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Setler [38]
3 years ago
13

there are four seventh grade math teachers at central middle school mr. abott teaches 120 students in 6 classes mrs. garcia teac

hes 48 students in 3 classes mrs.nguyen's 5 classes total 109 students and ms. turner has a total of 121 students in 6 classes what is the average number of students in each seventh grade math class at central middle school

Mathematics
1 answer:
Mumz [18]3 years ago
4 0
You see what the average is in each class. (Divid students and classes) and add them together and divid by four. You should get 18 if you round to the whole number.

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-7x − 3x + 2 = −8x – 8 .<br><br> SHOW WORK!!!
denis23 [38]

-7x-3x+2=-8x-8\\\\(-7x-3x)+2=-8x-8\\\\-10x+2=-8x-8\qquad|\text{subtract 2 from both sides}\\\\-10x=-8x-10\qquad|\text{add 8x to both sides}\\\\-2x=-10\qquad|\text{divide both sides by (-2)}\\\\\boxed{x=5}

8 0
3 years ago
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To rent a certain meeting room, a college charges a reservation fee of $49 and an additional fee of $5.40 per hour. The chemistr
elena-14-01-66 [18.8K]

Answer:

The maximum amount of money the chemistry club can spend without going over their buget is $81.4 ... With a total of 6 hours their allowed to use the room for.

Step-by-step explanation:

$49 is our constant and will not change. 5.40 however will. Let's use variable h for the amount hours they can rent the room for. Now we can make an inequality;

49+ 5.40h < 86.80

5.40h < 86.80

5.4h< 37.8

h< 7

Now that we know h will have to be less than seven for the chemistry club to spend less than their buget we can find that 6 is the maxium amount of hours they can spend in the room while staying within their budget.

(This answer does not consider decimals or half hours)

6 0
2 years ago
I need to know how to solve this
Greeley [361]
Vertical angles are congruent, so
107=3x-4
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8 0
3 years ago
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If log2 5 = k, determine an expression for log32 5 in terms of k.
lukranit [14]

Answer:

log_3_2(5)=\frac{1}{5} k

Step-by-step explanation:

Let's start by using change of base property:

log_b(x)=\frac{log_a(x)}{log_a(b)}

So, for log_2(5)

log_2(5)=k=\frac{log(5)}{log(2)}\hspace{10}(1)

Now, using change of base for log_3_2(5)

log_3_2(5)=\frac{log(5)}{log(32)}

You can express 32 as:

2^5

Using reduction of power property:

log_z(x^y)=ylog_z(x)

log(32)=log(2^5)=5log(2)

Therefore:

log_3_2(5)=\frac{log(5)}{5*log(2)}=\frac{1}{5} \frac{log(5)}{log(2)}\hspace{10}(2)

As you can see the only difference between (1) and (2) is the coefficient \frac{1}{5} :

So:

\frac{log(5)}{log(2)} =k\\

log_3_2(5)=\frac{1}{5} \frac{log(5)}{log(2)} =\frac{1}{5} k

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What is the volume of the prism?
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Jaden Smith can't act or rap.
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