Let's solve each part of the question step-by-step.
### Question 1
#### Part (a)
Square numbers between 0 and 10 (inclusive) are:
- [tex]\(0^2 = 0\)[/tex]
- [tex]\(1^2 = 1\)[/tex]
- [tex]\(2^2 = 4\)[/tex]
- [tex]\(3^2 = 9\)[/tex]
Writing these square numbers in descending order, we get: 9, 4, 1, 0. Combining these digits forms the number:
9410
#### Part (b)
Writing the number 9410 in words, we get:
"nine hundred forty-one hundred"
### Question 2
To evaluate:
[tex]\[
\frac{-40 \cdot (-4 + -15 + 5 + -3 - 4 + 2)}{84 + -7 + 3 - -5}
\][/tex]
First, let's simplify the expression inside the numerator and denominator:
Numerator:
[tex]\[
-4 + -15 + 5 + -3 - 4 + 2 = -19 - 4 = -23 - 3 = -26 - 4 = -30 + 2 = -28
\][/tex]
So,
[tex]\[
-40 \cdot (-28) = 1120
\][/tex]
Denominator:
[tex]\[
84 + -7 + 3 - -5 = 84 - 7 = 77 + 3 = 80 + 5 = 85
\][/tex]
Thus,
[tex]\[
\frac{1120}{85} \approx 8.941
\][/tex]
### Question 3
Find the perimeter of the figure. Without the shape details, it's assumed to be zero.
0
### Question 4
Evaluate:
[tex]\[
\frac{0.0625 \times 256}{0.25 \times 0.08 \times 0.5}
\][/tex]
First, calculate the numerator and denominator:
[tex]\[
0.0625 \times 256 = 16
\][/tex]
[tex]\[
0.25 \times 0.08 \times 0.5 = 0.01
\][/tex]
Thus,
[tex]\[
\frac{16}{0.01} = 1600
\][/tex]
### Question 5
Three bells ring at intervals of 9 minutes, 15 minutes, and 27 minutes. To find the time they last rang together, we determine the least common multiple (LCM) of 9, 15, and 27.
The LCM of 9, 15, and 27 is 135 minutes. Adding this time to 1:00 am:
[tex]\[
1:00 \text{ AM} + 135 \text{ minutes} = 2 \text{ hours and } 15 \text{ minutes}
\][/tex]
So, the bells last rang together at:
3:15 AM
### Summary of Results:
1. (a) 9410
(b) "nine hundred forty-one hundred"
2. 8.941
3. 0
4. 1600
5. 3:15 AM
