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Sia ? 6 years, 6 months ago
Let breadth of the rectangular park = {tex}x {/tex} m
Then, length of the rectangular park = ({tex}x+3{/tex}) m
Now, area of the rectangular park is = {tex}x(x+3)=(x^2+3x)m^2{/tex} {tex}[\because area=length\times breadth]{/tex}
Given, base of the triangular park = Breadth of the rectangular park
Therefore base of triangular park is = {tex}x{/tex} m
and it is given that altitude of triangular park is = {tex}12 {/tex} m
Therefore, area of the triangular park will be = {tex}\frac{1}{2}\times x\times 12= 6x {/tex} {tex}m^2{/tex} {tex}[\because area(Triangle)=\frac{1}{2}\times base\times altitude]{/tex}
As per the question area of rectangular park is = {tex}4 + {/tex} Area of triangular park
{tex}\Rightarrow x^2+3x=4+6x{/tex}
{tex}\Rightarrow x^2+3x-6x-4=0{/tex}
{tex}\Rightarrow x^2-3x-4=0{/tex}
{tex}\Rightarrow x^2-4x+x-4=0{/tex} [ by factorization ]
{tex}\Rightarrow x(x-4)+1(x-4)=0{/tex}
{tex}\Rightarrow (x-4)(x+1)=0{/tex}
{tex}\Rightarrow x-4=0 {/tex} or {tex}x+1=0{/tex} {tex}{/tex}
⇒ x = 4 or x = - 1
Since, breadth cannot be negative , so we will neglect {tex}x=-1{/tex} and choose {tex}x = 4{/tex}
Hence, breadth of the rectangular park will be = 4 m
and length of the rectangular park will be = {tex}x+3=4+3=7m{/tex}
Verification:
Area of rectangular park is = 28 m2
Area of triangular park is = 24 m2 = (28 – 4) m2
Posted by Aisha Massey 7 years, 9 months ago
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Posted by Priya Dharshini ? 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let an be the n th term of the A.P. Then,
an = Sn - Sn-1
{tex}\Rightarrow{/tex} an= (5n2 + 3n) - {5(n -1)2 + 3 (n- 1)} {tex}\left[ \begin{array} { l } { \text { Replacing } n \text { by } ( n - 1 ) \text { in } S _ { n } } \\ { \text { to get } S _ { n - 1 } = 5 ( n - 1 ) ^ { 2 } + 3 ( n - 1 ) } \end{array} \right]{/tex}
{tex}\Rightarrow{/tex}an = (5n2 + 3n) - {5(n2 - 2n+ 1) + 3n - 3}
{tex}\Rightarrow{/tex} an = (5n2 + 3n) - {5n2 -10n + 5 + 3n - 3}
{tex}\Rightarrow{/tex} an = (5n2 + 3n) - (5n2 - 7n + 2)
{tex}\Rightarrow{/tex} an = 10n - 2
Posted by Virat Kumar 5 years, 8 months ago
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Sia ? 6 years, 6 months ago
Let the present ages of Aftab and his daughter be x year and y year respectively. Then the algebraic representation is
Given by the following equations:
x - 7 = 7(y - 7)
{tex}\Rightarrow{/tex} x - 7y + 42 = 0 ...(1)
And x + 3 = 3(y + 3)
{tex}\Rightarrow{/tex} x - 3y - 6 = 0 ...(2)
To, represent this equation graphically, well find two solution for each equation, These solution are given below;
For Equation (1) x - 7y + 42 = 0
{tex}\Rightarrow{/tex} 7y = x + 42
{tex}\Rightarrow y = \frac{{x + 42}}{7}{/tex}
Table 1 of solutions
| x | 0 | 7 |
| y | 6 | 7 |
For Equation (2) x - 3y - 6 = 0
{tex}\Rightarrow{/tex} 3y = x - 6
{tex}\Rightarrow y = \frac{{x - 6}}{3}{/tex}
Table 2 of solutions
| x | 0 | 6 |
| y | -2 | 0 |
We plot the A(0, 6) and B(7, 7)
Corresponding to the solutions in table 1 on a graph paper to get the line AB representing the equation (1) and the points C(0, -2) and D(6, 0) corresponding to the solutions in table 2 on the same graph paper to get the line CD representing the equation (2), as shown in the figure
We observe in figure that the two lines representing the two equations are intersecting at the point P(42, 12).
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Sia ? 6 years, 6 months ago
g(x) = x2 + 3x + 1, f(x)= 3x4 + 5x3 - 7x2 + 2x + 4
If we divide f(x) by g(x) then remainder, r(x) = 2
Hence, r(x) {tex}\ne{/tex} 0.
If remainder is not 0, then g(x) is not a factor of p(x).
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