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Home Question Bank Math Algebra Archive 2024 December 28

Algebra Questions from Dec 28,2024

Browse the Algebra Q&A Archive for Dec 28,2024, featuring a collection of homework questions and answers from this day. Find detailed solutions to enhance your understanding.

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\( \left. \begin{array} { l } { ( x + y ) ^ { 3 } = z } \\ { ( y + z ) ^ { 3 } = x } \\ { ( z + x ) ^ { 3 } = y } \end{array} \right. \) Three quantities \( t, x \) and \( y \) are such that \( t \) varies directly as \( x \) and invisely as the square root of \( y \). Find the percentage decrease in \( t \) if \( x \) decreases by \( 44 \% \). Una montaña rusa tiene 15 carritos para pasajeros. Algunos carritos pueden llevar hasta 4 pasajeros y los otros carritos restantes pueden llevar hasta 6 . El número total máximo de personas que pueden usar el juego en cada vuelta es de 72 . ¿Cuántos carritos de 4 pasajeros hay y cuántos de 6 ? 2. (B-2) \( \sqrt{\sqrt[3]{x}}=x^{a} \) bo'lsa, \( a \) ning qiymatini toping. \( \begin{array}{llll}\text { A) } a=\frac{2}{3} & \text { B) } a=\frac{1}{5} & \text { C) } a=\frac{5}{6} & \text { D) } a=\frac{1}{6}\end{array} \) 212 / बीजगणित (Algebra) 24. यदि \( 4,2 x-4, \ldots \ldots, 64,21 x+2 \) एउटा गुणोत्तर अनुक्रमको पदहरू भए, If \( 4,2 x-4, \ldots \ldots, 64,21 x+2 \) are the terms of geometric sequence (i) \( x \) को मान पत्ता लगाउनुहोस् । (Find the value of \( x \).) (ii) समान अनुपात पत्ता लगाउनुहोस् । (Find the common ratio.) (iii) चौथो पद पत्ता लगाउनुहोस् । (Find the value of fourth term.) Calcula \( X \) tal que \( X-B^{2}=A \cdot B \), siendo que: \[ A=\left(\begin{array}{lll}1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 0 & 2\end{array}\right), \quad B=\left(\begin{array}{ccc}1 & 0 & -1 \\ 1 & 1 & 1 \\ 0 & 0 & 1\end{array}\right) \] \( \left. \begin{array} { l } { x ^ { 2 } - y ^ { 2 } + y } \\ { = 3 x ^ { 2 } y - y ^ { 3 } + x y } \end{array} \right. \) a) \( 3 a^{2}+(3 a+b)(b-a) \) b) \( 4 x(y-x)-(x+6 y)(1-4 x) \) c) \( (n-5)(7-5 n)-2(n+2)(n-6) \) d) \( -2(6-3 m)(m+1)+5(m-4)(m-5) \) 7. \( \frac{a^{2}+1}{3 a-6} \div\left(\frac{a^{3}+a}{6 a-12} * \frac{4 x+8}{x-3}\right. \) a) \( (5 x-2 y)(x+y)-5 x^{2} \) b) \( 2 b(7-b)-(a+2 b)(3-b) ; \) c) \( 3(x-4)(x+2)+(3 x-1)(5-x) \) d) \( 4(2 c+1)(5 c-3)-3(c+2)(c+ \)
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