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ALL OF GRADE 10 MATH IN ONLY 1 HOUR!!! | jensenmath.ca

JensenMath・2 minutes read

The video teaches grade 10 math in one hour with 22 practice questions, covering linear systems, triangles, circles, quadratics, factoring, and parabolas. It also includes applications, such as finding distances, heights, and angles using trigonometry and formulas.

Insights

The video compresses learning grade 10 math into one hour with 22 practice questions, aiming for efficiency and practical application.
Various methods like graphing, substitution, and elimination are demonstrated to solve linear systems, emphasizing a comprehensive understanding of mathematical problem-solving techniques.
Quadratic functions are explored in different forms, each serving distinct purposes such as identifying intercepts, vertex coordinates, and transformations, highlighting the versatility and complexity of mathematical functions.

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How can I learn grade 10 math quickly?
By watching a video with 22 practice questions.

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Summary

00:00

"Grade 10 Math: Learn in One Hour"

The Ministry of Education suggests 110 hours to learn grade 10 math, but the video aims to teach it in one hour with 22 practice questions.
The video serves as a review or introduction to grade-level math questions.
Example one demonstrates solving a linear system through graphing, substitution, and elimination methods.
Graphing involves isolating y in equations to plot and find the point of intersection.
Elimination involves adding equations to eliminate variables and find the solution.
Substitution requires isolating variables and substituting them into equations to find the solution.
Example two showcases a more complex linear system solved using elimination by adjusting coefficients to eliminate variables.
Example three presents an application problem involving selling Adidas and Air Jensen shoes, solved through elimination of variables.
The midpoint and distance between two points are calculated using the average of coordinates and the distance formula derived from the Pythagorean theorem.
The midpoint is found by averaging x and y coordinates, while the distance is determined by calculating the square root of the sum of squared differences in coordinates.

13:20

Geometry Problems: Triangles, Circles, and Distances

Example 5: Drawing a triangle with specified vertices and finding the equation of the median from C to AB.
The median is a line segment joining a vertex to the midpoint of the opposite side.
Calculating the midpoint of AB and the slope of the median.
Finding the y-intercept of the median and writing its equation as y = -x + 4.
Example 6: Determining the equation for the right bisector of the line segment with endpoints A and B.
The right bisector is perpendicular to the line connecting A and B.
Calculating the slope of AB and determining the perpendicular slope for the right bisector.
Finding the y-intercept of the right bisector and writing its equation as y = -x + 4.
Example 7: Classifying a triangle with given vertices as scalene, isosceles, or equilateral and checking for a right angle.
Using the distance formula to find the lengths of the triangle's sides.
Confirming it is a scalene triangle and determining it does not have a right angle.
Example 8: Calculating the radius of a circle with an equation of x^2 + y^2 = 36.
The radius is found to be 6 units.
Example 9: Finding the equation of a circle centered at the origin and passing through the point (-3, 4).
The equation is x^2 + y^2 = 25, and checking if point (5, 2) lies outside the circle.
Example 10: Determining the shortest distance from point (-3, 5) to the line y = 1/4x + 10.
Finding the equation of the line perpendicular to y = 1/4x + 10 passing through (-3, 5).
Solving for the intersection point of the two lines to find the shortest distance.

27:11

Understanding Quadratic Functions and Their Forms

The point of intersection is at (-4, 9).
The shortest distance is between (-3, 5) and (-4, 9), calculated using the distance formula.
Quadratic functions can be in standard, vertex, or factored form, each serving different purposes.
Standard form helps determine the y-intercept and the direction of opening based on the a value.
Vertex form is useful for identifying the coordinates of the vertex.
Factored form aids in finding the x-intercepts.
Transitioning between the three forms involves understanding the roles of a, h, and k.
The vertex form equation for a given quadratic is derived by substituting the vertex coordinates and solving for a.
Transformations of a quadratic can be described in terms of shifts, stretches or compressions, and reflections.
Factored form allows for easy identification of x-intercepts, vertex, and axis of symmetry, aiding in graphing the quadratic function.

41:10

Factoring Quadratic Equations for Solving Solutions

When the leading coefficient is not 1, the middle term of the equation needs to be split into two parts that satisfy the product and sum.
By rewriting the equation with the middle term split, common factors are taken out from the first two terms and the last two terms separately.
After factoring by grouping, a common binomial is identified, allowing for further factoring.
In cases where the leading coefficient is not 1, the equation needs to be factored using decomposition and grouping.
The product and sum of the numbers satisfying the equation are crucial in splitting the middle term for further factoring.
Special products like a difference of squares or perfect square trinomials have specific factoring methods.
To expand equations into standard form, the double distributive property is used, multiplying all terms and collecting like terms.
Converting standard form quadratic equations into vertex form involves specific steps like common factoring and adding/subtracting a special number inside the brackets.
Solving quadratic equations involves factoring them into standard form, setting each factor to zero, and solving for the variables.
In cases where equations are not factorable, the quadratic formula is used to find the solutions, considering the discriminant for the number of answers.

55:16

Parabola Vertex and Projectile Motion Analysis

X-intercepts are at -6 and -2.
The vertex is halfway between the x-intercepts due to parabolas' symmetry.
The x-coordinate of the vertex is -4, found by averaging the x-intercepts.
The y-coordinate of the vertex is also -4, determined by plugging -4 into the original equation.
The vertex is at (-4, -4).
Parabolas are symmetrical, so a point 4 units to the right of the vertex is mirrored 4 units to the left.
The object launched upward at 64 ft/s from an 80 ft platform.
The object lands on the ground after 5 seconds.
The max height of the object is 144 feet at 2 seconds.
The object is 100 feet off the ground at 0.33 and 3.66 seconds.

01:10:10

Trigonometry: Values and Laws Explained

The approximate value of cos 54 is 0.5878.
The value of s squared is approximately 66.71006468.
The square root of s squared is approximately 8.17 kilometers.
For right angle triangles, use SOHCAHTOA; for oblique triangles, use sine law and cosine law.

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