Another way of drawing the graph of y = mx + b
An object is thrown downwards with initial velocity v0
Brain Twister
y = 5x => y/x = 5
a. Point (3, 15) satisfies both equations: 15 = 3(5), 15/3 = 5, b. But why doesn’t (0, 0) satisfy y/x = 5 or y – 0 / x – 0 = 5?
Answer a. 15/3 = (15 – 0)/(3 – 0) = 5 The slope = rise/run.
Note y/x = 5 is the same as (y – 0)/(x – 0) = 5. There are two points, (0, 0) and (3, 15).
b. Note Since 0 = 0(3) and 0 = 0(5) you might conclude that 0/0 = 3 or 0/0 = 5.
From (0, 0) to (0, 0), (0 – 0)/(0 – 0), there is no run or rise. A line is defined by two points.
Area under a linear function
Perhaps you missed this.
The slope is obtained by dividing df(x) by dx, which is a function one degree lower than f(x).
The area is obtained by multiplying f(x) by dx, which is a function one degree higher than f(x).
Caution, things can get worse
Many years ago, in Venezuela I watch a TV program in which an economist was asked if the decline of the economy of a country could eventually stop. “No,” he replied, “when it hits bottom it starts digging.” In desperation, some countries have elected as president a person, who though has not been successful in life, promises to fully recover the economy. Do you
1. mx, the area of the gray rectangle, is the value of y. T/F 2. y = mx + b, where b represents _______
3. dA equals a. dy b. mdx 4. dx/dy = m T/F
Answer: 1. F, mx is the change in y 2. initial value of y 3. a. b. 4. F
The gray shaded area under the constant function m is the linear function mx.
m, the height of the rectangle, measures the rate of growth of the area
1. What is the unit of the area?
2. After the object has fallen t sec its velocity increases by ___
3. v(t) = gt + v0 T/F
4. dv = ______ is the increase in velocity in time dt.
5. _____ is the rate at which the area, the velocity, increases.
Answer 1. (m/sec2)(sec) = m/sec 2. gt 3. T 4. gdt 5. g
1. (1m/sec)(1sec) = ____ measures a unit of _______
2. vdt, the area of blue rectangle, measures _______
3. ds = (gt + v0)dt, ds/dt = _________
4. gt + v0 is the rate at which the area increases. T/F v0t is the area of the blue rectangle
5. The area of the white triangle is (1/2)(t)(___) .
v0t + (1/2)gt2 is the area of the blue rectangle plus the area of the white triangle.
6. v0t + (1/2)gt2is a. s, the position of the object b. s – s0, the change in position of the object.
s – s0 = v0t + (1/2)gt2 is the area under the linear function gt + v0, with base t
The slope of y = axn is dy/dx = anxn – 1. The area under y = ann is A = ( 𝑎
𝑛+1)xn+1+ c Answer 1. 1m, area
2. change in position 3. gt + v0 4. T 5. gt 6. b
think a person who owns a small shop could successfully manage a company which has a large number of supermarkets?
Some countries have tried it and the results have been a complete failure. No coach would randomly pick a fan from the stands to replace an injured player in the field. He may talk very nicely but can he play?
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Using your imagination
At points A and B of the line segment AB draw two intersecting circles with the same radii.
Q1 What property do all the points on the circle with center A have in common?
Q2 What property do the points where the circles intersect have in common?
Draw a line through the two intersecting points of the two circles.
Q3 What property do all the points on this line have in common?
Congruency SSS. If the three sides of two triangles are equal, SSS, they overlap, they are congruent.
Constructing a perpendicular bisector of a line segment AB
if <AOC = 1800, AB is a line segment. Now we only have one point, C. to draw the bisector of the 1800 angle, we need to find another point D. All the points on line CD are equidistant from A and B, and also bisects <ABD = 1800.
Construct an angle equal to a given <BAC ‒ another use of SSS
Congruency of triangles SAS
Congruency of two right angled triangles RHS ‒ right angle, hypothenuse, side.
If the triangle has a 900 angle and the hypothenuse and a leg are known, the other leg can be calculated using the formula a2 + b2 = c2.
Two triangles that overlap are said to be congruent; the sides and angles are equal. What is the minimum requirement for two triangles to be congruent? For example, two triangles may have the same angles, but the sides be different., so they are not congruent. However, if the angles and one side are equal, they are congruent; AAAS. If two angles are equal, say 600 and 400, the third angle 800 is also equal, so instead of AAAS we use AAS.
An interesting application of this congruency is illustrated when an angle is bisected. To bisect
<AOB with a compass, 1st draw an arc intersecting the sides at A and B.
Q1 How do we find C with the compass?
Q2 ΔOAC ≅ ΔOBC by SSS. T/F
We conclude that <COB = <COA, <AOB is bisected.
Q3 If <AOB = 1800, OC is perpendicular to AB and ΔABC is an _____ triangle.
Answer Q1 with the same radius, draw from A and B intersecting arcs at C Q2 T Q3 isosceles
By drawing an arc, the segments AB = AC define two sides of ΔABC. The third side is BC. By following this procedure, we can construct another congruent triangle with sides AB, AC and BC. Hence <BAC is duplicated.
To show that these two triangles are congruent, place line segment DE over AB. Since <A = <D, DE falls along AC. Since DF = AC, F falls on C.
Line FE overlaps on CB.
The triangles are congruent by SAS.
B
A C
a c
b
Distance from point B to a line BC.
From B we can draw lines BA and line BC perpendicular to AC.
We can show that BC is shorter than BA regardless of the position of A
Since c2 = b2 + a2, a2 = c2 ‒ b2. The smaller b is, the closer c is to a. When b = 0, c = a.
a is the smallest value of c. The shortest distance between C and AC is a, the perpendicular.
Lesson 62 Congruent
triangles
Theorem If a point C is equidistant from A and B (given), the point is on the right bisector of AB (prove).
We had previously shown that if a point is on the right bisector of AB, it is equidistant from A and B. Then we proved that if p is true then p is true, now we want to prove that if q is true, then p is true.
C
A M B
Brain Teaser
Q1 How many circles can be drawn through a given point P with a compass?
Q2 How many points on OP, the perpendicular bisector of AB, are equidistant from A and B?
Q3 Only two right bisectors are needed to find O, the circumcenter of a triangle. T/F Q4 Where is O located when the triangle is a) acute b) obtuse c) right angled triangle?
Q5 With a compass, draw three circles that pass-through O.
Q6 How many circles can be drawn through A and B Q7 Which is the shortest distance between O and AB?
Unit circle
Q1 In the theorem, what is given?
Q2 We have to prove that a line from C to the midpoint of AB is perpendicular to AB. T/F Q3 Why is ΔAMC ≅ ΔBMC?
Q4 Since <AMC = <BMC, <AMC = <BMC = 900. T/F
Answer Q1 a point C so that CA = CB Q2 T Q3 SSS Q4 <AMC + <BNC = 1800 The right bisector contains all the points equidistant from A and B.
Given <CBA, let’s bisect it, that is divide it in half, with a compass.
1st draw an arc EF. Next with a certain aperture of the compass and centers at E and F draw arcs intersecting at D.
Q1 How many sides of ΔFBD and ΔEBD are equal?
Q2 ΔFBD ≅ ΔEBD by SSS. T/F Answer Q1 3 Q2 T, <FBD = <EBD
If D is on the bisector of <ABC, it can be shown that D is at the same distance from the sides.
Q1 What is given?
Q2 What must we prove?
Next, I drew CD and DA perpendicular to BC and BA.
Q3 ΔABD ≅ ΔCBD AAS. T/F
Next suppose DC = DA. Can we prove that D is on the bisector of <ABC?
Q4 What do ΔABD and ΔCBD have in common?
Q5 ΔABD ≅ ΔCBD RHS. T/F
Answer Q1 D is on the bisector Q2 DC = DA Q3 T Q4 right angle and two sides Bisector of an angle and its properties
The angle α, at the center, is measured by an arc α because the circle has radius 1.
Q1 The arc length corresponding to 3600 is ______ .
The angle at P, which is on the circumference, is half the arc α.
Q2 The angles of the isosceles triangle are (1/2)α, (1/2) α and 1800 ‒ α. T/F Q3 When α goes from ‒ π/2 to π/2, the angle at P is a right angle. T/F Answer Q1 2π Q2 T Q3 T, π/2
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Q1 Since CP is a bisector, angle ____ equals angle _____ . Q2 If a point is on the bisector of an angle, it is equidistant from its Q3 According to the bisector of <A, Ic is equidistant from AC and _____ . Q4 According to bisector of <C, Ic is equidistant from ____ and _____ . Q5 Why is Ic on the bisector of angle <ABC?
Q6 A circle with center at Ic, called the incenter of the circle, touches the three sides. T/F Q7 When a circle touches a line, its radius is perpendicular to the line. T/F
Answer Q1 1 = 2 Q2 sides Q3 AB Q4 Ic is equidistant from CA and CB. Q5 Ic is equidistant from BC and BA it is on the bisector of <CBA Q6 T, it is at the same distance from the three sides. Q7 T
The Incenter is always inside the triangle, which is not true of the orthocenter.