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Zina [86]
3 years ago
6

The least common multiple of 3, 4, 6, and 8 is A. 24. B. 96. C. 72. D. 8.

Mathematics
1 answer:
garri49 [273]3 years ago
6 0
Try the lowest option first. 8 is not divisible by 3 so you can rule out D. The next lowest option is 24. 24 is divisible by all 4 numbers so that it is the answer.

ANSWER: A) 24
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The sum of two integers is 6and the difference is 4 find the intergers
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Prove (without using a calculator ) <img src="https://tex.z-dn.net/?f=%2055%21%5C%20%5Ctextless%20%5C%2028%5E%7B55%7D%20" id="Te
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4 0
4 years ago
For the level 3 course, exam hours cost twice as much as workshop hours, workshop hours cost twice as much as lecture hours. How
natulia [17]
<h2>Answer</h2>

Cost of lectures = $7.33 per hour

<h2>Explanation</h2>

Let e the cost of the exam hours

Let w be the cost of the workshop hours

Let l be the cost of the lecture hours.

We know from our problem that exam hours cost twice as much as workshop, so:

e=2w equation (1)

We also know that workshop hours cost twice as much as lecture hours, so:

w=2l equation (2)

Finally, we also know that 3hr exams 24hr workshops  and 12hr lectures cost $528, so:

3e+24w+12l=528 equation (1)

Now, lets find the value of l:

Step 1.  Solve for l in equation (3)

3e+24w+12l=528

12l=528-3e-24w equation (4)

Step 2. Replace equation (1) in equation (4) and simplify

12l=528-3e-24w

12l=528-3(2w)-24w

12l=528-6w-24w

12l=528-30w equation (5)

Step 3. Replace equation (2) in equation (5) and solve for l

12l=528-30w

12l=528-30(2l)

12l=528-60l

72l=528

l=\frac{528}{72}

l=\frac{22}{3}

l=7.33

Cost of lectures  = $7.33 per hour



3 0
3 years ago
Read 2 more answers
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