Practice Questions
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Simplify the expressions:
a) \( 2\sqrt{8} - 4\sqrt{2} \qquad \) b) \( \dfrac{2 + \sqrt{12}}{2} \qquad\) c) \( 2\sqrt{3} + \dfrac{6}{\sqrt{3}} \qquad \) d) \( 3\sqrt{5} - \dfrac{1}{\sqrt{5} - 2} \)View Video Solution
This problem involves simplifying expressions with square roots and radicals.
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Expand and simplify the polynomials below:
a) \( -(-2x - 2)(-x + 3) - (3x^2 + x - 2) \qquad \) b) \( 3(-x - 5)^2 - 4x(-x - 7) \)View Video Solution
This problem requires applying the distributive property and combining like terms.
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Simplify the expressions below:
a) \( (2x^2)^2 (3x)^2 \qquad \) b) \( (10x^4)^0 (3x^4)^3 \qquad \) c) \( \dfrac{(-4x^4y^2)^2}{(2xy)^3} \)View Video Solution
Apply the rules of exponents (product rule, quotient rule, and power rule) to simplify.
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Simplify the rational expressions:
a) \( \dfrac{-3x + 2}{x - 1} \cdot \dfrac{2x - 2}{3x - 2} \qquad \) b) \( \dfrac{(2n + 3)(3n + 15)}{(n - 5)(6n + 9)} \qquad \) c) \( \dfrac{2x^2 + x}{x(2x + 1)} \qquad \) d) \( \dfrac{x^2 + 2x - 3}{(x + 3)(2x - 2)} \)View Video Solution
Factor the numerators and denominators to find common terms that can be canceled out.
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Evaluate and simplify the expressions for the given value(s) of the variable(s):
a) \( \sqrt{x + 6} \), for \( x = -2 \qquad \) b) \( |x + y - 10| \), for \( x = -4 \) and \( y = 3 \qquad \) c) \( \dfrac{-x^3 + 1}{5} + \dfrac{x^2 - 2}{2} \), for \( x = -2 \)View Video Solution
Substitute the given values into the algebraic expressions and simplify using the order of operations.
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Factor the algebraic expressions:
a) \( 4x - 16 \qquad \) b) \( 6x^2 + 2x \qquad \) c) \( x^2 - 9\qquad \) d) \( 9x^2 - 16y^2 \qquad \) e) \( x^2 - 4x - 5 \qquad \) f) \( \dfrac{1}{2}x^2 + x + \dfrac{1}{2} \)View Video Solution
Find the greatest common factor (GCF) or use methods like grouping or difference of squares.
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Solve the following equations:
a) \( -2(x + 2) - x = 3(x + 2) + 2 \qquad \) b) \( \dfrac{x - 2}{3} = \dfrac{2x + 1}{2} \qquad \) c) \( \dfrac{3x + 1}{4} - 2 = \dfrac{-x + 3}{2} \)View Video Solution
Isolate the variable by performing inverse operations on both sides of the equation.
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Solve the following quadratic equations:
a) \( (x-1)(x+8) = 0 \)
b) \( x(x-1) = 1 \)
c) \( 2x^2 + 6x = 8 \)View Video Solution
For standard quadratic equations, set them to equal zero and factor, or use the quadratic formula.
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Solve the following equations:
a) \( \sqrt{x} = 2 \)
b) \( \sqrt{x - 2} = 4 \)
c) \( \sqrt{\dfrac{x}{10}} = 2 \)View Video Solution
Isolate the radical expression, then square both sides to eliminate the square root.
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Solve the following inequalities and represent the solution set using intervals, graphs on a number line, and inequality symbols:
a) \( 3(x - 2) + 2x < -(x+2) \)
b) \( \dfrac{x + 5}{3} \ge \dfrac{-x+2}{2} \)
c) \( -(2x - 2) \le 3x + 12 \)View Video Solution
Remember to flip the inequality sign if you multiply or divide by a negative number.
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For what value of the parameter \( a \) is the point with coordinates \( (a, 3) \) on the line whose equation is given by \( 2x - 3y = 4 \)?
View Video Solution
Substitute the coordinates into the equation and solve for the unknown parameter.
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What is the point of intersection of the lines given by the equations \( 2x + y = 5 \) and \( 3x - 2y = 4\)?
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Use either substitution or elimination methods to solve the system of linear equations.
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What is the slope of each of the lines given by the equations?
a) \( 4x = -2y + 4 \)
b) \( 3y = -9 \)
c) \( -5x = 10 \)View Video Solution
Rewrite each equation in slope-intercept form (\(y = mx + b\)) to identify the slope (\(m\)).
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Find the x and y intercepts, if possible, of the lines with equations:
a) \( -3x = 2y + 6 \)
b) \( 2y = 8 \)
c) \( -3x = 6 \)View Video Solution
Set \( x = 0 \) to find the y-intercept, and set \( y = 0 \) to find the x-intercept.
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a) Find the value of the parameter \( s \) so that the slope of the line through the points \( (-2 , s) \) and \( (-4, 5) \) is equal to \( -1 \).
b) Find the equation of the line.View Video Solution
Use the slope formula: \( m = \dfrac{y_2 - y_1}{x_2 - x_1} \).
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Find the equation of the line through the point \( (-2, 4) \) and perpendicular to the line whose equation is given by \( -2y + 4x = -2 \).
View Video Solution
Remember that perpendicular lines have negative reciprocal slopes.
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Find the equation of the line shown below given that the length of segment AB is 3.1 and the length of BC is 6.2 and point \( D \) is on the line.
View Video Solution
Use the given lengths to calculate the rise over run (slope), then deduce the equation.
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Which of the following lines whose equations are given below has a negative slope?
a) \( 2x - 2y = 0 \)
b) \( 3x + 6y = 9 \)
c) \( -y = 9 \)
d) \( -y = -x + 3 \)View Video Solution
Convert each to \( y = mx + b \) format to check the sign of \( m \).
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In the figure below, a building has the top of the first floor at point C and the top of the second floor at point D. The angles of elevation from point A, at a distance of 40 meters on the ground, to points C and D are \( 14^\circ \) and \( 21^\circ \) respectively and point B is at the base of the building. Find the height \( DC \) from the first to the second floor.
View Video Solution
Use tangent ratios in the two right triangles to find the vertical heights, then subtract.
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The population of a small town increased by 600 in a year. This represents a 5% increase over the year before. What was the population of the town the year before the increase?
View Video Solution
Set up a percentage equation: \( 0.05 \times \text{Original} = 600 \).
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Find the area of the shaded shape given that AE is parallel to CD which is the diameter of the semi-circle. BC is parallel to ED which is perpendicular to AE. All lengths in mm.
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Decompose the complex shape into simpler geometric figures (rectangles, triangles, semi-circles) to calculate total area.
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Find the area of a regular hexagon whose perimeter is 18 cm.
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A regular hexagon is composed of 6 equilateral triangles. Use the perimeter to find the side length.
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Find the shaded area enclosed by the quarter of a circle and a square of diagonal 10 mm.
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Use the diagonal to find the side length of the square, and interpret the geometry to calculate area subtractions.
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The volume of the right prism is equal to \( 1500 \text{ cm}^3 \).
a) Find the length \( L \) of the prism.
b) Find the total surface area (lateral and bases) of the prism.
c) What is the cost of making the prism with a material that costs $200 per square meter?View Video Solution
Calculate the base area first, divide the volume by the base area to find length \( L \), then calculate total surface area and cost.
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Among 100 students, 70 study math (among which some study physics as well), and 50 study physics (among which some study math as well). Assuming every student studies at least one of the two subjects:
a) How many students are enrolled in both math and physics?
b) How many are enrolled in math only?
c) How many are enrolled in physics only?View Step-by-Step Solution
Let \( M \) represent the set of students studying Math, and \( P \) represent the set of students studying Physics.
- Total number of students: \( n(M \cup P) = 100 \)
- Students studying Math: \( n(M) = 70 \)
- Students studying Physics: \( n(P) = 50 \)
a) Students in both (\( M \cap P \)):
Use the set theory formula:
\( n(M \cup P) = n(M) + n(P) - n(M \cap P) \)
\( 100 = 70 + 50 - n(M \cap P) \)
\( 100 = 120 - n(M \cap P) \)
\( n(M \cap P) = 20 \)b) Students in math only:
Subtract the students who study both from the total Math students:
\( 70 - 20 = 50 \)c) Students in physics only:
Subtract the students who study both from the total Physics students:
\( 50 - 20 = 30 \) -
In the figure below, BC is parallel to DE. Find the area of triangle ADE if the area of triangle ABC is equal to \( 30 \text{ cm}^2 \).
View Step-by-Step Solution
Because line \( BC \) is parallel to line \( DE \), triangle \( ABC \) and triangle \( ADE \) are similar triangles.
When two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding side lengths. Let the ratio of their corresponding sides be \( k = \dfrac{AD}{AB} = 9/3 = 3\).
The area formula relating the two triangles is:
\( \text{Area}(ADE) = \text{Area}(ABC) \times \left( \dfrac{AD}{AB} \right)^2 \)\( \text{Area}(ADE) = 30 \times 3^2 = 270 { cm}^2 \)
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One liter of water weighs 1 kilogram and 1 cubic decimeter (\( \text{dm}^3 \)) is equal to 1 liter. What is the weight of water contained in a right rectangular prism container with dimensions of 10 cm, 20 cm, and 200 cm?
View Step-by-Step Solution
Step 1: Find the volume in cubic centimeters (\( \text{cm}^3 \)).
Volume \( = \text{length} \times \text{width} \times \text{height} \)
\( V = 10 \text{ cm} \times 20 \text{ cm} \times 200 \text{ cm} = 40,000 \text{ cm}^3 \)Step 2: Convert cubic centimeters to cubic decimeters (liters).
We know that \( 1 \text{ dm} = 10 \text{ cm} \), so \( 1 \text{ dm}^3 = (10 \text{ cm})^3 = 1,000 \text{ cm}^3 \).
\( \text{Volume in liters} = \dfrac{40,000 \text{ cm}^3}{1,000 \text{ cm}^3/\text{L}} = 40 \text{ liters} \)Step 3: Calculate the weight.
Since 1 liter weighs 1 kg, the weight of the water is:
\( 40 \text{ liters} \times 1 \text{ kg/L} = 40 \text{ kg} \) -
Linda traveled at an average speed of 65 miles per hour going to a city and at an average speed of 50 miles per hour coming back using the same road. She drove a total of 6 hours away and back. What is the distance from Linda's house to the city she visited? (Round your answer to the nearest mile).
View Step-by-Step Solution
Let \( D \) be the one-way distance to the city.
Using the formula \( \text{Time} = \dfrac{\text{Distance}}{\text{Speed}} \):
Time going = \( \dfrac{D}{65} \)
Time returning = \( \dfrac{D}{50} \)The total driving time is 6 hours, so we set up the equation:
\( \dfrac{D}{65} + \dfrac{D}{50} = 6 \)Find a common denominator for 65 and 50. (The Least Common Multiple is 650).
Multiply the numerator and denominator of the first fraction by 10, and the second by 13:
\( \dfrac{10D}{650} + \dfrac{13D}{650} = 6 \)\( \dfrac{23D}{650} = 6 \)
Multiply both sides by 650:
\( 23D = 3900 \)Divide by 23:
\( D = \dfrac{3900}{23} \approx 169.565 \text{ miles} \)Rounded to the nearest mile, the distance is 170 miles.
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Which real number has its square equal to the sum of its half and its third?
View Step-by-Step Solution
Let the unknown real number be \( x \).
Translate the word problem into an algebraic equation:
\( x^2 = \dfrac{x}{2} + \dfrac{x}{3} \)Find a common denominator to add the fractions on the right side:
\( x^2 = \dfrac{3x}{6} + \dfrac{2x}{6} \)
\( x^2 = \dfrac{5x}{6} \)Bring all terms to one side to set the quadratic equation to zero:
\( x^2 - \dfrac{5x}{6} = 0 \)Factor out \( x \):
\( x\left(x - \dfrac{5}{6}\right) = 0 \)This gives two possible solutions:
\( x = 0 \) or \( x = \dfrac{5}{6} \)
Both 0 and 5/6 are real numbers that satisfy the condition. -
The ratio of boys to girls at a school is equal to 6:5. What is the ratio of boys to the total number of students?
View Step-by-Step Solution
The ratio of boys to girls is 6:5. This means that for every 6 boys, there are 5 girls.
The total number of "parts" in the ratio is the sum of the boy parts and the girl parts:
Total parts = \( 6 + 5 = 11 \)The ratio of boys to the total number of students is simply the boy parts divided by the total parts:
Ratio = 6:11 -
There are 200 balls in a container. The balls have one of three colors: red, blue, and green. The ratio of blue balls to red balls is 3:10, and the ratio of green to blue is 7:3. How many balls of each color are there?
View Step-by-Step Solution
Let's look at the given ratios:
- Blue : Red = 3 : 10
- Green : Blue = 7 : 3
Because the "Blue" portion is represented by the number 3 in both ratios, we can easily combine them into a single continuous ratio for Green : Blue : Red:
Green : Blue : Red = 7 : 3 : 10
Add up the total number of parts in this combined ratio:
Total parts = \( 7 + 3 + 10 = 20 \)Divide the total number of balls (200) by the total number of parts to find the value of one part:
Value of 1 part = \( 200 / 20 = 10 \text{ balls} \)Multiply the ratio numbers by the value of one part to find the quantity of each color:
- Green: \( 7 \times 10 = 70 \text{ balls} \)
- Blue: \( 3 \times 10 = 30 \text{ balls} \)
- Red: \( 10 \times 10 = 100 \text{ balls} \)