Answer:
Colin has <em>8 sheets </em>left for his third class.
Step-by-step explanation:
Given that:
Total Number of pieces of papers = 
Number of pieces of papers used for 1st class = 5 fewer than half of the pieces in the pad
Writing the equation:
Also, Given that number of pieces of papers used for the 2nd class are 2 more than that of papers used in the 1st class.
Now, number of pieces of papers left for the third class = Total number of pieces of papers in the pad - Number of pieces of papers used in the first class - Number of pieces of papers used in the first class
So, the answer is:
Colin has <em>8</em> <em>sheets </em>left for his third class.
2 | <u>5</u><u>5</u><u>0</u><u>,</u><u>7</u><u>5</u><u>0</u><u>,</u><u>9</u><u>0</u><u>0</u>
2 | <u>2</u><u>7</u><u>5</u><u>,</u><u>3</u><u>7</u><u>5</u><u>,</u><u>4</u><u>5</u><u>0</u>
3 |<u> </u><u>2</u><u>7</u><u>5</u><u>,</u><u>3</u><u>7</u><u>5</u><u>,</u><u>2</u><u>2</u><u>5</u>
<u>3</u><u> </u>| <u>2</u><u>7</u><u>5</u><u>,</u><u>1</u><u>2</u><u>5</u><u>,</u><u>7</u><u>5</u>
5 | <u>2</u><u>7</u><u>5</u><u>,</u><u>1</u><u>2</u><u>5</u><u>,</u><u>2</u><u>5</u>
5 | <u>55,25,5</u>
5 | <u>1</u><u>1</u><u>,</u><u>5</u><u>,</u><u>5</u>
11 | <u>1</u><u>1</u><u>,</u><u>1</u><u>,</u><u>1</u>
LCM:-2×2×3×3×5×5×5×11=49500
Answer:
3.8s
Step-by-step explanation:
Solve for h=0, you will get t=3.75
Answer:
the second one
Step-by-step explanation:
7 - (2m - 5)
Answer:
x=7.5
Step-by-step explanation:
