Answer :
Sure, let's tackle each point step-by-step.
### a) 5 added to a number is 9
Equation: Let the number be [tex]\( x \)[/tex]. The equation will be:
[tex]\[ x + 5 = 9 \][/tex]
Solution:
[tex]\[ x = 9 - 5 \][/tex]
[tex]\[ x = 4 \][/tex]
### b) 3 subtracted from a number is equal to 12
Equation: Let the number be [tex]\( y \)[/tex]. The equation will be:
[tex]\[ y - 3 = 12 \][/tex]
Solution:
[tex]\[ y = 12 + 3 \][/tex]
[tex]\[ y = 15 \][/tex]
### c) 5 times a number decreased by 2 is 4
Equation: Let the number be [tex]\( z \)[/tex]. The equation will be:
[tex]\[ 5z - 2 = 4 \][/tex]
Solution:
[tex]\[ 5z = 4 + 2 \][/tex]
[tex]\[ 5z = 6 \][/tex]
[tex]\[ z = \frac{6}{5} \][/tex]
[tex]\[ z = 1.2 \][/tex]
### d) 2 times the sum of the number [tex]\( x \)[/tex] and 7 is 13
Equation: Let the number be [tex]\( w \)[/tex]. The equation will be:
[tex]\[ 2(x + 7) = 13 \][/tex]
Solution:
[tex]\[ x + 7 = \frac{13}{2} \][/tex]
[tex]\[ x + 7 = 6.5 \][/tex]
[tex]\[ x = 6.5 - 7 \][/tex]
[tex]\[ x = -0.5 \][/tex]
### i. A number is 12 more than the other. Find the numbers if their sum is 48
Equation: Let the two numbers be [tex]\( a \)[/tex] and [tex]\( b \)[/tex], where [tex]\( a = b + 12 \)[/tex] and [tex]\( a + b = 48 \)[/tex].
Solution:
Substitute [tex]\( a = b + 12 \)[/tex] into the sum equation:
[tex]\[ (b + 12) + b = 48 \][/tex]
[tex]\[ 2b + 12 = 48 \][/tex]
[tex]\[ 2b = 36 \][/tex]
[tex]\[ b = 18 \][/tex]
Now, [tex]\( a \)[/tex] is:
[tex]\[ a = b + 12 \][/tex]
[tex]\[ a = 18 + 12 \][/tex]
[tex]\[ a = 30 \][/tex]
### ii. Twice the number decreased by 22 is 48. Find the number
Equation: Let the number be [tex]\( c \)[/tex]. The equation will be:
[tex]\[ 2c - 22 = 48 \][/tex]
Solution:
[tex]\[ 2c = 48 + 22 \][/tex]
[tex]\[ 2c = 70 \][/tex]
[tex]\[ c = \frac{70}{2} \][/tex]
[tex]\[ c = 35 \][/tex]
### iii. [tex]\(\frac{4}{5}\)[/tex] of a number is more than [tex]\(\frac{3}{4}\)[/tex] of the number by 5
Equation: Let the number be [tex]\( d \)[/tex]. The equation will be:
[tex]\[ \frac{4d}{5} - \frac{3d}{4} = 5 \][/tex]
Solution: Find a common denominator for the fractions, which is 20:
[tex]\[ \frac{16d}{20} - \frac{15d}{20} = 5 \][/tex]
[tex]\[ \frac{d}{20} = 5 \][/tex]
[tex]\[ d = 5 \times 20 \][/tex]
[tex]\[ d = 100 \][/tex]
### iv. In 25 Minutes, a train travels 20 km. How far will it travel in 5 minutes?
Solution:
First, find the distance per minute:
[tex]\[ \text{Distance per minute} = \frac{20 \, \text{km}}{25 \, \text{minutes}} \][/tex]
[tex]\[ \text{Distance per minute} = 0.8 \, \text{km/minute} \][/tex]
Now, for 5 minutes:
[tex]\[ \text{Distance in 5 minutes} = 0.8 \times 5 \][/tex]
[tex]\[ \text{Distance in 5 minutes} = 4 \, \text{km} \][/tex]
### v. What should be added to [tex]\(4x^3 - 10x^2 + 12x + 6\)[/tex] so that it becomes exactly divisible by [tex]\(2x + 1\)[/tex]?
To determine what constant should be added, perform polynomial division of [tex]\(4x^3 - 10x^2 + 12x + 6\)[/tex] by [tex]\(2x + 1\)[/tex] and find the remainder.
After performing polynomial division, the remainder is found to be -3. Therefore, to make the polynomial exactly divisible, you add the remainder as a positive number.
Solution:
[tex]\[ k = 3 \][/tex]
### a) 5 added to a number is 9
Equation: Let the number be [tex]\( x \)[/tex]. The equation will be:
[tex]\[ x + 5 = 9 \][/tex]
Solution:
[tex]\[ x = 9 - 5 \][/tex]
[tex]\[ x = 4 \][/tex]
### b) 3 subtracted from a number is equal to 12
Equation: Let the number be [tex]\( y \)[/tex]. The equation will be:
[tex]\[ y - 3 = 12 \][/tex]
Solution:
[tex]\[ y = 12 + 3 \][/tex]
[tex]\[ y = 15 \][/tex]
### c) 5 times a number decreased by 2 is 4
Equation: Let the number be [tex]\( z \)[/tex]. The equation will be:
[tex]\[ 5z - 2 = 4 \][/tex]
Solution:
[tex]\[ 5z = 4 + 2 \][/tex]
[tex]\[ 5z = 6 \][/tex]
[tex]\[ z = \frac{6}{5} \][/tex]
[tex]\[ z = 1.2 \][/tex]
### d) 2 times the sum of the number [tex]\( x \)[/tex] and 7 is 13
Equation: Let the number be [tex]\( w \)[/tex]. The equation will be:
[tex]\[ 2(x + 7) = 13 \][/tex]
Solution:
[tex]\[ x + 7 = \frac{13}{2} \][/tex]
[tex]\[ x + 7 = 6.5 \][/tex]
[tex]\[ x = 6.5 - 7 \][/tex]
[tex]\[ x = -0.5 \][/tex]
### i. A number is 12 more than the other. Find the numbers if their sum is 48
Equation: Let the two numbers be [tex]\( a \)[/tex] and [tex]\( b \)[/tex], where [tex]\( a = b + 12 \)[/tex] and [tex]\( a + b = 48 \)[/tex].
Solution:
Substitute [tex]\( a = b + 12 \)[/tex] into the sum equation:
[tex]\[ (b + 12) + b = 48 \][/tex]
[tex]\[ 2b + 12 = 48 \][/tex]
[tex]\[ 2b = 36 \][/tex]
[tex]\[ b = 18 \][/tex]
Now, [tex]\( a \)[/tex] is:
[tex]\[ a = b + 12 \][/tex]
[tex]\[ a = 18 + 12 \][/tex]
[tex]\[ a = 30 \][/tex]
### ii. Twice the number decreased by 22 is 48. Find the number
Equation: Let the number be [tex]\( c \)[/tex]. The equation will be:
[tex]\[ 2c - 22 = 48 \][/tex]
Solution:
[tex]\[ 2c = 48 + 22 \][/tex]
[tex]\[ 2c = 70 \][/tex]
[tex]\[ c = \frac{70}{2} \][/tex]
[tex]\[ c = 35 \][/tex]
### iii. [tex]\(\frac{4}{5}\)[/tex] of a number is more than [tex]\(\frac{3}{4}\)[/tex] of the number by 5
Equation: Let the number be [tex]\( d \)[/tex]. The equation will be:
[tex]\[ \frac{4d}{5} - \frac{3d}{4} = 5 \][/tex]
Solution: Find a common denominator for the fractions, which is 20:
[tex]\[ \frac{16d}{20} - \frac{15d}{20} = 5 \][/tex]
[tex]\[ \frac{d}{20} = 5 \][/tex]
[tex]\[ d = 5 \times 20 \][/tex]
[tex]\[ d = 100 \][/tex]
### iv. In 25 Minutes, a train travels 20 km. How far will it travel in 5 minutes?
Solution:
First, find the distance per minute:
[tex]\[ \text{Distance per minute} = \frac{20 \, \text{km}}{25 \, \text{minutes}} \][/tex]
[tex]\[ \text{Distance per minute} = 0.8 \, \text{km/minute} \][/tex]
Now, for 5 minutes:
[tex]\[ \text{Distance in 5 minutes} = 0.8 \times 5 \][/tex]
[tex]\[ \text{Distance in 5 minutes} = 4 \, \text{km} \][/tex]
### v. What should be added to [tex]\(4x^3 - 10x^2 + 12x + 6\)[/tex] so that it becomes exactly divisible by [tex]\(2x + 1\)[/tex]?
To determine what constant should be added, perform polynomial division of [tex]\(4x^3 - 10x^2 + 12x + 6\)[/tex] by [tex]\(2x + 1\)[/tex] and find the remainder.
After performing polynomial division, the remainder is found to be -3. Therefore, to make the polynomial exactly divisible, you add the remainder as a positive number.
Solution:
[tex]\[ k = 3 \][/tex]
