First: Choose the correct answer (first 12 questions):
Which of the following is equivalent to $-x \le -2$ ?
Answer: D $x \ge 2$$x(x+2) =$ ..........
Answer: B $x^2 + 2x$If the lengths of two diagonals of a rhombus are 8 cm and 10 cm, then its area is ........ square centimeters.
Answer: C $40$If: $(x-3)(x+3)=x^2-k$, what is the value of $k$?
Answer: B $9$$(x^2 + x) \div x =$ ..........
Answer: B $x+1$Which of the following is equal to $0.0000052$?
Answer: B $5.2 \times 10^{-6}$A rhombus has a side length of $10$ cm, and the lengths of its two diagonals are $12$ cm and $16$ cm. Its height = ........ cm.
Answer: C $9.6$The multiplicative inverse of $\sqrt{\dfrac{25}{16}}$ is ..........
Answer: B $\dfrac{4}{5}$The area of a square whose its diagonal $12$ cm is ........ square centimeters.
Answer: B $72$If: $x^3=-27$, what is the value of $x$?
Answer: B $-3$If: $2^4 \times a = 2^{20}$, what is the value of $a$?
Answer: A $2^{16}$If: $2.5 \times 10^n = 0.000025$, what is the value of $n$?
Answer: A $-5$Second: Answer the following (3 questions):
Find the solution set of the following inequality in $N$: $2x-3 \le 1$.
Answer:$2x-3 \le 1$
Add $3$ to both sides:
$\Rightarrow 2x \le 4$
Divide by $2$:
$\Rightarrow x \le 2$
Since $x \in N$
$\therefore$ S.S. $=\{1,2\}$
Find the solution set of the following inequality in $Z$: $4-2x<-6$.
Answer:$4-2x<-6$
Subtract $4$ from both sides:
$\Rightarrow -2x<-10$
Divide by $-2$ (reverse the inequality sign):
$\Rightarrow x>5$
Since $x \in Z$
$\therefore$ S.S. $=\{6,7,8,\dots\}$
Calculate the area of a rectangle whose length is $5$ units greater than its width, and whose width is equal to $x$ units of length.
Answer:Width $=x$
Length $=x+5$
Since: Area of rectangle $=$ length $\times$ width
$\Rightarrow A=x(x+5)$
$\therefore$ $A=x^2+5x$
If the area of a rectangle is $(4x^4+8x^3+12x^2)$ square units, and the length of one of its dimensions is $(4x^2)$ units of length, find the length of the other dimension in terms of $x$.
Answer:Since $A=l \times w$
$\Rightarrow 4x^4+8x^3+12x^2=4x^2 \times w$
$\Rightarrow w=\dfrac{4x^4+8x^3+12x^2}{4x^2}$
$\therefore w=x^2+2x+3$
Simplify in the simplest form: $\dfrac{6x^3(3x^2-6x-9)}{9x^2}$, Then find the numerical value of the result when $x=1$.
Answer:$\dfrac{6x^3(3x^2-6x-9)}{9x^2}$
Since $3x^2-6x-9=3(x^2-2x-3)$
$\Rightarrow \dfrac{6x^3 \times 3(x^2-2x-3)}{9x^2}$
$\Rightarrow \dfrac{18x^3(x^2-2x-3)}{9x^2}$
$\Rightarrow 2x(x^2-2x-3)$
Since $x=1$
$\Rightarrow 2(1)\left(1^2-2(1)-3\right)$
$\Rightarrow 2(1-2-3)$
$\Rightarrow 2(-4)$
$\therefore -8$
Find in simplest form: $(3x-2)(x^2-2x+5)$.
Answer:$(3x-2)(x^2-2x+5)$
$\Rightarrow 3x(x^2-2x+5)-2(x^2-2x+5)$
$\Rightarrow 3x^3-6x^2+15x-2x^2+4x-10$
$\Rightarrow 3x^3-8x^2+19x-10$
Which has a larger area: a rhombus whose diagonals lengths are $6$ cm and $8$ cm or a square whose diagonal length is $8$ cm?
Answer:Since the area of a rhombus $=\dfrac{1}{2} \times d_1 \times d_2$
$\Rightarrow A=\dfrac{1}{2}\times6\times8$
$\Rightarrow A=24 \text{ cm}^2$
Since the area of a square $=\dfrac{d^2}{2}$
$\Rightarrow A=\dfrac{8^2}{2}$
$\Rightarrow A=\dfrac{64}{2}$
$\Rightarrow A=32 \text{ cm}^2$
$\therefore$ The square has the larger area.
Find the length of diagonal of a square whose area is equal to the area of a rhombus whose diagonals lengths are $4$ meters and $25$ meters.
Answer:Since the area of a rhombus $=\dfrac{1}{2} \times d_1 \times d_2$
$\Rightarrow A=\dfrac{1}{2}\times4\times25$
$\Rightarrow A=50 \text{ m}^2$
Since the area of a square $=\dfrac{d^2}{2}$
$\Rightarrow \dfrac{d^2}{2}=50$
$\Rightarrow d^2=100$
$\Rightarrow d=10 \text{ m}$
Simplify in the simplest form: $(x+2)^2 - x(x+4)$.
Answer:$(x+2)^2 - x(x+4)$ $= x^2+4x+4 - (x^2+4x)$
$= x^2+4x+4 - x^2 -4x$ $= 4$
Find the quotient of: $(x^2+9x+20)$ divided by $(x+4)$ where $x \ne 0$.
Answer:By using a long division: $\dfrac{x^2+9x+20}{x+4}$ $=x+5$
Find in simplest form: $\sqrt{\dfrac{81}{49}}+\left(\dfrac{3}{4}\right)^0+\sqrt[3]{\dfrac{125}{343}}$.
Answer:$\sqrt{\dfrac{81}{49}}+\left(\dfrac{3}{4}\right)^0+\sqrt[3]{\dfrac{125}{343}}$
Since $\sqrt{\dfrac{81}{49}}=\dfrac{9}{7}$
Since $\left(\dfrac{3}{4}\right)^0=1$
Since $\sqrt[3]{\dfrac{125}{343}}=\dfrac{5}{7}$
$\therefore$ $\sqrt{\dfrac{81}{49}}+\left(\dfrac{3}{4}\right)^0+\sqrt[3]{\dfrac{125}{343}}$ = $\dfrac{9}{7}+1+\dfrac{5}{7}$
$= \dfrac{14}{7}+1$ $= 2+1$ $= 3$
Find in simplest form: $\dfrac{(-3)^5\times(-3)^6}{(-3)^3\times(-3)^5}$.
Answer:$\dfrac{(-3)^5\times(-3)^6}{(-3)^3\times(-3)^5}$ $=\dfrac{(-3)^{5+6}}{(-3)^{3+5}}$ $=\dfrac{(-3)^{11}}{(-3)^8}$ $=(-3)^{11-8}$ $=(-3)^3$ $=-27$
If $(2x-1)$ is one factor of the expression $(2x^2+7x-4)$, find the other factor.
Answer:By using a long division: $\dfrac{2x^2+7x-4}{2x-1}=x+4$
Find the value of $x$ in each of the following:
$1)\;2x^2-5=13$ $\;\;\;\;\;\;\;2)\;3x^3+15=96$
$1)\;2x^2-5=13$
$2x^2=18$
$x^2=9$
$x=\pm3$
$2)\;3x^3+15=96$
$3x^3=81$
$x^3=27$
$x=3$
Write the result of the following in scientific notation: $(4.5\times10^7)\times(4\times10^8)$.
Answer:$(4.5\times10^7)\times(4\times10^8)$ $=(4.5\times4)\times(10^7\times10^8)$
$=18\times10^{15}$ $=1.8\times10^{16}$
Find the solution set of the inequality: $5x+6\le4x+2 \quad (x\in Q)$.
Answer:$5x+6\le4x+2$
$5x-4x\le2-6$
$x\le-4$
$\therefore$ S.S. $=\{x\in Q: x\le-4\}$
Find in simplest form: $\dfrac{4x^5}{x^5}+\dfrac{x^4}{x^3}+\dfrac{x^3}{-x^3}$.
Answer:$\dfrac{4x^5}{x^5}+\dfrac{x^4}{x^3}+\dfrac{x^3}{-x^3}$ $=4+x-1$ $=x+3$
If the expression $(x^3-x^2-4x-m)$ is divisible by $(x-3)$, find the value of $m$.
Answer: By using a long division:
How to study for the March Math Test (Prep 1):
1. Master the laws of exponents: product of powers, power of a power, quotient of powers, zero and negative exponents.
2. Practice simplifying expressions involving powers and exponents step by step.
3. Understand the difference between base and exponent, and how to evaluate numerical expressions.
4. Review word problems that require exponent rules.
5. Practice solving basic equations involving exponents.
6. Review the first 3 multiple-choice and 3 essay questions in this revision guide as a starting point.