Multiply 12 and 25.
We get 300. Both, in 1.2 and 2.5, there is 1 digit to the right of the decimal point.
So, count 1 + 1 = 2 digits from the right most digit (i.e., 0) in 300 and move towards left.
We get 3.00 or 3.

No, we know that, opposite faces of a die always have a total of seven dots on them. In the given net, one pair of opposite faces will have 1 and 4 on them whose total is 5 (≠7) and another pair of opposite faces will have 3 and 6 on them, whose total is 9 (≠7). So, given figure cannot be a net for a die.

Initial population of the city = 25000
Decreased population of the city = 24500
Thus, decrease in population = 25000 - 24500 = 500
\(\therefore\)  Percentage decrease in population
\(=\left({Decrease \ in \ population\over initial \ population}\times 100\right)\%\)
\(=\left({500\over 25000}\times 100)\%=({50\over25}\right) \%=2\%\)
Hence, the percentage decrease in population is 2%.

It is impossible that moon will not revolve around the earth.

\(\frac { 3 }{ 5 } \)

\(1-\frac{1}{5}=\frac{5}{5}-\frac{1}{5}=\frac{5-1}{5}=\frac{4}{5}\)

\(\Rightarrow \quad \frac { q }{ 4 } =5-7\)
\(\Rightarrow \quad \frac { q }{ 4 } =-2\)
\(\Rightarrow\) q = -2 \(\times\) 4
\(\therefore\) q = -8

We have, \(\frac { 25\times { 5 }^{ 2 }\times { t }^{ 8 } }{ { 1 }0^{ 3 }\times { t }^{ 4 } } =\frac { { 5 }^{ 2 }\times { 5 }^{ 2 }\times { t }^{ 8 } }{ { \left( 2\times 5 \right) }^{ 3 }\times { t }^{ 4 } } \)
[\(\because\)25=5x5=5² and 10=2x5]
\(=\frac { { 5 }^{ 2+2 }\times { t }^{ 8 } }{ { 2 }^{ 3 }\times { 5 }^{ 3 }\times { t }^{ 4 } } \)                                 [ \(\because\)a^m x aⁿ = a^mn  and (ab)^m = a^m x b^m ]
\(=\frac { 5^{ 4-3 }\times { t }^{ 8-4 } }{ { 2 }^{ 3 } } =\frac { 5\times { t }^{ 4 } }{ 2\times 2\times 2 } =\frac { 5{ t }^{ 4 } }{ 8 } \)    [\(\because\)a^m\(\div\)aⁿ=a^m-n]

On putting m = 2 in 3m² - 2m - 7, we get
3m² - 2m - 7 = 3(2)² - 2(2) - 7
= 3 X 4 - 4 - 7 = 12 -11 = 1

We have, 43.07 X 100 = 4307

Given, \(\frac{-3}{5} \) as a rational number
with numerator 9 
\(\frac { -3\times \left( -3 \right) }{ 5\times \left( -3 \right) } =\frac { 9 }{ -15 } \)

We have, if \(\angle \)4 + \(\angle \)5 = 180°, then a||b.
If a transversal intersects two parallel lines, then the interior angles on the same side of the transversal are supplementary.

36

10m

\(\frac{5}{6}\)

We have, x+(y+z)
\(=\frac { -4 }{ 9 } +\left( \frac { 5 }{ 12 } +\frac { 7 }{ 18 } \right) =\frac { -4 }{ 9 } +\left( \frac { 5\times 3+7\times 2 }{ 36 } \right) \\ =\frac { -4 }{ 9 } +\left( \frac { 15+14 }{ 36 } \right) \\ =\frac { -4 }{ 9 } +\frac { 29 }{ 36 } =\frac { -4\times 4+29 }{ 36 } =\frac { 13 }{ 36 } \)

135°

\(\frac{9^n\times 3^2\times 3^n-(27)^2}{(3^3)^5\times 2^3}=\frac{1}{27}\)
\(\Rightarrow \frac{(3\times 3)^n\times 3^2\times3^n-(3\times3\times3)^n}{3^{15}\times 2^3}=\frac{1}{3\times3\times3}\)
\(\Rightarrow \frac{(3^2)^n\times3^{n+2}-(3^3)^n}{3^{15}\times2^3}=\frac{1}{3^3}\)
\(\Rightarrow \frac{3^{2n}\times 3^{n+2}-3^{2n}}{3^{15}\times 2^3}=3^{-3}\)
\(\Rightarrow \frac{3^{2n+n+2}-3^{3n}}{3^{15}\times 2^3}=3^{-3}\)
\(\Rightarrow \frac{3^{2n+n+2}-3^{3n}}{3^{15}\times2^3}=3^{-3}\)
\(\Rightarrow \frac{3^{3n}(3^2-1)}{3^{15}\times 2^3}=3^{-3}\)
\(\Rightarrow \frac{3^{3n-15}(9-1)}{2^3}=3^{-3}\)
\(\Rightarrow 3^{3n-15}\times \frac{8}{8}=3^{-3}\)
\(\Rightarrow 3^{3n-15}=3^{-3}\)
As base are same on both sides, so
3n - 15 = -3
\(\Rightarrow \) 3n = -3 + 15
\(\Rightarrow \) 3n = 12
thus,  \(n=\frac{12}{3}=4\)

Given, \(\frac { 2x-3 }{ 5 } +\frac { x+3 }{ 4 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow \quad \frac { 4(2x-3) }{ 5\times 4 } +\frac { 5(x+3) }{ 5\times 4 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow \quad \frac { 8x-12 }{ 20 } +\frac { 5x+15 }{ 20 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow \quad \frac { 8x-12+5x+15 }{ 20 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow \quad \frac { 13x+3 }{ 20 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow\) 7(13x + 3) = 20(4x + 1)
\(\Rightarrow\) 91x + 21 = 80x + 20
\(\Rightarrow\) 91x - 80x = 20 - 21
\(\Rightarrow\) 11x = -1
Thus, \(x=-\frac { 1 }{ 11 } \)