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- Question 7 and 8, Exercise 5.2
- ves: \begin{align*} f(x) &= \left(x - \frac{1}{2}\right)(2x^{2} - 14x + 20)\\ &=\left(x - \frac{1}{2}\right)(2x^{2} - 14x + 20) \\ &= \left(x - \frac{1}{2}\right)(2x^{2} - 10x - 4x + 20 )\\ &= \left(x - \frac{1}{2}\right)[2x(x - 5) - 4(x - 5) ]\\ &=\frac{1}{2}\left(2x -
- Question 4 & 5, Review Exercise
- ) &= 6y^{3} - y^{2} - 5y + 2\\ f\left(\frac{2}{3}\right) &= 6\left(\frac{2}{3}\right)^{3} - \left(\frac{2}{3}\right)^{2} - 5\left(\frac{2}{3}\right) + 2 \\ &= 6\left(\frac{8}{27}\right) - \left(\frac{4}{9}\right) - 5\lef
- Question 2 & 3, Review Exercise
- akistan. =====Question 2===== $\left(64 y^{3}-8\right) \div(4 y-2) \quad$ ** Solution. ** \begin{ali... {align*} =====Question 3===== $\left(125 y^{3}-8\right) \div(5 y-2)$ ** Solution. ** \begin{align*} \f... (5 y-2)} &= \frac{(5y - 2)\left(25y^{2} + 10y + 4\right)}{5y - 2} \\ & = 25y^{2} + 10y + 4. \end{align*} ... e-ex-p1|< Question 1 ]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit05:Re-
- Question 6 & 7, Review Exercise
- at the remainder upon dividing $\left(x^{2}+8 x+k\right)$ by $(x-4)$ is zero. ** Solution. ** Let \( p(... ex-p3|< Question 3 &4]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit05:Re-
- Question 6, Review Exercise
- at the remainder upon dividing $\left(x^{2}+8 x+k\right)$ by $(x-4)$ is zero. ** Solution. ** ... Re-ex-p5|< Question 5]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit05:Re-
- Question 1, Exercise 5.1
- emainder is 35. ====Go to ==== <text align="right"><btn type="success">[[math-11-nbf:sol:unit05:ex5
- Question 2 and 3, Exercise 5.1
- x5-1-p1|< Question 1 ]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit05:ex5
- Question 4 and 5, Exercise 5.1
- -p2|< Question 2 & 3 ]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit05:ex5
- Question 6 and 7, Exercise 5.1
- -p3|< Question 4 & 5 ]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit05:ex5
- Question 8 and 9, Exercise 5.1
- -p4|< Question 6 & 7 ]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit05:ex5
- Question 10, Exercise 5.1
- oom has a volume of $\left(x^{3}+11 x^{2}+34 x+24\right)$ cubic feet. The height of the room is $(x+1)$ f
- Question 1 and 2, Exercise 5.2
- - 1)(x + 1). \] ====Go to ==== <text align="right"><btn type="success">[[math-11-nbf:sol:unit05:ex5
- Question 3 and 4, Exercise 5.2
- -p1|< Question 1 & 2]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit05:ex5
- Question 5 and 6, Exercise 5.2
- -p2|< Question 3 & 4]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit05:ex5
- Question 1, Exercise 5.3
- 12 cm$\\ GOOD ====Go to ==== <text align="right"><btn type="success">[[math-11-nbf:sol:unit05:ex5