pa tulong po math grade 10 2nd quarter

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pa tulong po math grade 10 2nd quarter​

annesenit08 · Answer

Answer:To solve these equations, we need to find the values of $x$ that satisfy each equation. Let's solve them one by one:1) $x^{2}-12-x=0$ can be rearranged as $x^{2}-x-12=0$. This is a quadratic equation. Factoring or using the quadratic formula, we can find that the solutions are $x = -3$ and $x = 4$.2) $x^{2}-16=0$ can be rewritten as $x^{2}-4^{2}=0$, which is a difference of squares. This equation can be factored as $(x-4)(x+4)=0$. The solutions are $x = -4$ and $x = 4$.3) $2x^{2}+x^{3}-5x-6=0$ is a cubic equation. Factoring or using other methods, we can find that one of the solutions is $x = -2$. We can then divide the equation by $x+2$ to get $x^{2}-x-3=0$, which is a quadratic equation. We can factor this as $(x-3)(x+1)=0$. So the solutions to the original equation are $x = -2$, $x = -1$, and $x = 3$.4) $x^{3}+6x^{2}+11x+6=0$ is a cubic equation. By using synthetic division or other methods, we can find that one of the solutions is $x = -1$. Dividing the equation by $x+1$, we get $x^{2}+5x+6=0$. This quadratic equation can be factored as $(x+2)(x+3)=0$. So the solutions to the original equation are $x = -1$, $x = -2$, and $x = -3$.5) $x^{4}+2x^{3}-3x^{2}-8x-4=0$ is a quartic equation. By using factoring or other methods, we can find that $x = -1$ is a solution. Dividing the equation by $x+1$, we get $x^{3}+x^{2}-4x-4=0$. By synthetic division or other methods, we find that $x = -1$ and $x = 2$ are solutions. Dividing the equation by $x-2$, we get $x^{2}+3x+2=0$, which can be factored as $(x+1)(x+2)=0$. So the solutions to the original equation are $x = -1$, $x = -2$, and $x = 2$.6) $2x^{2}-3x-2=0$ is a quadratic equation. We can factor this as $(2x+1)(x-2)=0$. So the solutions are $x = -\frac{1}{2}$ and $x = 2$.7) $-17x^{2}+18x+3x^{3}+8=0$ can be rearranged as $3x^{3}-17x^{2}+18x+8=0$. By using synthetic division or other methods, we find that $x = -\frac{2}{3}$ is a solution. Dividing the equation by $x+\frac{2}{3}$, we get $3x^{2}-20x+12=0$. This quadratic equation can be factored as $(3x-2)(x-6)=0$. So the solutions to the original equation are $x = -\frac{2}{3}$, $x = \frac{2}{3}$, and $x = 6$.8) $2x^{3}-18x=0$ can be factored as $2x(x^{2}-9)=0$. This equation has three solutions: $x = 0$, $x = -3$, and $x = 3$.9) $x^{4}-7x^{2}+6x=0$ can be factored as $x(x^{3}-7x+6)=0$. This equation has two solutions: $x = 0$ and $x = 1$.10) $x^{5}-3x^{4}-11x^{3}+27x^{2}+10x-24=0$ is a quintic equation. By using synthetic division or other methods, we find that $x = 1$ is a solution. Dividing the equation by $x-1$, we get \(x^{

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