank \: you}

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3 years ago
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