7m (m - 5) + 1 (5 - m)
Taking out the common binomial factor (m - 5)
= 7m (m - 5) + (-1)(-5 + m)
= 7m (m - 5) - 1 (m - 5)
Taking out the common binomial factor (m - 5) = (m - 5) (7m - 1)

\(=\frac{4 m^{2} n+9 n^{2} m+3 m n}{4 m n}\)

\(=m^{2-1}+\frac{9}{4} n^{2-1}+\frac{3}{4}=m+\frac{9}{4} n+\frac{3}{4}\)

Length of the log = 3a + 4b-2
Remaining length of the log = (3a + 4b - 2)-(2a - b)
= (3a - 2a) + [4b - (-b)]- 2
= (3 - 2) a + (4 + 1) b - 2
= a + 5b-2

\(\frac { -7 }{ 11 } =\frac { -70 }{ 110 } ;\frac { 6 }{ -11 } =\frac { -60 }{ 110 } \)
We have \(\frac { -61 }{ 110 } ,\frac { -62 }{ 110 } ....,\frac { -69 }{ 110 } \) between \(\frac{-7}{11}\) and \(\frac{6}{-11}\)

\(0,-\frac { 1 }{ 2 } ,\frac { 1 }{ 2 } ,\frac { 3 }{ 4 } ,\frac { 6 }{ 7 } ,-5,6\)

(10 x 5) + (9 x 4)

\(LHS=\left( \frac { \frac { 7 }{ 9 } -5 }{ \frac { 4 }{ 3 } } \right) \div \frac { 3 }{ 2 } +\frac { 4 }{ 9 } -\frac { 1 }{ 3 } =\left( \frac { \frac { 7-(5\times 9) }{ 9 } }{ \frac { 4 }{ 3 } } \right) \div \frac { 3 }{ 2 } +\frac { 4 }{ 9 } -\frac { 1 }{ 3 } \)
\(=\left( \frac { \frac { 7 }{ 9 } -5 }{ \frac { 4 }{ 3 } } \right) \div \frac { 3 }{ 2 } +\frac { 4 }{ 9 } -\frac { 1 }{ 3 } =\left( \frac { \frac { -38 }{ 9 } }{ \frac { 4 }{ 3 } } \right) \div \frac { 3 }{ 2 } +\frac { 4 }{ 9 } -\frac { 1 }{ 3 } \)
\(=\left( \frac { -38 }{ 9 } \times \frac { 3 }{ 4 } \right) \div \frac { 3 }{ 2 } +\frac { 4 }{ 9 } -\frac { 1 }{ 3 } =\frac { -19 }{ 6 } \div \frac { 3 }{ 2 } +\frac { 4 }{ 9 } -\frac { 1 }{ 3 } \)
\(=\frac { -19 }{ 6 } \times \frac { 2 }{ 3 } +\frac { 4 }{ 9 } -\frac { 1 }{ 3 } =\frac { -19 }{ 9 } +\frac { 4 }{ 9 } -\frac { 1 }{ 3 } =\frac { -19+4-(1\times 3) }{ 9 } \)
\(=\frac { -15-3 }{ 9 } =\frac { -18 }{ 9 } =-2=RHS\)

identity fails 5 – 0 ≠ 0 – 5

\(\frac { 2 }{ 3 } ,\frac { 4 }{ 5 } \)
LCM of 3 and 5 is 15.
\(\frac { 2 }{ 3 } =\frac { 2\times 5 }{ 3\times 5 } =\frac { 10 }{ 15 } \)
\(\frac { 4 }{ 5 } =\frac { 4\times 3 }{ 5\times 3 } =\frac { 12 }{ 15 } \)
\(\frac { 10 }{ 15 } <\frac { 12 }{ 15 } \)
\(\therefore \frac { 2 }{ 3 } <\frac { 4 }{ 5 } \)

The rational number for the point marked on the number line is 1\(\frac{3}{4}=\frac{7}{4}\)

–1.2 and -2.3
\(-1.2=\frac { -1.2\times 10 }{ 1\times 10 } =\frac { -12 }{ 10 } \)
\(-2.3=\frac { -2.3\times 10 }{ 1\times 10 } =\frac { -23 }{ 10 } \)
∴ Five rational numbers between -1.2\(\\ \\ (=\frac { -12 }{ 10 } )\) and -2.3 \((=\frac { -23 }{ 10 } )\) are \(\frac { -21 }{ 10 } ,\frac { -20 }{ 10 } ,\frac { -15 }{ 10 } ,\frac { -14 }{ 10 } ,\frac { -13 }{ 10 } \)

\(\frac{-1}{2}\) and \(\frac{3}{5}\)
LCM of 2 and 5 = 2 x 5 = 10
\(-\frac { 1 }{ 2 } =\frac { -1\times 5 }{ 2\times 5 } =\frac { -5 }{ 10 } \)
\(\frac { 3 }{ 5 } =\frac { 3\times 2 }{ 5\times 2 } =\frac { 6 }{ 10 } \)
∴ Five rational numbers between \(-\frac { 1 }{ 2 } (=\frac { -5 }{ 10 } )\) and \(\frac{3}{5}\) \(\\ \\ \\ (=\frac { 6 }{ 10 } )\) are \(\frac { -3 }{ 10 } ,\frac { -1, }{ 10 } ,0,\frac { 1 }{ 10 } ,\frac { 2 }{ 10 } ,\frac { 5 }{ 10 } \)

c³-27b³a³ = c³-33b³a³
= c³-(3ba)³
Comparing this with a³-b³ we have a = x and b = 3ba
a³ - b³ = (a-b)(a²+ab+b²)
\(\therefore\) c³-(3ba)³ = (c-3ba)(c²+(c)(3ba)+(3ba)²)
= (c-3ba) (c²+3bac+3²b²a²)
c³-27b³a³ = (c-3ab)(c²+3bac+9a²b²)

64m³+n³ = (4³m³)+n³
= (4m)³+n³
= (4m)³+n³
Comparing this with a³+b³ we have a = 4m; b = n
a³+b³ = (a+b)(a²-ab+b²)
(4m)³ + n³ = (4m+ n) [(4m)²- (4m) (n) + n2]
= (4m + n) [4²m²-4mn + n²]
= (4m + n) [16m² - 4mn + n²]
64m³ + n³ = (4m + n) (16m²-4mn + n²)

2a³+16
(2 x a³) + (2 x8) = 2(a³ + 8) = 2 (a³ + 2³)
\(\therefore\) 2a³+16 = 2(a³+2³)
2(a³+2³) = [2(a+2)(a²-(a)(2)+2²)] = 2[(a+2)(a²-2a+4)]
2a³+16 = 2(a+2) (a²-2a+4)

Area of the house = Area of a square of side 6 cm + Area of a rectangle with 1 = 8cm, b = 6 cm + Area of a Δ with b = 6 cm and h = 4 cm + Area of a parallelogram with b = 8cm, h = 4cm
= (side x side) + (l x b) +( \(\frac { 1 }{ 2 } \)x b x h) + bh cm²
= (6 x 6) + (8 x 6) + (\(\frac { 1 }{ 2 } \) x 6 x 4) + (8 x 4) cm²
= 36 + 48 + 12 + 32 cm²
= 128 cm²
Required Area = 128 cm²

Proof: from the \(\triangle \)SUN and \(\triangle \) RAY
SU = 10
UN = 12
SN = 14
RA = 5
AY = 6
RY = 7
We have
 \(\frac { SU }{ RA } =\frac { 10 }{ 5 } =\frac { 2 }{ 1 } \\ \frac { UN }{ AY } =\frac { 12 }{ 6 } =\frac { 2 }{ 1 } \\ \frac { SN }{ RY } =\frac { 14 }{ 7 } =\frac { 2 }{ 1 } \)
From (1), (2) and (3) we have
\(\frac { SU }{ RA } =\frac { UN }{ AY } =\frac { SN }{ RY } =\frac { 2 }{ 1 } \)
The sides are proportional
∴ \(\triangle \)SUN ~ \(\triangle \)RAY

Cuboid

\(\text { In } \triangle \mathrm{ODC} \text { and } \triangle \mathrm{EDC} \text {, }\)
CD = CD
\(\angle \mathrm{CDO}=\angle \mathrm{CDE}=90^{\circ}\)
OD = ED (given)
\(\therefore \triangle \mathrm{ODC} \equiv \Delta \mathrm{EDC}(\mathrm{SAS})\)
Hence proved

Given in\( \triangle \mathrm{WHY} and \ \triangle \mathrm{WET}, \)
\(\angle \mathrm{W} =\angle \mathrm{W} \)
\(\angle \mathrm{WYH} =\angle \mathrm{WTE} \)
\(\angle \mathrm{WHY} =\angle \mathrm{WET} \\ \therefore \Delta \mathrm{WHY} \sim \Delta \mathrm{WET} \)
Also \(\triangle \)WHY ~ \(\triangle \)WET
∴ Corresponding sides are proportionated
\(\frac { WH }{ WE } =\frac { HY }{ ET } =\frac { WY }{ WT } \)
\(\\ \frac { 6 }{ 6+HE } =\frac { 4 }{ ET } =\frac { 4 }{ 16 }\)
\( \\ \frac { 6 }{ 6+HE } =\frac { 4 }{ 16 }\)
\( \\ 6+HE=\frac { 6 }{ 4 } \times16\)
\(\\ 6+HE=24\)
\(\\ \therefore HE=24-6\\ HE=18\)
\(\\ Again\quad \frac { 4 }{ ET } =\frac { 4 }{ 16 } \)
\(\\ ET=\frac { 4 }{ 4 } \times 16\)
\(\\ ET=16\)

iv, v, ii, iii, i

From the figure
diameter = 10 cm
radius = 5 cm
Area of the shaded part
= Area of the square - Area of the four circular quadrant
\( =a^{2}-\left(4 \times \frac{\theta}{360} \times \pi r^{2}\right) \)
\(=(10 \times 10)-\left(4 \times \frac{90}{360} \times 3.14 \times 5 \times 5\right) \)
= 100 - 78.5 = 21.5 cm² (approximately)

(2x + 3) (2x - 4) 
= 4x² - 8x + 6x - 12 
= 4x² - 2x - 12

The first number can be selected in 10 different ways.
The second number can be selected in 10 different ways.
The third number can be selected in 10 different Ways.
The fourth number can be selected in 10 different ways.
The four digit code is selected in 10 x 10 x 10 x 10 different ways.
That is 10000 ways.

The student has 2 school bags and 3 water bottles.

Therefore he can carry both a school bag and a water bottle in 6 (2 x 3) different possible ways.