q5 1. here are the first four terms of a number sequence. 8 14 20 26 (a) write down the next term of the number sequence. (b) explain how you found your answer. q7 express the following into its prime - factors 1. 120 2. 36 3. 72 4. 96 q8 1. a rectangle measures 12cm by 36cm both measurements are given correct to the nearest centimeter. find the upper and lower bound for (a) the perimeter find the upper and lower bound for (b) the area

Question

Accepted Answer

### a.
# Explanation:
## Step1: Find the common difference
Subtract consecutive terms: $14 - 8=6$, $20 - 14 = 6$, $26 - 20=6$. The common difference is 6.
## Step2: Calculate the next term
Add the common difference to the last - given term. $26+6 = 32$.
# Answer:
32

### b.
# Explanation:
By subtracting consecutive terms ($14 - 8$, $20 - 14$, $26 - 20$), we found that the common difference in the sequence is 6. To get the next term, we added this common difference of 6 to the last - given term 26.
# Answer:
The common difference in the sequence is 6. We added 6 to the last - given term 26 to get the next term.

### Q.7
#### 1. For 120:
# Explanation:
## Step1: Start with the smallest prime number
Divide 120 by 2: $120\div2 = 60$.
## Step2: Keep dividing by 2
$60\div2=30$, $30\div2 = 15$.
## Step3: Divide by the next prime number
Since 15 is odd, divide by 3: $15\div3=5$. And 5 is a prime number.
So, $120=2	imes2	imes2	imes3	imes5=2^{3}	imes3	imes5$.
# Answer:
$2^{3}	imes3	imes5$

#### 2. For 36:
# Explanation:
## Step1: Divide by 2
$36\div2 = 18$.
## Step2: Divide by 2 again
$18\div2=9$.
## Step3: Divide by 3
$9\div3 = 3$.
So, $36=2	imes2	imes3	imes3=2^{2}	imes3^{2}$.
# Answer:
$2^{2}	imes3^{2}$

#### 3. For 72:
# Explanation:
## Step1: Divide by 2
$72\div2 = 36$.
## Step2: Divide by 2 again
$36\div2=18$.
## Step3: Divide by 2 once more
$18\div2 = 9$.
## Step4: Divide by 3
$9\div3=3$.
So, $72=2	imes2	imes2	imes3	imes3=2^{3}	imes3^{2}$.
# Answer:
$2^{3}	imes3^{2}$

#### 4. For 96:
# Explanation:
## Step1: Divide by 2
$96\div2 = 48$.
## Step2: Divide by 2 again
$48\div2=24$.
## Step3: Divide by 2 once more
$24\div2 = 12$.
## Step4: Divide by 2 again
$12\div2=6$.
## Step5: Divide by 2 one last time
$6\div2 = 3$.
So, $96=2	imes2	imes2	imes2	imes2	imes3=2^{5}	imes3$.
# Answer:
$2^{5}	imes3$

### Q.8 a.
The lower bound of 12 cm is 11.5 cm and the upper bound is 12.5 cm. The lower bound of 36 cm is 35.5 cm and the upper bound is 36.5 cm.
The perimeter of a rectangle is $P = 2(l + w)$.
# Explanation:
## Step1: Calculate the lower - bound perimeter
$P_{lower}=2(11.5 + 35.5)=2	imes47 = 94$ cm.
## Step2: Calculate the upper - bound perimeter
$P_{upper}=2(12.5+36.5)=2	imes49 = 98$ cm.
# Answer:
Lower bound: 94 cm, Upper bound: 98 cm

### Q.8 b.
The area of a rectangle is $A=l	imes w$.
# Explanation:
## Step1: Calculate the lower - bound area
$A_{lower}=11.5	imes35.5 = 408.25$ $cm^{2}$.
## Step2: Calculate the upper - bound area
$A_{upper}=12.5	imes36.5=456.25$ $cm^{2}$.
# Answer:
Lower bound: $408.25$ $cm^{2}$, Upper bound: $456.25$ $cm^{2}$

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QUESTION IMAGE

Question

Explanation:

a.

Step1: Find the common difference

Step2: Calculate the next term

Step1: Start with the smallest prime number

Step2: Keep dividing by 2

Step3: Divide by the next prime number

Answer:

b.