:
$\to$ 6m + 10 − 10m
$\to$ (6m - 10m) + 10
$\to$ - 4m + 10 is the simplified form.

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Steps: 2 Marks
Answer: 1 Mark
$(a+b)(2a-3b+c)-(2a-3b)c$
$=\left[\right(a+b\left)\right(2a-3b+c\left)\right]-\left[\right(2a-3b\left)c\right]$
$=[2a^2-3ba+ac+2ab-3b^2+bc]-[2ac-3bc]$
$=2a^2-3ba+ac+2ab-3b^2+bc-2ac+3bc$
$=2a^2-3b^2-ab-ac+4bc$
The simplified expresion is
$=2a^2-3b^2-ab-ac+4bc$.

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Forming the equation: 1 Mark
Steps: 1 Marks
Equation for perimeter: 1 Mark
Perimeter: 1 Mark
Perimeter of the triangle = sum of the sides of the triangle
$6p^2-4p+9=p^2-2p+1+3p^2-5p+3+3^{rd}side$
$3^{rd}side=6p^2-4p+9-\left[\right(p^2-2p+1)+(3p^2-5p+3\left)\right]$
$3^{rd}side=6p^2-4p+9-(4p^2-7p+4)$
$3^{rd}side=2p^2+3p+5$
The $3^{rd}$ side of the triangle has length is$2p^2+3p+5$
Giventhat
p = 3
The perimeter of thetriangle is $6p^2-4p+9$ .
On substituting the values we get:
$=6\times 3^2-4\times 3+9$
$=54-12+9$
$=51$
The perimeter of the triangle is 51 units.

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Forming equation: 1 Mark
Equating equation according to question: 1 Mark
Steps: 1 Mark
Answer: 1 Mark
Given that
Tatesville has $x$ inches of rainfall in April, $(x+1.3)inches$ of rainfall in May, and $(2x+0.5)inches$ in June.
The total rainfall during these months will be the sum of these expressions.
Total amount of rainfall
$=x+(x+1.3)+(2x+0.5)inches$
$=x+x+1.3+2x+0.5inches$
$=4x+1.8inches$
Given that, rainfall in May is 2.5 inches.
Rainfall in May
$(x+1.3)=2.5inches$
$x+1.3=2.5$
$x=2.5-1.3$
$x=1.2inches$
On substituting the value of$x=1.2inches$ in the ecpression for total rainfall.
Total Rainfall
$=4x+1.8=4\left(1.2\right)+1.8=4.8+1.8=6.6inches$
So, the total rainfall over the three months is $6.6inches$

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Answer: 1 Mark
Example: 1 Mark
The product of a monomial and a binomial expression is a binomial.
E.g. $a(a^2-b^2)=a^3-a{b}^2$the expression has only two terms $a^3$and $a{b}^2$.

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Steps: 1 Mark
Answer: 1 Mark
As per the question:
We have to find out the sum of the numerical coefficients of the expression$2x^2+31y-5xy$
The numerical coefficient of$2x^2$ = $2$
The numerical coefficient of$31y$ = $31$
The numerical coefficient of$-5xy$ = $-5$
Sum of the coefficients = 2 + 31 - 5 = 28
Hence, the sum of the numerical coefficients of the expression$2x^2+31y-5xy$ is $28$.

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Formula: 1 Mark
Steps: 1 Mark
Answer: 1 Mark
Given that:
Two adjacent sides of arectangleare $5x^2-3y^2$and $x^2-2xy$.
The perimeter of a rectangle is given by:
Perimeter of a rectangle = 2 (Length +Breadth)
$=2\left[\right(5x^2-3y^2)+(x^2-2xy\left)\right]$
$=[10x^2-6y^2)+(2x^2-4xy)]$
$=12x^2-6y^2-4xy$
So, the perimeter of the rectangle is$12x^2-6y^2-4xy$.

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Forming the equation: 1 Mark
Steps: 2 Marks
Result: 1 Mark
Let the numbers be $x$ and $x+5$
Therefore, $x+x+5=55$
$\Rightarrow 2x+5=55$
$\Rightarrow 2x=55-5$
$\Rightarrow 2x=50$
$\Rightarrow x=\frac{50}{2}$
$\Rightarrow x=25$
Therefore, the requirednumbers are $x=25,andx+5=30$
Therefore, the two numbers with a difference of 5 whose sum is 55 are 25 and 30.

:
Each part: 1.5 Marks
(i) Value of the expression for $x=-4$
$2x^2+12x+3$
$2(-4{)}^2+12(-4)+3$
$2\left(16\right)-48+3$
$32-48+3=-15$
Hence, the value of expression $2x^2+12x+3$ at $x=-4$ is lesser than 15.
(ii) As per question
$x^2$ is added to $2x^2+12x+3$ and 15 is subtracted from it
So, the final expression is:
$2x^2+12x+3+x^2-15$
$3x^2+12x-12$
On putting x = -4 in the above equation we get:
$3\times (-4{)}^2+12\times (-4)-12$
$=48-48-12$
$=-12$

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Forming the equation: 1 Mark
Steps: 2 Marks
Answer: 1 Mark
Given that
A number is 12 more than the other number.
Let the number be $x$, thentheother number will be $x+12$.
As per question
$x+(x+12)=48$
$\Rightarrow 2x+12=48$
$\Rightarrow 2x=48-12$
$\Rightarrow x=\frac{36}{2}=18$
Then, the other number is $x+12=18+12=30$.
Therefore, the numbers are 18 and 30.
The product of these numbers $=18\times 30=540$.