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An airline allows each passenger 33 kilograms of luggage. Carl has to lower the weight of his suitcase by 25% to stay within the

damaskus [11]

1- 0.25= 0.75 (multiplier)

33÷0.75=44

The suitcase originally weighed 44kg

Hope this helps!

5 0

2 years ago

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Select yes or no to indicate whether a zero must be written in the dividend to find the quotient

neonofarm [45]

Answer:

2.25 divided by 0.6 : Yes and 5.2 divided by 8, 3.63 divided by 3, 71.1 divided by 9 : No

Step-by-step explanation:

We have that when two numbers are divided, they can be written in the form $\frac{p}{q}$ , where p is the dividend and q is the divisor.

Now, when we divide 5.2 by 8, we can write $\frac{5.2}{8}$

Also, the division 3.63 by 3, can be written as $\frac{3.63}{3}$ = 1.21

Further, when we divide 71.1 by 3, we can write $\frac{71.1}{3}$ = 23.7

We can see that above divisions are easy divisions and do not require to write any type of 0 in the numerator.

Moreover, when we divide 2.25 by 0.6, we can write $\frac{2.25}{0.6}$ = $\frac{2.25\times 10}{6}$ .

So, we get that in the last division, we need to write a zero in order to multiply the numerator with 10 for easy division.

Hence, 2.25 divided by 0.6 : Yes and 5.2 divided by 8, 3.63 divided by 3, 71.1 divided by 9 : No

3 0

3 years ago

53w + 13 < 56w + 16 solve for w

Sergio039 [100]

Answer:

w>-1

Step-by-step explanation:

1 Move terms

53w-56w<16-13

2 Collect like terms then subtract

-3w< 3

3 Divide both sides by -3

Answer: w>-1

4 0

3 years ago

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How many of the numbers from 10 through 92 have the sum of their digits equal to a perfect square?

horrorfan [7]

All the numbers in this range can be written as $10d_1+d_0$ with $d_1\in\{1,2,\ldots,9\}$ and $d_2\in\{0,1,\ldots,9\}$ . Construct a table like so (see attached; apparently the environment for constructing tables isn't supported on this site...)

so that each entry in the table corresponds to the sum of the tens digit (row) and the ones digit (column). Now, you want to find the numbers whose digits add to perfect squares, which occurs when the sum of the digits is either of 1, 4, 9, or 16. You'll notice that this happens along some diagonals.

For each number that occupies an entire diagonal in the table, it's easy to see that that number $n$ shows up $n$ times in the table, so there is one instance of 1, four of 4, and nine of 9. Meanwhile, 16 shows up only twice due to the constraints of the table.

So there are 16 instances of two digit numbers between 10 and 92 whose digits add to perfect squares.