Question

plz thoda jldi....urgent h​

Answer 1

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$\huge\underline\mathcal\red{Answer}$

Q1) To draw a graph of 3x+10y=-57 . Find y when x=1 .

$\: \: \boxed{ \bf{solution \to}}$

$\: \: \\ \implies \displaystyle \tt{3x + 10y = - 54...given \: equation}$

$\: \\ \implies \displaystyle \tt{ \color{brown} \: to \: find \: the \: value \: of \: y \: at \: given \: value \: x = 1}$

$\: \: \implies \displaystyle \tt{3x + 10y = - 54}$

$\: \implies \displaystyle \tt{3(1) + 10y = - 54}$

$\: \implies \displaystyle \tt{3 + 10y = - 54}$

$\: \: \implies \displaystyle \tt{3 + 10y + 54 = 0}$

$\: \implies \displaystyle \tt{10y + 57 = 0}$

$\: \implies \displaystyle \tt{y = \frac{ - 57}{10} }$

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Q2)Compare the Quadratic equation (x-5)(x+7)=0 to the general form and write values of a and c.

$\: \: \divideontimes \underline{ \boxed{ \bf \: solution \to}}$

$\: \: \implies \displaystyle \tt{(x - 5)(x + 7) = 0}$

$\: \implies \displaystyle \tt{(x(x + 7) - 5(x + 7)) = 0}$

$\: \implies \displaystyle \tt{ {x}^{2} + 7x - 5x - 35 = 0}$

$\: \implies \displaystyle \tt{ {x}^{2} + 2x - 35 = 0}$

$\: \\ \dashrightarrow \underline{\bf{comparing \: this\: quadratic \: equation \: with \: a {x}^{2} + bx + c = 0}}$

$\: \: \\ \implies \displaystyle \tt{a = 1 \: and \: c = - 35 \: is \: the \: answer}$

$\: \bigstar\boxed{ \bf{a = 1 \: and \: c = - 35 \: is \: the \: value \: }}$

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Q3) In the Ap, -34,-27,-20,-13,-6.............

,Find the common difference and the value of t4

$\: \: \underline{ \boxed{ \bf{solution \to}}}$

$\implies\displaystyle\tt{-34,-27,-20,-13,-6 \: are \: the \: series \: of \: Ap}$

$\: \\ \implies \displaystyle \tt{ \: - common \: difference (d)= - 27 - ( - 34)}$

$\: \\ \implies \displaystyle \tt{common \: difference(d) = - 27 + 34}$

$\: \\ \implies \displaystyle \tt{common \: difference(d) = 7}$

$\: \boxed{ \divideontimes { \displaystyle \tt{ \color{lime} \: common \: difference \: (d) = 7}}}$

$\displaystyle\bf{To\: Find\to\: The \: value \: of \: t_4}$

$\large\underline{\sf{Given\checkmark}}$

$\bullet \bf{a=-34 \: and \: d=7 }$

$\implies\displaystyle\tt{tn=a+(n-1)d}$

$\implies\displaystyle\tt{t_4=-34+(4-1)\times7}$

$\implies\displaystyle\tt{t_4=-34+3\times7}$

$\implies\displaystyle\tt{t_4=-34+21}$

$\implies\displaystyle\tt{t_4=-13}$

$\divideontimes\underline{\boxed{\displaystyle\bf{\color{red} \: the \: value \: of \: t_4 \: is \: -13}}}$