Whats a multiplication problem with a product of 5 to the 13th power
1 answer:
You might be interested in
<u>Part A
</u><u />To estimate this, we should first look at our fractions and see if they can be combined to form a whole number. In this case,
and 
equal approximately 1. We can add this "1" to the other to full gallons to estimate that the painter needs about 3 gallons.
<u>Part B
</u><u /><u />To find the exact amount, we should first change the mixed numbers to improper fractions. We do this by multiplying the denominator by the whole number, adding the numerator, and placing that value over the denominator.
Now, we need to find the least common denominator. This is the lowest value that both denominators will divide evenly into. In this case, that number is 15.
Next, we should multiply both fractions so that the denominator is that number. Remember that we must also multiply the numerator for the fraction to remain equivalent to its original value.
Now, we can simply add our numerators.
We know that he needs
gallons of paint, but this is not in the most simplified format. To simplify, we need to turn our improper fraction back to a mixed number. To do this, we need to divide our numerator by the denominator to create our whole number, and the remainder becomes our new numerator.
Using that logic, we can see that the painter needs exactly
gallons of paint.
</u><u />To estimate this, we should first look at our fractions and see if they can be combined to form a whole number. In this case,
<u>Part B
</u><u /><u />To find the exact amount, we should first change the mixed numbers to improper fractions. We do this by multiplying the denominator by the whole number, adding the numerator, and placing that value over the denominator.
Now, we need to find the least common denominator. This is the lowest value that both denominators will divide evenly into. In this case, that number is 15.
Next, we should multiply both fractions so that the denominator is that number. Remember that we must also multiply the numerator for the fraction to remain equivalent to its original value.
Now, we can simply add our numerators.
We know that he needs
Using that logic, we can see that the painter needs exactly