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RS Aggarwal Solutions: Coordinate Geometry- 1
# RS Aggarwal Solutions: Coordinate Geometry- 1 - RS Aggarwal Solutions for Class 10 Mathematics
## Exercise: 16a
Q.1. Find the distance between the points:
A(9, 3) and B(15, 11)
In this question, we have to use the distance formula to find the distance between two points which is given by, say for points P(x1,x2) and Q(y1,y2) then
PQ = √(x2 – x1)2 + (y2 – y1)2
AB = √{(15 – 9)2 + (11 – 3)2}
= √{(6)2 + (8)2}
= √{36 + 64}
= √100
∴ AB = 10 units.
Q.2. Find the distance between the points:
A(7, – 4) and B(– 5, 1)
In this question, we have to use the distance formula to find the distance between two points which is given by, say for points P(x1,x2) and Q(y1,y2) then
PQ = √(x2 – x1)2 + (y2 – y1)2
AB = √{(– 5 – 7)2 + (1 – (– 4))2}
= √{(– 12)2 + (5)2}
= √{144 + 25}
= √169
∴ AB = 13 units
Q.3. Find the distance between the points:
A(– 6, – 4) and B(9, – 12)
In this question, we have to use the distance formula to find the distance between two points which is given by, say for points P(x1,x2) and Q(y1,y2) then
PQ = √(x2 – x1)2 + (y2 – y1)2
AB = √{(9 – (– 6))2 + (– 12 – (– 4))2}
= √{(15)2 + (– 8)2}
= √{225 + 64}
= √289
∴ AB = 17 units
Q.4. Find the distance between the points:
A(1, – 3) and B(4, – 6)
In this question, we have to use the distance formula to find the distance between two points which is given by, say for points P(x1,x2) and Q(y1,y2) then
PQ = √(x2 – x1)2 + (y2 – y1)2
AB = √{(4 – 1)2 + (– 6 – (– 3))2}
AB = √{(4 – 1)2 + (– 6 – (– 3))2}
= √{9 + 9}
= √18
∴ AB = 3√2 units
Q.5. Find the distance between the points:
P(a + b , a – b) and Q(a – b, a + b)
In this question, we have to use the distance formula to find the distance between two points which is given by, say for points P(x1,x2) and Q(y1,y2) then
PQ = √(x2 – x1)2 + (y2 – y1)2
AB = √{((a – b) – (a + b))2 + ((a + b) – (a – b))2}
= √{(– 2b)+ (2b)2}
= √{4b+ 4b2}
= √8b2
∴ AB = 2√2b units
Q.6. Find the distance between the points:
P(a sin a, acos a) and Q(a cos a, – a sin a)
In this question, we have to use the distance formula to find the distance between two points which is given by, say for points P(x1,x2) and Q(y1,y2) then
PQ = √(x2 – x1)2 + (y2 – y1)2
PQ = √{(a cos a – a sin a)2 – (– a sin a – a cos a)2}
= √{(a2 cos2 a + a2 sin2 a – 2a2 sina.cosa + a2 cos2 a + a2 sin2 a + 2a2 sina.cosa }
= √{ a2 (cos2 a + sin2 a) + (a2 (cos2 a + sin2 a))}
= √a2(1) + a2 (1)
= √a2(1 + 1)
∴ PQ = a√2 units
Q.7. Find the distance of each of the following points from the origin:
A(5, – 12)
Since it is given that the distance is to be found from origin so in this question we have to use the distance formula keeping one – point fix i.e. O (0,0), as shown below:
OA = √{(5 – 0)2 + (– 12 – 0)2}
= √{(5)2 + (– 12)2}
= √{25 + 144}
= √169
∴ OA = 13 units
Q.8. Find the distance of each of the following points from the origin:
B(– 5, 5)
Since it is given that the distance is to be found from origin so in this question we have to use the distance formula keeping one – point fix i.e. O (0,0), as shown below:
OB = √{(– 5 – 0)2 + (5 – 0)2}
= √{(– 5)2 + (5)2}
= √{25 + 25}
= √50
∴ OB = 5√2 units
Q.9. Find the distance of each of the following points from the origin:
C(– 4, – 6).
Since it is given that the distance is to be found from origin so in this question we have to use the distance formula keeping one – point fix i.e. O (0,0), as shown below:
OC = √{(– 4 – 0)2 + (– 6 – 0)2}
= √{(– 4)2 + (– 6)2}
= √{16 + 36}
∴ OC = √52 units
Q.10. Find all possible values of × for which the distance between the points A(x, – 1) and B(5, 3) is 5 units.
Given:
Distance AB = 5 units
By distance formula, as shown below:
AB = √{(5 – x)2 + (3 – (– 1))2}
5 = √{(5 – x)2 + (4)2}
5 = √{25 + x2 – 10x + 16}
5 = √{41 + x2 – 10x}
Squaring both sides we get
25 = 41 + x2 – 10x
⇒ 16 + x2 – 10x = 0
⇒ (x – 8)(x – 2) = 0
⇒ × = 8 or × = 2
∴ The values of × can be 8 or 2
Q.11. Find all possible values of y for which the distance between the points A(2, – 3) and B(10, y) is 10 units.
Given, the distance AB = 10 units
By distance formula, as shown below:
AB = √{(10 – 2)2 + (y – (– 3))2}
10 = √{(8)2 + (y + 3)2}
10 = √{64 + y2 + 6y + 9}
10 = √{73 + y2 + 6y}
Squaring both sides we get
100 = 73 + y2 + 6y
On solving the equation, 100 = 73 + y2 + 6y
⇒ 27 + y2 + 6y = 0
⇒ y2 + 6y + 27 = 0
⇒ (y – 3)(y + 9) = 0
⇒ y = 3 or y = – 9
∴ The values of y can be 3 or – 9
Q.12. Find the values of x for which the distance between the points P(x, 4) and Q(9, 10) is 10 units.
Given the distance PQ = 10 units
By distance formula, as shown below:
PQ = √{(9 – x)2 + (10 – 4)2}
10 = √{(9 – x)2 + (6)2}
10 = √{81 + x2 – 18x + 36}
10 = √{117 + x2 – 18x}
Squaring both sides we get
⇒ 100 = 117 + x2 – 18x
⇒ x2 – 18x + 17x = 0
⇒ (x – 1)(x – 17)
⇒ × = 1 or × = 17
Q.13. If the point A(x, 2) is equidistant from the points B(8, – 2) and C(2, – 2), find the value of x. Also, find the length of AB.
Given that point A is equidistant from points B and C , so AB = AC
By distance formula, as shown below:
AB = √{(8 – x)2 + (– 2 – 2)2}
= √{(8 – x)2 + (– 4)2}
= √{64 + x2 – 16x + 16}
= √{80 + x2 – 16x}
AC = √{(2 – x)+ (– 2 – 2)2}
= √{(2 – x)2 + (4)2}
= √{4 + x2 – 4x + 16}
= √{20 + x2 – 4x}
Now, AB = AC
Squaring both sides, we get,
(80 + x2 – 16x) = (20 + x2 – 4x)
60 = 12x
x = 5
⇒ AB = √{80 + x2 – 16x}
⇒ AB = √(80 + 52 – 16× 5)
= 5 units
Q.14. If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find the value of p. Also, find the length of AB.
Given that point A is equidistant from points B and C, so AB = AC
By distance formula, as shown below:
AB = √{(3 – 0)2 + (p – 2)2}
= √{(3)2 + (p – 2)2}
= √{9 + p2 – 4p + 4}
⇒ AB = √{13 + p2 – 4p}
AC = √{(p – 0)2 + (5 – 2)2}
= √{(p)2 + (3)2}
⇒ AB = √{9 + p2}
Now, AB = AC
Squaring both sides, we get,
(13 + p2 – 4p) = (9 + p2)
⇒ 4 = 4p
⇒ p = 1
Now, AB = √{13 + p2 – 4p}
⇒ AB = √(13 + 1 – 4)
= √10 units
Therefore, the distance of AB = √10 units.
Q.15. Find the point on the x – axis which is equidistant from the points (2, – 5) and (– 2, 9).
Let the point be X(x,0) and the other two points are given as A(2, – 5) and B(– 2,9)
Given XA = XB
By distance formula, as shown below:
XA = √{(2 – x)2 + (– 5 – 0)2}
= √{(2 – x)+ (– 5)2}
= √{4 + x2 – 4x + 25}
⇒ XA = √{29 + x2 – 4x}
XB = √{(– 2 – x)2 + (9 – 0)2}
= √{(– 2 – x)2 + (9)2}
= √{4 + x2 + 4x + 81}
⇒ XB = √{85 + x2 + 4x}
Now since
XA = XB
Squaring both sides, we get,
(29 + x2 – 4x) = (85 + x2 + 4x)
56 = – 8x
x = – 7
The point on × axis is (– 7, 0).
Q.16. Find points on the x – axis, each of which is at a distance of 10 units from the point A(11, – 8).
Let the point be X(x, 0)
XA = 10
By distance formula, as shown below:
XA = √{(11 – x)2 + (– 8 – 0)2}
10 = √{(11 – x)2 + (– 8)2}
10 = √{121 + x2 – 22x + 64}
10 = √{185 + x2 – 22x}
Squaring both sides we get
100 = (185 + x2 – 22x)
⇒ 85 + x2 – 22x = 0
⇒ x– 22x + 85 = 0
⇒ (x – 5)(x – 17)
⇒ × = 5 or × = 17
The points are (5, 0) and (17, 0).
Q.17. Find the point on the y – axis which is equidistant from the points A(6, 5) and B(– 4, 3).
Let the point be Y(0,y) and the other two points given as A(6,5) and B(– 4,3)
Given YA = YB
By distance formula, as shown below:
YA = √{(6 – 0)2 + (5 – y)2}
= √{(6)2 + (5 – y)2}
= √{36 + 25 + y2 – 10y}
⇒ YA = √{61 + y2 – 10y}
YB = √{(– 4 – 0)2 + (3 – y)2}
= √{(– 4)+ (9 + y– 6y)}
= √{16 + 9 + y2 – 6y}
⇒ YB = √{25 + y2 – 6y}
Now, YA = YB
Squaring both sides, we get,
(61 + y2 – 10y) = (25 + y2 – 6y)
36 = 4y
⇒ y = 9
The point is (0, 9).
Q.18. If the point P(x, y) is equidistant from the points A(5, 1) and B(– 1, 5), prove that 3x = 2y.
The point P(x, y) is equidistant from the points A(5, 1) and B(– 1, 5), means PA = PB
By distance formula, as shown below:
PA = √{(5 – x)2 + (1 – y)2}
= √{(25 + x2 – 10x) + (1 + y2 – 2y)}
⇒ PA = √{26 + x2 – 10x + y2 – 2y}
PB = √{(– 1 – x)2 + (5 – y)2}
= √{(1 + x2 + 2x + 25 + y2 – 10y)}
⇒ PB = √{(26 + x2 + 2x + y2 – 10y)}
Now, PA = PB
Squaring both sides, we get
26 + x2 – 10x + y2 – 2y = 26 + x2 + 2x + y2 – 10y
⇒ 12x = 8y
⇒ 3x = 2y
Hence proved.
Q.19. If P(x, y) is a point equidistant from the points A(6, – 1) and B(2, 3), show that × – y = 3.
By distance formula, as shown below:
PA = √{(6 – x)2 + (– 1 – y)2}
= √{(36 + x2 –12x) + (1 + y2 + 2y)}
⇒ PA = √{37 + x2 – 12x + y2 + 2y}
PB = √{(2 – x)2 + (3 – y)2}
= √{(4 + x2 – 4x + 9 + y2 – 6y)}
⇒ PB = √{(13 + x2 – 4x + y2 – 6y)}
Given: PA = PB
Squaring both sides, we get
(37 + x2 – 12x + y2 + 2y) = (13 + x2 – 4x + y2 – 6y)
24 = 8x – 8y
Dividing by 8
x – y = 3
Hence proved.
Q.20. Find the coordinates of the point equidistant from three given points A(5, 3), B(5, – 5) and C(1, – 5).
Let the point be P(x, y), then since all three points are equidistant therefore
PA = PB = PC
By distance formula, as shown below:
We have, PA = √{(5 – x)2 + (3 – y)2}
= √{25 + x2 – 10x + 9 + y2 – 6y}
⇒ PA = √{34 + x2 – 10x + y2 – 6y}
PB = √{(5 – x)2 + (– 5 – y)2}
= √{25 + x2 – 10x + 25 + y2 + 10y}
⇒ PB = √{50 + x2 – 10x + y2 + 10y}
PC = √{(1 – x)2 + (– 5 – y)2}
= √{1 + x2 – 2x + 25 + y2 + 10y}
⇒ PC = √{26 + x2 – 2x + y2 + 10y}
Squaring PA and PB we get
{34 + x2 – 10x + y2 – 6y} = {50 + x2 – 10x + y2 + 10y}
⇒ – 16 = 16y
⇒ y = – 1
Squaring PB and PC we get
{50 + x2 – 2x + y+ 10y} = {26 + x2 – 10x + y2 + 10y}
24 = – 8x
x = – 3
P(– 3, – 1)
Q.21. If the points A(4, 3) and B(x, 5) lie on a circle with the centre O(2, 3), find the value of x.
OA = √{(4 – 2)2 + (3 – 3)2}
= √4
= 2
OB = √{(x – 2)2 + 4 }
= √{x2 + 4 – 4x + 4}
√{ 8 + x2 – 4x}
OA= OB2
4 = 8 + x2 – 4x
⇒ x2 – 4x + 4 = 0
⇒ x2 – 2x – 2x + 4 = 0
⇒ x(x– 2) – 2(x – 2) = 0
⇒ (x – 2) (x – 2) = 0
x = 2
Q.22. If the point C(– 2, 3) is equidistant from the points A(3, – 1) and B(x, 8), find the values of x. Also, find the distance BC.
By distance formula
AC = √{(3 – (– 2))2 + (– 1 – 3)2}
= √{(5)2 + (– 4)2}
= √{25 + 16}
= √{41}
BC = √{(x –(– 2))2 + (8 – 3)2}
= √{(x + 2)2 + 52 }
= √{x2 + 4 + 2x + 25}
= √{x+ 2x + 29}
AB = BC
√{x2 + 2x + 29} = √{41}
× = 2 or × = – 6
Since, AB = BC
BC = √41 units
Q.23. If the point P(2, 2) is equidistant from the points A(– 2, k) and B(– 2k, – 3), find k. Also, find the length of AP.
AP = BP
AP = √{(– 2 – 2)2 + (k – 2)2}
= √{16 + k2 – 4k + 4}
= √(k2 – 2k + 20)
BP = √{(– 2k – 2)+ (– 3 – 2)2}
= √{4k2 + 8k + 4 + 25}
= √(4k2 + 8k + 29)
Squaring AP and BP and equating them we get
k2 – 4k + 20 = 4k2 + 8k + 29
3k2 + 12k + 9 = 0
(k + 3)(k + 1) = 0
⇒ k = – 3
⇒ AP = √41units
Or k = – 1
⇒ AP = 5 units
Q.24. If the point (x, y) is equidistant from the points (a + b, b – a) and (a – b, a + b), prove that bx = ay.
Let point P(x,y) , A(a + b,a – b) , B(a – b,a + b)
Then AP = BP
AP = √{((a + b) – x)2 + ((a – b) – y)2}
= √{(a + b)2 + x2 – 2(a + b)x + (a – b)2 + y– 2(a – b)y}
= √(a2 + b2 + 2ab + x2 – 2(a + b)x + b+ a2 – 2ab + y2 – 2(a – b)y)
BP = √{((a – b) – x)+ ((a + b) – y)2}
= √{(a – b)2 + x2 – 2(a – b)x + (a + b)+ y2 – 2(a + b)y}
= √(a2 + b2 – 2ab + x2 – 2(a – b)x + b2 + a2 + 2ab + y2 – 2(a + b)y)
Squaring and Equating both we get
a2 + b2 + 2ab + x2 – 2(a + b)x + b2 + a2 – 2ab + y2 – 2(a – b)y = a2 + b– 2ab + x– 2(a – b)x + b2 + a2 + 2ab + y2 – 2(a + b)y
– 2(a + b)x – 2(a – b)y = – 2(a – b)x – 2(a + b)y
ax + bx + ay – by = ax – bx + ay + by
Hence
bx = ay
Q.25. Using the distance formula, show that the given points are collinear:
(i) (1, – 1), (5, 2) and (9, 5)
(ii) (6, 9), (0, 1) and (– 6, – 7)
(iii) (– 1, – 1), (2, 3) and (8, 11)
(iv) (– 2, 5), (0, 1) and (2, – 3).
Three or more points are collinear, if slope of any two pairs of points is same. With three points A, B and C if Slope of AB = slope of BC = slope of AC
then A, B and C are collinear points.
Slope of any two points is given by:
(y2 – y1)/(x2 – x1).
(i) Slope of AB = (2 – (– 1))/(5 – 1) = 3/4
Slope of BC = (5 – 2)/(9 – 5) = 3/4
Slope of AB = slope of BC
Hence collinear.
(ii) Slope of AB = (1 – 9)/(0 – 6) = 8/6 = 4/3
Slope of BC = (– 6 – 0)/(– 7 – 1) = 6/6 = 1
Slope of AC = (– 7 – 9)/(– 6 – 6) = – 16/ – 12 = 4/3
Slope of AB = slope of AC
Hence collinear.
(iii) Slope of AB = ((3 – (– 1))/((2 – (– 1)) = 4/3
Slope of BC = (11 – 2)/(8 – 3) = 9/5 = 1
Slope of AC = ((11 – (– 1))/((8 – (– 1)) = 12/9 = 4/3
Slope of AB = slope of AC
Hence collinear.
(iv) Slope of AB = (1 – 5)/((0 – (– 2)) = – 4/2 = – 2
Slope of BC = (– 3 – 1)/(2 – 0) = – 4/2 = – 2
Slope of AB = slope of AB
Hence collinear.
Q.26. Show that the points A(7, 10), B(– 2, 5) and C(3, – 4) are the vertices of an isosceles right triangle.
In an isosceles triangle any two sides are equal.
AB = √{(– 2 – 7)2 + (5 – 10)2}
= √{(– 9)2 + (– 5)2}
= √{81 + 25}
= √{106}
BC = √{(– 4 – 5)2 + (3 – (– 2))2}
= √{(– 9)2 + (5)2}
= √{81 + 25}
= √{106}
AB = BC
∴ It is an isosceles triangle.
Q.27. Show that the points A(3, 0), B(6, 4) and C(– 1, 3) are the vertices of an isosceles right triangle.
In an isosceles triangle any two sides are equal.
AB = √{(6 – 3)2 + (4 – 0)2}
= √{(3)2 + (4)2}
= √{9 + 16}
= √{25} = 5 units
BC = √{(– 1 – 6)2 + (3 – 4)2}
= √{(– 7)2 + (– 1)2}
= √{49 + 1}
= √{50}
AC = √{(– 1 – 3)2 + (3 – 0)2}
= √{(– 4)2 + (3)2}
= √{16 + 9}
= √{25} = 5 units
AB = AC
∴ It is an isosceles triangle.
Q.28. If A(5, 2), B(2, – 2) and C(– 2, t) are the vertices of a right triangle with ∠B = 90°, then find the value of t.
A(5, 2), B(2, – 2) and C(– 2, t) are the vertices of a right triangle with ∠B = 90°
The value of t.
From the fig we have ∠B = 90°,
so by Pythagoras theorem we have AC2 = AB2 + BC2
AC= (– 2 – 5)2 + (t – 2)2
= (– 7)2 + t2 + 4 – 2t
= 49 +t+ 4 - 2t
= 53 + t2 – 2t
AB2 = (2 – 5)+ (– 2 – 2)2
=(-3)2 + (–4)2
= 9 + 16
= 25
BC2 = (– 2 – 2)2 + (t + 2)2
= (– 4)2 + (t + 2)2
= 16 + t2 + 4 + 2t
= 20 + t2 + 2t
AB+ BC2 = 25 + 20 + t2 + 2t
= 45 + t2 + 2t
AC2 = 53 + t– 2t
⇒ 53 + t2 – 2t = 45 + t2 + 2t
⇒ 53 - 45 = 4t
⇒ 8 = 4t
⇒ t = 2
Q.29. Prove that the points A(2, 4), B(2, 6) and C(2 + √3, 5) are the vertices of an equilateral triangle.
For an equilateral triangle
AB = BC = AC
AB = √{(6 – 4)2 + (2 – 2)2}
= √{(2)2 + 0}
= √{4 + 0}
= √{4} = 2 units
BC = √{(2 + √3 – 2)+ (5 – 6)2}
= √{3 + (– 1)2}
= √{4} = 2 units
AC = √{(2 + √3 – 2)2 + (5 – 4)2}
= √{3 + (– 1)2}
= √{4} = 2 units
Hence , AB = BC = AC
∴ ABC is an equilateral triangle.
Q.30. Show that the points (– 3, – 3), (3, 3) and (– 3√3, 3√3) are the vertices of an equilateral triangle.
Let the points be 3 (–3, –3), B (3, 3) and C (–3√3, 3√3)
Then, AB = √(3 + 3)2+( 3 + 3)2
=√(-6)2+(6)2
= √36+36
= √72
= 3√8
BC=√(-3√3+3)2+(3√3-3)2
= √(1-√3)232+(√3+1)232
= 3√[ 1+3-2√3+3+1+2√3]
= 3√8
CA = √(-3√3-3)2+(3√3-3)2
= √(-√3-1)232+(√3-1)232
= 3√[3+1+2√3+3+1-2√3]
= 3√8
∵ AB = BC = CA
⇒ A, B, C are the vertices of an equilateral triangle.
Q.31. Show that the points A(– 5, 6), B(3, 0) and C(9, 8) are the vertices of an isosceles right – angled triangle. Calculate its area.
AB = √{(0 – 6)2 + (3 – (– 5))2}
= √{(– 6)2 + (8)2}
= √{36 + 64}
= √{100} = 10 units
BC = √{(9 – 3)2 + (8 – 0)2}
= √{(6)2 + (8)2}
= √{36 + 64}
= √{100} = 10 units
AC = √{(9 – (– 5))2 + (8 – 6)2}
= √{(14)2 + (2)2}
= √{196 + 4}
= √{200}
For the right angled triangle
AC2 = AB+ BC2
AC2 = 200
AB2 + AC= 100 + 100 = 200
Since AB = BC
∴ ABC is an isosceles triangle.
Area = 1/2 (AB) (BC)
= 1/2 (10) (10)
= 1/2 (100)
= 50 sq units
Q.32. Show that the points 0(0, 0), A(3, √3) and B(3, – √3) are the vertices of an equilateral triangle. Find the area of this triangle.
OA = √{(√3)2 + (3 – 0)2}
= √{(3) + (3)2}
= √{3 + 9}
= √{12}
AB = √{(– √3 – √3)2 + (3 – 3)2}
= √{ – 2√3)2}
= √{12}
OB = √{(3 – 0)2 + (– √3 – 0)2}
= √{9 + 3}
= √{12}
Since OA = AB = OB , ∴ equilateral triangle.
Area = 1/2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= 1/2[ – 3√3 – 3√3 ]
= – 3√3 sq units
Q.33. Show that the following points are the vertices of a square:
A(3, 2), B(0, 5), C(– 3, 2) and D(0, – 1)
AB = √{(0 – 3)2 + (5 – 2)2} = √{9 + 9} = √18 units
BC = √{(– 3 – 0)+ (2 – 5)2} = √{9 + 9} = √18 units
CD = √{(0 – (– 3))2 + (– 1 – 2)2} = √{9 + 9} = √18 units
DA = √{(0 – 3)2 + (– 1 – 2)2} = √{9 + 9} = √18 units
AC = √{(– 3 – 3)2} = √36 = 6 units
BD = √{(– 1 – 5)2} = √36 = 6 units
Since AB = BC = CD = DA and AC = BD
∴ ABCD is a square.
Q.34. Show that the following points are the vertices of a square:
A(6, 2), B(2, 1), C(1, 5) and D(5, 6)
AB = √{(2 – 6)2 + (1 – 2)2} = √{16 + 1} = √17 units
BC = √{(1 – 2)2 + (5 – 1)2} = √{1 + 16} = √17 units
CD = √{(5 – 1)2 + (6 – 5)2} = √{16 + 1} = √17 units
DA = √{(5 – 6)2 + (6 – 2)2} = √{16 + 1} = √17 units
AC = √{(1 – 6)2 + (5 – 2)2} = √{25 + 9} = √34 units
BD = √{(5 – 2)2 + (6 – 1)2} = √{25 + 9} = √34units
Since AB = BC = CD = DA and AC = BD
∴ ABCD is a square.
Q.35. Show that the following points are the vertices of a square:
A(0, – 2), B(3, 1), C(0, 4) and D(– 3, 1)
AB = √{(3 – 0)2 + (1 – (– 2))2} = √{9 + 9} = √18 units
BC = √{(0 – 3)2 + (4 – 1)2} = √{9 + 9} = √18 units
CD = √{(– 3 – 0)+ (1 – 4)2} = √{9 + 9} = √18 units
DA = √{(– 3 – 0)2 + (1 – (– 2))2} = √{9 + 9} = √18 units
AC = √{ (4 – (– 2))2} = √{36} = 6 units
BD = √{(– 3 – 3)2 + (1 – 1)2} = √{36} = 6units
Since AB = BC = CD = DA and AC = BD
∴ ABCD is a square.
Q.36. Show that the points A(– 3, 2), B(– 5, – 5), C(2, – 3) and D(4, 4) are the vertices of a rhombus. Find the area of this rhombus. HINT Area of a rhombus = 1/2 × (product of its diagonals).
AC = √{(2 – (– 3))2 + (– 32)2} = √{25 + 25} = √50 units
BD = √{(4 – (– 5))2 + (4 – (– 5))2} = √{81 + 81} = √162 units
Area = 1/2× (product of diagonals)
= 1/2 × √50 × √162
= 45 sq units
Q.37. Show that the points A(3, 0), B(4, 5), C(– 1, 4) and D(– 2, – 1) are the vertices of a rhombus. Find its area.
AB = √{(4 – 3)2 + (5 – 0)2} = √{1 + 25} = √26 units
BC = √{(– 1 – 4)2 + (4 – 5)2} = √{25 + 1} = √26 units
CD = √{(– 2 – (– 1))2 + (– 1 – 4)2} = √{1 + 25} = √26 units
DA = √{(– 2 – 3)2 + (0 – 1)2} = √{25 + 1} = √26 units
AC = √{ (– 1 – 3)2 + (4 – 0)2} = √{32}
BD = √{(– 2 – 4)2 + (– 1 – 5)2} = √{36 + 36} = 6√2units
Since AB = BC = CD = DA
Hence, ABCD is a rhombus
Area = 1/2 × (product of diagonals)
= 1/2 × 4√2 × 6√2
= 24 sq units
Q.38. Show that the points A(6, 1), B(8, 2), C(9, 4) and D(7, 3) are the vertices of a rhombus. Find its area.
AB = √{(8 – 6)2 + (2 – 1)2} = √{4 + 1} = √5 units
BC = √{(9 – 8)2 + (4 – 2)2} = √{1 + 4} = √5 units
CD = √{(7 – 9)2 + (3 – 4)2} = √{4 + 1} = √5 units
DA = √{(7 – 6)2 + (3 – 1)2} = √{1 + 4} = √5 units
AC = √{ (9 – 6)2 + (4 – 1)2} = √(9 + 9) = 3√2 units
BD = √{(7 – 8)2 + (3 – 2)2} = √{1 + 1} = √2 units
Since AB = BC = CD = DA
Hence, ABCD is a rhombus
Area = 1/2 × (product of diagonals)
= 1/2 × 3√2 × √2
= 3 sq units
Q.39. Show that the points A(2, 1), B(5, 2), C(6, 4) and D(3, 3) are the angular points of a parallelogram. Is this figure a rectangle?
AB = √{(5 – 2)2 + (2 – 1)2} = √{9 + 1} = √10 units
BC = √{(6 – 5)2 + (4 – 2)2} = √{1 + 4} = √5 units
CD = √{(3 – 6)2 + (3 – 4)2} = √{9 + 1} = √10 units
DA = √{(3 – 2)2 + (3 – 1)2} = √{1 + 4} = √5 units
Since AB = CD and BC = DA
∴ ABCD is Parallelogram
AC = √{(6 – 2)2 + (4 – 1)2} = √{16 + 9} = 5 units
For a Rectangle
AC2 = AB2 + BC2
Here AC2 = 25
But AB2 + BC2 = 15
∴ ABCD is not a rectangle
Q.40. Show that A(1, 2), B(4, 3), C(6, 6) and D(3, 5) are the vertices of a parallelogram. Show that ABCD is not a rectangle.
AB = √{(4 – 1)2 + (3 – 2)2} = √{9 + 1} = √10 units
BC = √{(6 – 4)2 + (6 – 3)2} = √{4 + 9} = √13 units
CD = √{(6 – 3)2 + (5 – 6)2} = √{9 + 1} = √10 units
DA = √{(3 – 1)2 + (5 – 2)2} = √{4 + 9} = √13 units
AB = CD and BC = DA
∴ ABCD is a parallelogram
∴ AC = √{(6 – 1)2 + (6 – 2)2} = √{25 + 16} = √41 units
For a Rectangle
AC2 = AB2 + BC2
Here AC2 = 41
But AB+ BC2 = 23
∴ ABCD is not a rectangle
Q.41. Show that the following points are the vertices of a rectangle:
A(– 4, – 1), B(– 2, – 4), C(4, 0) and D(2, 3)
A(– 4, – 1), B(– 2, – 4), C(4, 0) and D(2, 3)
AB = √{(– 2 – (– 4))2 + (– 4 – (– 1))2}
= √{4 + 9} = √13units
BC = √{(4 – (– 2))2 + (0 – (– 4))2}
= √{36 + 16} = √52units
CD = √{(2 – 4)2 + (3 – 0)2}
= √{4 + 9} = √13 units
DA = √{(2 – (– 4))2 + (3 – (– 1))2}
= √{36 + 16} = √52units
AB = CD and BC = DA
AC = √{(4 – (– 4))2 + (0 – (– 1))2}
= √{64 + 1} = √65 units
For a Rectangle
AC= AB2 + BC2
Here AC2 = 65
But AB+ BC= 13 + 52 = 65
∴ ABCD is a rectangle
Q.42. Show that the following points are the vertices of a rectangle:
A(2, – 2), B(14, 10), C(11, 13) and D(– 1, 1)
AB = √{(14 – 2)2 + (10 – (– 2))2}
= √{144 + 144} = √288
BC = √{(11 – 14)+ (10 – 13)2}
= √{9 + 9} = √18 units
CD = √{(– 1 – 11)2 + (1 – 13)2}
= √{144 + 144}
= √288 units
DA = √{(– 1 – 2)+ (1 – (– 2))2}
= √{9 + 9} = √18units
AB = CD and BC = DA
AC = √{(11 – 2)2 + (13 – (– 2))2}
= √{81 + 225}
= √306 units
For a Rectangle
AC2 = AB2 + BC2
Here AC2 = 306
But AB+ BC2 = 288 + 18 = 306
∴ ABCD is a rectangle
Q.43. Show that the following points are the vertices of a rectangle:
A(0, – 4), B(6, 2), C(3, 5) and D(– 3, – 1)
AB = √{(6 – 0)2 + (2 – (– 4))2}
= √{36 + 36}
= √72units
BC = √{(3 – 6)2 + (5 – 2)2}
= √{9 + 9}
= √18units
CD = √{(3 – (– 3))2 + (– 1 – 5)2}
= √{36 + 36}
= √72 units
DA = √{(– 3 – 0)2 + (– 1 – (– 4))2}
= √{9 + 9}
= √18units
AB = CD and BC = DA
AC = √{(3 – 0)2 + (5 – (– 4))2}
= √{9 + 81}
= √90 units
For a Rectangle
AC2 = AB2 + BC2
Here AC2 = 90
But AB2 + BC2 = 72 + 18 = 90
∴ ABCD is a rectangle
The document RS Aggarwal Solutions: Coordinate Geometry- 1 | RS Aggarwal Solutions for Class 10 Mathematics is a part of the Class 10 Course RS Aggarwal Solutions for Class 10 Mathematics.
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# How to solve the linear inequalities in one variable
## Steps to solve the Linear inequalities in one variable.
• Obtain the linear inequation
• Pull all the terms having variable on one side and all the constant term on another side of the inequation
• Simplify the equation in the form given above
$ax> b$
or
$ax \geq b$
or
$ax< b$
or $ax \leq b$
• Divide the coefficent of the variable on the both side.If the coefficent is positive,direction of the inequality does not changes,but if it is negative, direction of the inequation changed
• Put the result of this equation on number line and get the solution set in interval form
Example:
$x- 2 > 2x+15$
Solution
$x-2 > 2x+15$
Subtract x from both the side $x-2 -x > 2x+15 -x$
$-2 >x +15$
Subtract 15 from both the sides $-17 > x$
Solution set
$(-\infty,-17)$
Some Problems to practice
• $2x > 9$
• $x + 5 > 111$
• $3x < 4$
• $2(x + 3) < x+ 1$
## Steps to solve the linear inequality of the fraction form
$\frac {ax+b}{cx+d} > k$ or similar type
• Take k on the LHS
• Simplify LHS to obtain the inequation in the form
$\frac {px+q}{ex+f} > 0$
Make the coefficent positive if not
• Find out the end points solving the equatoon $px+q=0$ and $ex+f=0$
• Plot these numbers on the Number line. This divide the number into three segment
• Start from LHS side of the number and Substitute some value in the equation in all the three segments to find out which segments satisfy the equation
• Write down the solution set in interval form
Or there is more method to solves these
• Take k on the LHS
• Simplify LHS to obtain the inequation in the form
• $\frac {px+q}{ex+f} > 0$
Make the coefficent positive if not
• For the equation to satisfy both the numerator and denominator must have the same sign
• So taking both the part +, find out the variable x interval
• So taking both the part -, find out the variable x interval
• Write down the solution set in interval form
Lets take one example to clarify the points
Question
$\frac{x - 3}{x + 5} > 0$
Solution
Method A
1. Lets find the end points of the equation
Here it is clearly
x=3 and x=-5
2. Now plots them on the Number line
3. Now lets start from left part of the most left number
i.e
Case 1
$x < -5$ ,Let takes x=-6 then $\frac {-6-3}{-6+5} > 0$
$3 > 0$
So it is good
Case 2
Now take x=-5
as x+5 becomes zero and we cannot have zero in denominator,it is not the solution
Case 3
Now x > -5 and x < 3, lets take x=1 then $\frac {1-3}{1+5} > 0$
$\frac {-1}{6} > 0$
Which is not true
Case 4
Now take x =3,then
0> 0 ,So this is also not true
case 5
x> 3 ,Lets x=4
$\frac {4-3}{4+5} > 0$
$\frac {1}{9} > 0$
So this is good
4. So the solution is
x < -5 or x > 3
or
$(-\infty,-5)\cup (3,\infty)$
Method B
1. the numerator and denominator must have the same sign. Therefore, either
1) $x - 3 > 0$ and $x + 5 > 0, or 2)$x - 3 < 0$and$x + 5 < 0\$
2. Now, 1) implies x > 3 and x > -5.
Which numbers are these that are both greater than 3 and greater than -5?
Clearly, any number greater than 3 will also be greater than -5. Therefore, 1) has the solution
x > 3.
3. Next, 2) implies
x < 3 and x < -5.
Which numbers are these that are both less than 3 and less than -5?
Clearly, any number less than -5 will also be less than 3. Therefore, 2) has the solution
x < -5.
4. The solution, therefore, is
x < -5 or x > 3
|
# Matrices
#### Information
An example of a matrix is as follows:
2 4 3 5 4 -1
This 3x2 matrix has 3 rows and 2 columns and contains a total of 6 elements.
To add or subtract matrices you need to have the same size (order) matrices e.g 3x2.
1 -2 4 1 5 -3
+
-3 4 3 2 7 -1
=
-2 2 7 3 12 -4
1 -2 4 1 5 -3
-
2 1 -4 5 2 -4
=
-1 -3 8 -4 3 1
When a Matrix has the same number of rows as columns it is known as a square matrix e.g 2x2.
1 -2 4 1
or
2 1 3 -4 5 4 2 -4 2
If you multiply a matrix by a integer or fraction then you can easily find the new matrix by multiplying all of the elements in the matrix by the value outside of the matrix which then produces the new matrix as demonstrated below.
2
-2 4 5 2 3 -1
=
-4 8 10 4 6 -2
¼
-4 16 32 0 -8 -20
=
-1 4 8 0 -2 -5
In order to multiply Matrices Matrix A must have the same number of columns as Matrix B has rows.
A new Matrix is then formed with the amount of columns as matrix B and the same number of rows as Matrix A.
e.g (2x2) X (2x3) = (2x3) but (2x3) X (2x2) is not possible.
2 4 5 1 0 6 4 -2 5
x
-4 3 0 6 -4 7
=
m1 m2 m3 m4 m5 m6
=
-12 53 -20 39 -4 11
This calculation is done by multiplying the rows and columns together.
m1 = (2x4) + (4x0) + (5x-4) = 8 + 0 – 20 = -12
m2 = (2x-3) + (4x6) + (5x7) = -6 +24 +35 = 53
m3 = (1x4) + (0x0) + (6x-4) = 4 + 0 – 24 = -20
m4 = (1x-3) + (0x6) + (6x7) = -3 + 0 + 42 = 39
m5 = (4x4) + (-2x0) + (5x-4) = 16 + 0 – 20 = -4
m6 = (4x-3) + (-2x6) + (5x7) = -12 - 12 + 35 = 11
The identity matrix is the matrix where when multiplied by another matrix it is equal to the same value. It is always a square matrix and has a diagonal value of 1 and all other elements are equal to 0:
1 0 0 1
or
1 0 0 0 1 0 0 0 1
1 0 0 1
x
2 1 -7 6
=
2 1 -7 6
In order to get the Inverse of a matrix you first need to find the determinant.
The Determinant of a 2x2 matrix is defined by the top left and bottom right cells of the matrix multiplied minus the remaining two cells multiplied together.
If the determinant is equal to 0 then there is no inverse matrix to be found.
The determinant of a matrix can be represented by the symbol △,
7 2 4 9
e.g. △=(7x9)-(2x4)=36-8=28
or simply by writing det before the matrix
det
2 6 1 4
=(2x4)-(6x1)=2
In order to find the inverse of a matrix we first must find the determinent and then recipricate it
For example the reciprical of the determinant above is ½
We then need to rearrange the matrix such as below
a b c d
becomes
d -b -c a
We then need to rearrange the matrix such as below
Finally all we need to do is multiply the reciprical of the determinant by the new matrix
1/△
d -b -c a
Here is an example of finding the inverse of a matrix
Inverse of
2 6 1 4
4 -6 -1 2
=
2 -3 -½ 1
#### Quiz
These Questions auto regenerate so refresh the page if you want more practice
1 1 1 1
1 1 1 1
=
1 1 1 1
1 1 1 1
=
1 1 1 1
+
1 1 1 1
=
1 1 1 1
-
1 1 1 1
=
1
1 1 1 1
=
det
1 1 1 1
=
det
1 1 1 1
=
Inverse of
1 1 1 1
= 1/
|
Lesson Objective: 4.01a Students will know how to solve word problems using slope.
Presentation on theme: " Lesson Objective: 4.01a Students will know how to solve word problems using slope."— Presentation transcript:
Lesson Objective: 4.01a Students will know how to solve word problems using slope
In 2005, Joe planted a tree that was 3 feet tall. In 2010, the tree was 13 feet tall. Assuming the growth of the tree is linear, what was the rate of growth of the tree?
What does “rate of growth” mean? Slope!
Remember the slope equation: In order to find the slope we need two points
In 2005, Joe planted a tree that was 3 feet tall. In 2010, the tree was 13 feet tall. Assuming the growth of the tree is linear, what was the rate of growth of the tree? Remember, x is always years, so replace x 2 with the second year and x 1 with the first
In 2005, Joe planted a tree that was 3 feet tall. In 2010, the tree was 13 feet tall. Assuming the growth of the tree is linear, what was the rate of growth of the tree? Replace y 2 with the height from the second year and y 1 with the first
Simplify the top Simplify the bottom Simplify the fraction to get m. Keep it as a fraction if it can’t be simplified
In 2005, Joe planted a tree that was 3 feet tall. In 2010, the tree was 13 feet tall. Assuming the growth of the tree is linear, what was the rate of growth of the tree? The slope is 2, so the tree grows 2 feet per year
In 1995 a public library had 16,000 books on its shelves. In 1999 the library had 19,000 books. Assuming a linear increase, how many books were added to the library each year?
We’re looking for how many for each year, usually the word “each” means slope.
In 1995 a public library had 16,000 books on its shelves. In 1999 the library had 19,000 books. Assuming a linear increase, how many books were added to the library each year? Years are always x so replace the x’s with the years Replace the y’s with the number of books for each year
Replace y 2 with 19000 and y 1 with 16000 Replace x 2 with 1999 and x 1 with 1995 Simplify the top and the bottom Reduce the fraction
In 1995 a public library had 16,000 books on its shelves. In 1999 the library had 19,000 books. Assuming a linear increase, how many books were added to the library each year? m = 750, therefore the library adds 750 books each year
Monica feeds her dog the same amount of dog food each day from a very large bag. ON the 3 rd day, she has 44 cups left in the bag, and on the 11 th day she has 28 cups left. How many cups of food does she feed her dog a day?
Wendy bought a car for \$25,000 and its value depreciated linearly. After 3 years the value was \$21,250. What was the amount of yearly depreciation?
Jamal’s parents give him \$20 to spend at camp. Jamal spends the same amount of money on snacks each day. After 4 days he has \$12 left. How much money is he spending each day?
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# Symmetry in Graphs of Quadratic Functions
Videos and solutions to help Algebra I students learn how to examine quadratic equations in two variables represented graphically on a coordinate plane and recognize the symmetry of the graph. They explore key features of graphs of quadratic functions: y-intercept and x-intercepts, the vertex, the axis of symmetry, increasing and decreasing intervals, negative and positive intervals, and end behavior. They sketch graphs of quadratic functions as a symmetric curve with a highest or lowest point corresponding to its vertex and an axis of symmetry passing through the vertex.
Related Topics:
Lesson Plans and Worksheets for Algebra I
Math Lesson Plans and Worksheets for all Grades
Common Core For
Algebra I
#### New York State Common Core Math Module 4, Algebra I, Lesson 8
Graph Vocabulary
Axis of Symmetry: Given a quadratic function in standard form, f(x) = ax2 + bx + c, the vertical line given by the graph of the equation, x = -b/(2a) , is called the axis of symmetry of the graph of the quadratic function.
Vertex: The point where the graph of a quadratic function and its axis of symmetry intersect is called the vertex.
End Behavior of a Graph: Given a quadratic function in the form f(x) = ax2 + bx + c (or f(x) = a(x-h)2 + k ), the quadratic function is said to open up if a > 0 and open down if a < 0.
• If a > 0, then f has a minimum at the x-coordinate of the vertex, i.e. f, is decreasing for x-values less than (or to the left of) the vertex, and f is increasing for x values greater than (or to the right of) the vertex.
• If a > 0, then f has a maximum at x-coordinate of the vertex, i.e. f, is increasing for x-values less than (or to the left of) the vertex, and f is decreasing for x-values greater than (or to the right of) the vertex.
Exploratory Challenge 3
Below you see only one side of the graph of a quadratic function. Complete the graph by plotting three additional points of the quadratic function. Explain how you found these points then fill in the table on the right.
a. What are the coordinates of the -intercepts?
b. What are the coordinates of the -intercept?
c. What are the coordinates of the vertex? Is it a minimum or a maximum?
d. If we knew the equation for this curve, what would the sign of the leading coefficient be?
e. Verify that the average rate of change for interval, -3 ≤ x ≤ -2, [-3, -2] is 5. Show your steps.
f. What is the axis of symmetry?
Lesson 8 Summary
Quadratic functions create a symmetrical curve with its highest (maximum) or lowest (minimum) point corresponding to its vertex and an axis of symmetry passing through it when graphed. The x-coordinate of the vertex is the average of the x-coordinates of the zeros or any two symmetric points on the graph.
When the leading coefficient is a negative number, the graph opens down and its end behavior is that both ends move towards negative infinity. If the leading coefficient is positive, the graph opens up and both ends move towards positive infinity.
Lesson 8 Problem Set Sample Solutions
3. Consider the following key features discussed in this lesson for the four graphs of quadratic functions below: x-intercepts, y-intercept, line of symmetry, vertex, and end behavior.
a. Which key features of a quadratic function do graphs and have in common? Which features are not shared?
b. Compare the graphs and and explain the differences and similarities between their key features.
c. Compare the graphs and and explain the differences and similarities between their key features.
d. What do all four of the graphs have in common?
4. Use the symmetric properties of quadratic functions to sketch the graph of the function below, given these points and given that the vertex of the graph is the point (0,5).
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# When a number cube is rolled once, the possible numbers that could show face up are
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1 C3 Chapter 12 Understanding Probability Essential question: How can you describe the likelihood of an event? Example 1 Likelihood of an Event When a number cube is rolled once, the possible numbers that could show face up are. Each time you roll the cube, a number lands face up. This is called an event. Below is a list of 9 different events. Work with a partner to order the events from those least likely to happen to the ones that are most likely to happen when you roll the number cube one time. Use the space next to each event to write any notes that might help you order them. Some may be equally as likely as others; give them the same number if that occurs. Rolling a number less than 7 Rolling an 8 Rolling a 1, 2, or 3 Rolling a 5 Rolling a number other than 6 Rolling an even number Rolling a number greater than 5 Rolling an odd number Rolling a prime number Now answer the following questions: 1) How did you sort the events? 2) Are any of the events impossible? Why were they impossible?
2 An experiment is an activity involving chance in which results are observed. Each observation of an experiment is a trial, and each result is an outcome. A set of one or more outcomes is an event. The probability of an event, written P(event), measures the likelihood that the event will occur. Probability is a measure between 0 and 1 as shown on the number line and can be written as a fraction, a decimal, or percent. If the event is not likely to occur very many times, the probability of the event is close to 0. Likewise, if an event is likely to occur many times, the event s probability is closer to 1. Example 2 Describing Events Determine whether each event is impossible, unlikely, as likely as not, likely, or certain. Include the associated fraction, decimal or percent. A) You flip a coin. The coin lands tails up. B) You roll two number cubes and the sum of the numbers is 11. C) A bowl contains 14 red marbles and 3 green marbles. You pick a red marble. D) A spinner has 10 equal sections marked 1 through 10. You spin and land on a number greater than 0. Now You Try It! Describe each event as impossible, unlikely, as likely as not, likely, or certain. Include the associated fraction, decimal or percent. 1) A hat contains pieces of paper marked with the numbers 1 through 20. You pick an even number. 2) A spinner has 8 equal sections marked 1 through 8. You spin and land on 0.
3 3) The probability of event A is. The probability of event B is. What can you conclude about the two events? The complement of an event is the set of all outcomes not included in the event. For example, consider the event that you roll a number cube and get a 3. The complement is the event that you do not roll a 3. The complement is rolling a 1, 2, 4, 5, or 6. The sum of the probabilities of an event and its complement equals 1. P(event) + P(complement) = 1 Example 3 Using the Complement of an Event Describe a standard deck of cards: In a standard deck of cards, the probability of choosing a card at random and getting an Queen is. What is the probability of not getting an Queen? P(event) + P(complement) = P(Queen) + P ( ) = 1 + P( ) = 1 P(not getting a Queen) = 1 = Now You Try It! 1) A jar contains marbles marked with the numbers 1 through 10. The probability that you pick a number at random and get a 5 is. A) What is the complement of this event? B) What is the probability of the complement?
4 2) You roll a six-sided number cube. The probability that you roll an odd number is. A) What is the complement of this event? B) What is the probability of the complement? 3) Why do the probability of an event and the probability of its complement add up to 1? 4) Give an example of a real-world event and its complement. PRACTICE: 1) Define each of the following in your own words: Experiment Trial Outcome Event Probability Probability is a measure between and. Complement of an event P(event) + P(complement) =
5 2) In a hat, you have cards with the numbers 1 through 20 written on them. You pick one card at random. Order the events from least likely to happen to most likely to happen. greater than 0. You pick an even number. that is at least 8. that is at most 0. Determine whether each event is impossible, unlikely, as likely as not, likely, or certain. Include the associated fraction, decimal or percent. 3) randomly picking a purple card from a standard deck of playing cards 4) randomly picking a black card from a standard deck of playing cards 5) picking a number less that 20 from a jar with papers labeled from 1 to 10 6) picking a number that is divisible by 5 from a jar with papers labeled from 1 to 15 7) The probability of rolling a 5 on a number cube is. What is the probability of not rolling a 5? 8) The probability that a coin will land heads up when flipping a coin is. What is the probability of getting tails? 9) The probability of spinning a 4 on a spinner with 8 equal sections marked 1 through 8 is. What is the complement of this event? What is the probability of the complement? 10) The probability of picking a King from a standard deck of cards is. What is the complement of this event? What is the probability of the complement? 11) Describe an event that has a probability of 0% and an event that has a probability of 100%.
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That's a really great question, and this principle is one of the reasons why counting and combinatorics can be so tricky. It's wonderful that you're trying to delve into the meaning of why we multiply the number of choices together. The basic idea is that multiplication represents groups, and the number of ways to combine different choices can also be represented using groups.
There is another question from Professor Loh's original Mock Mathcounts Sprint Round relating to this concept. There is a free video explanation of one of the questions, which happens to relate to the multiplication principle. In the video, there are three mini-questions, one of which asks to find the number of ways to choose 2 people out of 8 to stand side-by-side for a photo:
https://daily.poshenloh.com/courses/take/2020-mathcounts-mock/lessons/10033528-welcome
I'll copy the explanation again here:
The important idea here is to use multiplication, not addition, to find the ways to arrange the two people. Multiplication can be used to express the number of things that are arranged into groups.
Let's first draw 8 people and name them Person 1, Person 2, Person 3, Person 4, Person 5, Person 6, Person 7 and Person 8.
30e6fec3-3557-4edd-bbbe-e454e2be2149-image.png
The ways to pick two of them to stand side-by-side for a picture can be arranged into eight groups, each of which has seven ways in it:
6cbe068d-b19f-496b-86d8-14e413bf46fa-image.png
In the picture above, a ways described as "48" would mean that Person 4 stands on the left, and Person 8 stands on the right. Order matters here, because the photo would look different if Person 8 were on the left and Person 4 were on the right, which is described as "84."
There are eight groups because there are eight people who can stand on the left, and given that choice of person on the left, there are seven choices for the person on the right.
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# PROBABILITY FOR BEGINNERS 1.0
Written by Temitope Sodipo · 1 min read
Theory of Probability
Probability is a mathematical subject with its own set of rules, definitions, and laws for determining the likelihood of specific outcomes, occurrences, and combinations of outcomes and events.
Specifically, the study of probability is about quantitative decision-making.
The probabilities of the original outcomes and events must be added, subtracted, multiplied, or divided to determine probability. Some mixes are so common that they’ve developed their own set of rules and formulas. The better you comprehend the concepts behind the formulas, the easier it will be to recall and apply them correctly.
Probability values are always assigned on a scale from 0 to 1. Whilst a probability near zero indicates an event is quite unlikely to occur, a probability near one indicates an event is almost certain to occur.
How to Solve Probability Questions
To solve probability problems, you must first construct the problem and then subsequently deconstruct the data. The process of problem-solving is rarely simple, and mastering it takes time.
Predicting probabilities becomes a lot easier with some practice and a few pointers.
When completing a probability word problem, one vital tip to remember is to look for the keyword, which will help you figure out which rule of probability to use. “And,” “or,” and “not” are the keywords.
When dealing with problems that include the keyword “and,” a multiplication rule is the probability rule to employ. When dealing with scenarios using the term “or,” an addition rule should be applied. The complement rule is the probability rule to employ when dealing with the keyword “not.”
Probabilities are found in everyday life – from weather predictions to football outcomes; from consumer buying behaviour to probabilities of gain or loss associated with investment decisions, product design, etc. Probability theory is also applicable in the field of medical science, pharmacology, livestock breeding (Agriculture).
Assigning Probabilities
1. Classical Method: This is the traditional way of determining probability and it is used when all possible outcomes are known in advance and all outcomes are equally likely. A practical example of the classical method of probability is rolling a die.
2. Relative Frequency Method: The relative frequency technique is applied, when all possible outcomes are unknown ahead of time and are not all equally likely. This method makes use of historical data, but similar situation. A manufacturing company that increases its production capacity based on last year’s sales would utilize the relative frequency technique.
3. Subjective Method: We define probability in the subjective approach as the degree of belief we have in the occurrence of an event. As a result, attributing probability based on the subjective method relies on judgment.
In general, combining estimates from the classical or relative frequency approach with subjective judgment yields the best probability estimates.
Marginal, conditional, and joint probabilities
Marginal Probability
This is the likelihood of an event occurring regardless of the outcome of another variable. It is an unconditional probability. It is not contingent on another event happening.
Conditional Probability
This is the probability of an event occurring, given that another event has occurred. It is calculated by adopting the ‘Multiplication Law’, that is multiplying the probability of the prior event by the probability of the conditional event.
Joint Probability
It is the probability of events A and B occurring together. Joint probability is the chance that two or more events may intersect.
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# NCERT Class 8 Mathematics Solutions: Chapter 6 – Squares and Square Roots Exercise 6.3 Part 1
Square and square root possible unit digit number
Question: 1 What could be the possible one’s digits of the square root of each of the following numbers:
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025
Unit’s digits of square of numbers are 0, 1, 4, 5, 6 and 9. Therefore, the possible unit’s digits of the given numbers are:
(i) 9801
If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9.
Therefore, one’s digit of the square root of 9801 is either 1 or 9.
(ii) 99856
If the number ends with 6, then the one’s digit of the square root of that number may be 4 or 6.
Therefore, one’s digit of the square root of 99856 is either 4 or 6.
(iii) 998001
If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9.
Therefore, one’s digit of the square root of 998001 is either 1 or 9.
(iv) 657666025
If the number ends with 5, then the one’s digit of the square root of that number will be 5. Therefore, the one’s digit of the square root of 657666025 is 5.
Question: 2 Without doing any calculation find the numbers which are surely not perfect squares:
(i) 153
(ii) 257
(iii) 408
(iv) 441
All perfect square numbers contain their unit’s place digits 0, 1, 4, 5, 6 and 9.
(i) 153
Given, number 153 has its unit digit 3.
So it is not a perfect square number.
(ii) 257
Given, number 257 has its unit digit 7.
So it is not a perfect square number.
(iii) 408
Given, number 408 has its unit digit 8.
So it is not a perfect square number.
(iv) 441
Given, number 441 has its unit digit 1.
So it would be a perfect square number
Question: 3 Find the square roots of 100 and 169 by the method of repeated subtraction.
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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A
These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A.
Other Exercises
Question 1.
Calculate the co-ordinates-of the point P which divides the line segment joining:
(i) A (1, 3) and B (5, 9) in the ratio 1 : 2
(ii) A (-4, 6) and B (3, -5) in the ratio 3 : 2
Solution:
(i) Let co-ordinates of P be (x,y)
Question 2.
In what ratio is the line joining (2, -3) and (5, 6) divided by the x-axis ?
Solution:
Let the point P (x, 0) divides in the ratio of m1 : m2 line joining the points A (2, -3) and B (5, 6)
Question 3.
In what ratio is the line joining (2, -4) and (-3, 6) divided by the y-axis ?
Solution:
Let the point P (0, y) divides the line joining the points A (2, -4) and (-3, 6) in the ratio of m1 : m2
Question 4.
In what ratio does the point (1, a) divide the join of (-1, 4) and (4, -1)? Also, find the value of ‘a’.
Solution:
Let the point P (1, a) divides the line joining the points (-1, 4) and (4, -1) in the ratio of m1 : m2
Question 5.
In what ratio does the point (a, 6) divide the join of (-4, 3) and (2, 8) ? Also, find the value of ‘a’.
Solution:
Let the point P (a, 6) divides the line joining the points A (-4, 3), B (2, 8) in the ratio of m1 : m2
Question 6.
In what ratio is the join of (4, 3) and (2, -6) divided by the x-axis. Also, find the co-ordinates of the point of intersection.
Solution:
Question 7.
Find the ratio in which the join of (-4, 7) and (3, 0) is divided by the y-axis. Also, find the co-ordinates of the point of intersection.
Solution:
Let, the points (0, y) be the point of intersection which divides the line joining the points A (-4, 7) and B (3, 0)
Question 8.
Points A, B, C and D divide the line segment joining the point (5, -10) and the origin in five equal parts. Find the co-ordinates of A, B, C and D.
Solution:
Points A, B, C and D divide the line segment joining the points (5, -10) and origin (0, 0) in five equal parts
Let co-ordinates of A be (x, y) which divides PO in the ratio of 1 : 4
Question 9.
The line joining the points A (-3, -10) and B (-2, 6) is divided by the point P such that = $$\frac { PA }{ PB }$$ = $$\frac { 1 }{ 5 }$$, find the co-ordinates of P.
Solution:
Let the co-ordinates of P be (x, y) which divides the line joining the points A (-3,-10) and B (-2,6) in the ratio of AP : PB i.e. (5 – 1) : 1 or 4 : 1
Question 10.
P is a point on the line joining A (4, 3) and B (-2, 6) such that 5AP = 2BP. Find the co-ordinates of P.
Solution:
Question 11.
Calculate the ratio in which the line joining the points (-3, -1) and (5, 7) is divided by the line x = 2. Also, find the co-ordinates of the point of intersection.
Solution:
Let the point P (2, y) divides the line joining the points A (-3, -1) and B (5, 7) in the ratio of m1 : m2
Question 12.
Calculate the ratio in which the line joining A (6, 5) and B (4, -3) is divided by the line y = 2.
Solution:
Let the point P (x, 2) divides the line joining the points A (6, 5) and B (4, -3) in the ratio of m1 : m2
Question 13.
The point P(5, -4) divides the line segment AB, as shown in the figure, in the ratio 2 : 5. Find the co-ordinates of points A and B.
Solution:
From the figure, the line AB intersects x-axis at A and y-axis at B.
Let the co-ordinates of A (x, 0) and B (0, y) and P (5, -4) divides it in the ratio of 2 : 5
Question 14.
Find the co-ordinates of the points of trisection of the line joining the points (-3, 0) and (6, 6).
Solution:
Question 15.
Show that the line segment joining the points (-5, 8) and (10, -4) is trisected by the co-ordinate axes.
Solution:
Let the points A (-5, 8) and B (10, -4).
Let P and Q be the two points on the axis which trisect the line joining the points A and B.
AP = PQ = QB
AP : PB = 1 : 2 and AQ : QB = 2 : 1
Question 16.
Show that A (3, -2) is a point of trisection of the line-segment joining the points (2, 1) and (5, -8). Also, find the co-ordinates of the other point of trisection.
Solution:
Let A and B are the points of trisection of the line segment joining the points P (2, 1) and Q (5, -8), then
PA = AB = BQ.
PA : AQ = 1 : 2 and PB : BQ = 2 : 1
Question 17.
If A = (-4, 3) and B = (8, -6)
(i) find the length of AB
(ii) In what ratio is the line joining A and B, divided by the x-axis ?
Solution:
Question 18.
The line segment joining the points M (5, 7) and N (-3, 2) is intersected by the y-axis at point L. Write down the abscissa of L. Hence, find the ratio in which L divides MN. Also, find the co-ordinates of L.
Solution:
Question 19.
A (2, 5), B (-1, 2) and C (5, 8) are the co-ordinates of the vertices of the triangle ABC. Points P and Q lie on AB and AC respectively, such that:
AP : PB = AQ : QC = 1 : 2.
(i) Calculate the co-ordinates of P and Q.
(ii) Show that PQ = $$\frac { 1 }{ 3 }$$ BC.
Solution:
Question 20.
A (-3, 4), B ( 3, -1) and C (-2, 4) are the vertices of a triangle ABC. Find the length of line segment AP, where point P lies inside BC, such that BP: PC = 2 : 3.
Solution:
Question 21.
ThelinesegmentjoiningA(2, 3)andB(6, -5) is intercepted by x-axis at the point K. Write down the ordinate of the point K. Hence, find the ratio in which K divides AB. Also, find the co-ordinates of the pointK. [1990, 2006]
Solution:
Let the line segment Intersect the x-axis at the point P
Co-ordinates of P are (x, 0)
Let P divide the line segment in the ratio K : 1 then
Question 22.
The line segment joining A (4, 7) and B (-6, -2) is intercepted by the y-axis at the point K. Write down the abscissa of the point K. Hence, find the ratio in which K divides AB. Also, find the co-ordinates of the point K.
Solution:
Points A (4, -7), B (-6, -2) are joined which intersects y-axis at K. abscissa of K will be 0
Let the coordinates of K be (0, y) and K divides AB line segment in the ratio m1 : m2
Question 23.
The line joining P (-4, 5) and Q (3, 2), intersects they axis at point R. PM and QN are perpendiculars from P and Q on the x-axis. Find:
(i) The ratio PR: RQ.
(ii) The co-ordinates of R.
(iii) The areas of the quadrilateral PMNQ. [2004]
Solution:
(i) Let divides the line joining the points P (-4, 5) and Q (3, 2) in the ratio k : 1
Question 24.
In the given figure, line APB meets the x- axis at point A and y-axis at point B. P is the point (-4, 2) and AP : PB = 1 : 2. Find the co-ordinates of A and B.
Solution:
Let the co-ordinates of A be (x1, 0) (as it lies on x-axis)
and co-ordinates of B be (0, y2)
and co-ordinates of P are (-4, 2)
AP : PB = 1 : 2 i.e. m1 = 1, m2 = 2
Now, P divides AB in the ratio m1 : m2 or 1 : 2
Question 25.
Given a line segment AB joining the points A (-4, 6) and B (8, -3). Find:
(i) the ratio in which AB is divided by the y-axis.
(ii) find the coordinates of the point of intersection.
(iii) the length of AB.
Solution:
(i) Let the y-axis divide AB in the ratio m : 1
Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
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# Triangular Pyramid Formula
When you think about a Pyramid, the first thing that comes to your mind is the Great Pyramids of Egypt. If you observe the shape of the Great Pyramid of Egypt, then you will notice that the pyramid has a base, that is square shaped. Hence, those pyramids are called Square Pyramid. Similarly, in the triangular pyramid the base of the same is triangular.
When a base is constructed to connect to the apex, a pyramid is formed. A pyramid that had a triangular base is a triangular pyramid. It is also known as the tetrahedron which has equilateral triangles for each of its faces. The illustration below will make it clear what the triangular pyramid looks like.
There are majorly two formulas for triangular pyramid:
$\large Volume\;of\;a\;triangular\;pyramid=\frac{1}{3}Base\;Area\times Height$
$\large Surface\;area=Base\;Area+\frac{1}{2}Perimeter\times Side\;length$
### Examples of Triangular Pyramid
Question 1: Find the volume of a triangular pyramid when base area is $9\;cm^{2}$ and height is 4 cm ?
Solution:
Given,
Base area = $9\;cm^{2}$
Height = 4 cm
As we know,
$Volume\;of\;a\;triangular\;pyramid=\frac{1}{3}\;Base\;Area\times Height$
According to the formula: $\frac{1}{3}\times9\times4=12\,cm^{2}$
Related Formulas T Test Formula Volume Formulas Volume of an Ellipsoid Formula Difference of Squares Formula Mean Median Mode Formula Probability Formulas Euler Maclaurin Formula Percentage Change Formula
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# Numeracy, Maths and Statistics
Velocity, acceleration and distance (instantaneous rest) additional Mathematics 2021
Velocity, acceleration and distance (instantaneous rest) additional Mathematics 2021
The SUVAT equations from equations of motion can only be used when an object is moving with constant acceleration. When the acceleration varies, this is when we must use calculus.
The diagram above shows the relationship between acceleration $a$, velocity $v$ and displacement $x$.
A particle $P$ is moving along the $x$-axis. The displacement $x\mathrm{m}$ from $O$ can we written as a function of $t$ $x = t^5 – 12t^2 +9.$ Find the formula for acceleration.
To find acceleration from displacement, we must differentiate twice. Differentiating once gives $v = \dfrac{\mathrm{d}x}{\mathrm{d}t} = \left(5t^4 – 24t\right)\mathrm{ms^{-1} }.$ This is the equation for the velocity of the particle. Differentiating again gives us the formula for the acceleration of the particle $a = \dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = \left(20t^3 – 24\right)\mathrm{ms^{-2} }.$
A particle $P$ is moving along the $x$-axis. The acceleration $a\mathrm{ms^{-2} }$ from $O$ can be written as a function of $t$ $a = t + 19.$ Suppose we know that when $t=0,$ we have $x=0$ and when $t=4$ we have $x=10$. Find the displacement when $t=10$.
First we can find the velocity by integrating the acceleration equation $v = \int t + 19 \:\mathrm{d}t = \dfrac{1}{2}t^2 + 19t + c.$ Now that we have the velocity we can integrate again to find the displacement $x = \int \dfrac{1}{2}t^2 + 19t + c \:\mathrm{d}t = \dfrac{1}{6}t^3 + \dfrac{19}{2}t^2 + ct + k.$ From the question we know that the particle has not moved ($x=0$) when $t=0$. We can use this information to find the value of $k$ by substituting $x=0$ and $t=0$ into the displacement equation above \begin{align} x &= \dfrac{1}{6}t^3 + \dfrac{19}{2}t^2 + ct + k,\\ 0 &= 0 + k,\\ \ k&=0. \end{align} Now, we also know that when $t=4,$ we have $x=10$. Substituting these values and setting $k=0$ in the displacement equation enables us to find the value of $c$. \begin{align} x &= \dfrac{1}{6}t^3 + \dfrac{19}{2}t^2 + ct,\\ 10 &= \dfrac{4^3}{6} + \dfrac{19\times 4^2}{2} + 4c,\\ 10& – \dfrac{32}{3} – 152 = 4c,\\ \ c &=-\dfrac{229}{6}. \end{align} Substituting this value of $c$ into the displacement equation gives us an equation in only $x$ and $t$ $x = \dfrac{1}{6}t^3 + \dfrac{19}{2}t^2 -\dfrac{190}{3}.$ Substituting in $t=10$, gives \begin{align} x &= \dfrac{1}{6}t^3 + \dfrac{19}{2}t^2 -\dfrac{190}{3},\\ &= \dfrac{1}{6}\times 10^3 + \dfrac{19}{2}\times 10^2 -\dfrac{190}{3},\\ &= 1053.33 \mathrm{m} \text{ (to }2\text{ d.p.)} \end{align}
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# Math Snap
## Question What is an equation of the line that passes through the point $(-7,-1)$ and is parallel to the line $x+y=5 ?$
#### STEP 1
Assumptions1. The given point is $(-7,-1)$. The given line is $x+y=5$ 3. We need to find the equation of a line that is parallel to the given line and passes through the given point.
#### STEP 2
The equation of a line in the slope-intercept form is $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.
#### STEP 3
The slope of a line parallel to the given line $x+y=5$ is the same as the slope of the given line.
#### STEP 4
To find the slope of the given line, we first need to rearrange the equation $x+y=$ into the slope-intercept form $y = mx + b$. $y = -x +$
#### STEP 5
From the rearranged equation, we can see that the slope $m$ of the given line is $-1$.
#### STEP 6
Since the slope of a line parallel to the given line is the same as the slope of the given line, the slope of the line we are trying to find is also $-1$.
#### STEP 7
Now, we can use the point-slope form of the equation of a line to find the equation of the line we are looking for. The point-slope form is $y - y1 = m(x - x1)$, where $(x1, y1)$ is a point on the line and $m$ is the slope of the line.
#### STEP 8
Substitute the given point $(-7,-1)$ and the slope $-1$ into the point-slope form. $y - (-1) = -1(x - (-7))$
#### STEP 9
implify the equation. $y + = -(x +7)$
#### STEP 10
istribute the $-$ on the right side of the equation. $y + = -x -7$
##### SOLUTION
Finally, rearrange the equation into the standard form $Ax + By = C$. $x + y = -8$The equation of the line that passes through the point $(-7,-)$ and is parallel to the line $x+y=5$ is $x + y = -8$.
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diferira7c
2022-01-03
Use symmetry to evaluate the following integrals.
${\int }_{-2}^{2}\left({x}^{2}+{x}^{3}\right)dx$
Mason Hall
Step 1
Given: An integral ${\int }_{-2}^{2}\left({x}^{2}+{x}^{3}\right)dx$
By symmetry an integral of form ${\int }_{-a}^{a}f\left(x\right)dx$ can be written as ${\int }_{-a}^{0}f\left(x\right)dx+{\int }_{0}^{a}f\left(x\right)dx$
Therefore, ${\int }_{-2}^{2}\left({x}^{2}+{x}^{3}\right)dx$ can be written as:
${\int }_{-2}^{2}\left({x}^{2}+{x}^{3}\right)dx={\int }_{-2}^{0}\left({x}^{2}+{x}^{3}\right)dx+{\int }_{0}^{2}\left({x}^{2}+{x}^{3}\right)dx$
Step 2
Now integrating the above function with respect to x.
${\int }_{-2}^{2}\left({x}^{2}+{x}^{3}\right)dx={\int }_{-2}^{0}\left({x}^{2}+{x}^{3}\right)dx+{\int }_{0}^{2}\left({x}^{2}+{x}^{3}\right)dx$
$={\left[\frac{{x}^{3}}{3}+\frac{{x}^{4}}{4}\right]}_{-2}^{0}+{\left[\frac{{x}^{3}}{3}+\frac{{x}^{4}}{4}\right]}_{0}^{2}$
$=\left[\left(0+0\right)-\left(\frac{-8}{3}+4\right)\right]+\left[\left(\frac{8}{3}+4\right)-\left(0+0\right)\right]$
$=\frac{8}{3}-4+\frac{8}{3}+4$
$=\frac{16}{3}$
Thus, the value of ${\int }_{-2}^{2}\left({x}^{2}+{x}^{3}\right)dx$ is $\frac{16}{3}$
Lynne Trussell
$\int \left({x}^{2}+{x}^{3}\right)dx$
Lets
karton
Given:
${\int }_{-2}^{2}\left({x}^{2}+{x}^{3}\right)dx\phantom{\rule{0ex}{0ex}}\int {x}^{2}+{x}^{3}dx\phantom{\rule{0ex}{0ex}}\int {x}^{2}dx+\int {x}^{3}dx\phantom{\rule{0ex}{0ex}}\frac{{x}^{3}}{3}+\frac{{x}^{4}}{4}\phantom{\rule{0ex}{0ex}}\left(\frac{{x}^{3}}{3}+\frac{{x}^{4}}{4}\right){|}_{-2}^{2}\phantom{\rule{0ex}{0ex}}\frac{{2}^{3}}{3}+\frac{{2}^{4}}{4}-\left(\frac{\left(-2{\right)}^{3}}{3}+\frac{\left(-2{\right)}^{4}}{4}\right)\phantom{\rule{0ex}{0ex}}Answer:\phantom{\rule{0ex}{0ex}}\frac{16}{3}$
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We have all been there before, trying to solve a quadratic equation that seems to impossible. Let me guess, you've tried: simplifying the equation, factoring, completing the square, and maybe even graphing, but nothing is working. In comes the quadratic formula:
Looks scary! Well it isn't all that bad, and once you memorize that formula you can solve any quadratic equation in the world! But where do all those variables come from?
That above is a one-variable quadratic polynomial. Yes, that is a mouthful. Let's take a second and dissect the exact meaning beginning with “one-variable.” Simple, this means that there can only be one variable in the entire equation, named x (in the above equation). But aren't the a, b, and c variables? No, they are not, we call them the coefficients. In layman terms a coefficient is simply something that multiples with the variable, generally a constant that must NOT involve a variable. Now let's take a look at the meaning of “quadratic polynomial.” This means that we do not raise any variable to a power greater than 2. Pretty simple so far, now let's connect the two ideas.
The quadratic formula has an a, b, and c just like the quadratic equation so all we have to do is substitute the numbers from the quadratic equation into the quadratic formula and solve. Remember to include negatives and zeros as well! You will see in the examples below what I am talking about.
Once you solve the quadratic formula it will give you 2 answers. They can either be real, or complex numbers. If they are both answers are equal then that means the graph of the equation is tangent to the x-axis at that point. If both answers are complex numbers, then that means the graph of the equation never crosses the x-axis. Finally, if both answers are real numbers, then the graph of the equation crosses the x-axis 2 times, once at the first solution and another time at the second solution. Now let's work some problems!
Example 1
Notice how the a is equal to one even though there is no coefficient in front of the x squared. This is because multiplying by 1 produces no effect, but we must include it when solving the quadratic formula! And once again, take note of how we include the negative sign when determining c. Now let's substitute these numbers into the quadratic formula and find out what we get.
Aha, the solutions are 1 and -3, therefore the graph of the equation must cross the x-axis at those two points. Does it?
It does, great! Whenever you work your own problems, it is a good idea to confirm your results by graphing the equation. Now let's have you solve two yourself. Feel free to post questions in the comment section.
Example 2
hint:
Example 3
Remember to confirm your solutions by graphing! Using the quadratic formula is an invaluable skill to posses. It is useful across a lot of fields such as chemistry, finance, business, and even art.
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# GCD of 599, 184, 43, 116 Calculator
Make use of GCD Calculator to determine the Greatest Common Divisor of 599, 184, 43, 116 i.e. 1 largest integer that divides all the numbers equally.
GCD of 599, 184, 43, 116 is 1
GCD(599, 184, 43, 116) = 1
Ex: 10, 15, 20 (or) 24, 48, 96,45 (or) 78902, 89765, 12345
GCD of
GCD of numbers 599, 184, 43, 116 is 1
GCD(599, 184, 43, 116) = 1
## Finding GCD of 599, 184, 43, 116 using Factoring
Given Input numbers are 599, 184, 43, 116
To find the GCD of numbers using factoring list out all the divisors of each number
Divisors of 599
List of positive integer divisors of 599 that divides 599 without a remainder.
1, 599
Divisors of 184
List of positive integer divisors of 184 that divides 184 without a remainder.
1, 2, 4, 8, 23, 46, 92, 184
Divisors of 43
List of positive integer divisors of 43 that divides 43 without a remainder.
1, 43
Divisors of 116
List of positive integer divisors of 116 that divides 116 without a remainder.
1, 2, 4, 29, 58, 116
Greatest Common Divisior
We found the divisors of 599, 184, 43, 116 . The biggest common divisior number is the GCD number.
So the Greatest Common Divisior 599, 184, 43, 116 is 1.
Therefore, GCD of numbers 599, 184, 43, 116 is 1
### Finding GCD of 599, 184, 43, 116 using Prime Factorization
Given Input Data is 599, 184, 43, 116
Make a list of Prime Factors of all the given numbers initially
Prime Factorization of 599 is 599
Prime Factorization of 184 is 2 x 2 x 2 x 23
Prime Factorization of 43 is 43
Prime Factorization of 116 is 2 x 2 x 29
The above numbers do not have any common prime factor. So GCD is 1
### Finding GCD of 599, 184, 43, 116 using LCM Formula
Step1:
Let's calculate the GCD of first two numbers
The formula of GCD is GCD(a, b) = ( a x b) / LCM(a, b)
LCM(599, 184) = 110216
GCD(599, 184) = ( 599 x 184 ) / 110216
GCD(599, 184) = 110216 / 110216
GCD(599, 184) = 1
Step2:
Here we consider the GCD from the above i.e. 1 as first number and the next as 43
The formula of GCD is GCD(a, b) = ( a x b) / LCM(a, b)
LCM(1, 43) = 43
GCD(1, 43) = ( 1 x 43 ) / 43
GCD(1, 43) = 43 / 43
GCD(1, 43) = 1
Step3:
Here we consider the GCD from the above i.e. 1 as first number and the next as 116
The formula of GCD is GCD(a, b) = ( a x b) / LCM(a, b)
LCM(1, 116) = 116
GCD(1, 116) = ( 1 x 116 ) / 116
GCD(1, 116) = 116 / 116
GCD(1, 116) = 1
GCD of 599, 184, 43, 116 is 1
### GCD of Numbers Calculation Examples
Here are some samples of GCD of Numbers calculations.
### FAQs on GCD of numbers 599, 184, 43, 116
1. What is the GCD of 599, 184, 43, 116?
GCD of 599, 184, 43, 116 is 1
2. Where do I get the detailed procedure to find GCD of 599, 184, 43, 116?
You can find a detailed procedure to find GCD of 599, 184, 43, 116 on our page.
3. How to find GCD of 599, 184, 43, 116 on a calculator?
You can find the GCD of 599, 184, 43, 116 by simply giving the inputs separated by commas and click on the calculate button to avail the Greatest Common Divisor in less time.
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# Equation of Ellipse
## Definition and Equation of an Ellipse with Vertical Axis
An ellipse is the set of all points $M(x,y)$ in a plane such that the sum of the distances from $M$ to fixed points $F_1$ and $F_2$ called the foci (plurial of focus) is equal to a constant.
$\overline{MF_1} + \overline{MF_2} = \sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = 2 a$
For $a \ge b \ge 0$, eliminating the square roots by squaring and simplifying using the relationship $a^2 = b^2 + c^2$, we can end up with the standard equation of an ellispe given by:
$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$
The width of the ellipse is $2 a$ and the height is $2 b$
Point $O(0,0)$ is the center of the ellipse.
Points $V_1(a,0)$ and $V_2(-a,0)$ are called the vertices of the ellipse.
The foci are at $F_1(c,0)$ and $F_2(-c,0)$
Example 1
An ellipse centered at $(0,0)$ has x intercepts at $(7,0)$ and $( -7 ,0)$ and y intercepts at $(0,4)$ and $( 0,-4)$. Find the equation of the ellipse and the foci $F_1$ and $F_2$
Solution to Example 1
The equation of an ellipse whose center is at the origin is given by
$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ , $a \gt 0$ and $b \gt 0$
Find the x axis by setting y = 0 in the above equation
$\dfrac{x^2}{a^2} + \dfrac{0^2}{b^2} = 1$
Simplify
$\dfrac{x^2}{a^2} = 1$
Solve for $x$
$x = a$ and $x = - a$
The x intercepts are given by $(7,0)$ and $( -7 ,0)$ which gives $a = 7$.
Find the y axis by setting x = 0 in the general given above equation
$\dfrac{0^2}{a^2} + \dfrac{y^2}{b^2} = 1$
Simplify
$\dfrac{y^2}{b^2} = 1$
Solve for $y$
$y = b$ and $y = - b$
The y intercepts are given by $(0,4)$ and $( 0 , -4)$ which gives $b = 4$.
In order to find the foci, we first need to find parameter $c$ which is related to $a$ and $b$ by
$a^2 = b^2 + c^2$
Substitute $a$ and $b$ by their values and solve for $c$
$7^2 = 4^2 + c^2$
$c = \sqrt{49 - 16} = \sqrt{33}$
The coordinates of the two foci F1 and F2 are given by
$F_1(c,0) = F_1(\sqrt{33} , 0)$ and $F_1( - c,0) = F_1( - \sqrt{33} , 0)$
## General Equation of an Ellipse
We can generalize and write the equation of an ellipse whose center is at $O(h,k)$ as follows
$\dfrac{(x-h)^2}{a^2} + \dfrac{(y-k)^2}{b^2} = 1$
with foci at $F_1(c+h,k)$ and $F_2(-c+h,k)$ and vertices at $V_1(a+h,k)$ and $V_2(-a+h,k)$.
Example 2
Find the center, foci and vertices of the ellipse given by the equation $(x - 1)^2 + 4(y-2)^2 = 16$ then use a graphing calculator to graph the given equation and check your answers.
Solution to Example 2
Rewrite the given equation in standard form by dividing all terms by 16.
$\dfrac{(x - 1)^2}{16} + \dfrac{4(y-2)^2}{16} = \dfrac{16}{16}$ .
Simplify and write the denominators of the terms on the left as squares
$\dfrac{(x - 1)^2}{4^2} + \dfrac{(y-2)^2}{2^2} = 1$
Compare the above equation of an ellipse in standard for to the general equation $\dfrac{(x-h)^2}{a^2} + \dfrac{(y-k)^2}{b^2} = 1$ and identify the parameters $a, b, h$ and $k$.
$a = 4, b = 2, h = 1$ and $k = 2$
The center is at the point $O(h,k) = O(1,2)$
Find the parameter $c$ using the relationship $a^2 = b^2 + c^2$
$c = \sqrt{a^2 - b^2} = \sqrt{16 - 4} = 2 \sqrt{3}$
The foci are at the points
$F_1(c+h,k) = F_1(2 \sqrt{3}+1,2)$ and $F_2(-c+h,k) = F_2(-2 \sqrt{3}+1,2)$
The vertices are at the points
$V_1(a+h,k) = V_1(5,2)$ and $V_2(-a+h,k) = V_2(-3,2)$
The graph of the given equation $(x - 1)^2 + 4(y-2)^2 = 16$ is shown below and it is that of an ellipse with center at $O(1,2)$ and vertices at $V_1(5,2)$ and $V_2(-3,2)$ as calculated above.
## Interactive Turorial on Equation of an Ellipse
An app to explore the equation of a parabola and its properties is now presented. The equation used is the standard equation that has the form
$\dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1$
The exploration is carried out by changing the parameters $a, b, h$ and $k$ included in the above equation. The default values, when you open this page, are: $a = 4, b = 2, h = 2$ and $k = 3$. Some activities are listed below but many other activities may be generated.
Click on the button "Plot Equation" to start.
$a$ = 4 $b$ = 2 $h$ = 2 $k$ = 3
Hover the mousse cursor on the graph or plotted point to read the coordinates.
Click the button above "Plot Equation". You may hover the mousse cursor over the graph of the ellipse and the points to read the coordinates. You may also Hover the mousse cursor on the top right of the graph to have the options of zooming, downloading the graph as a png file, ...
1 - Select a point M on the graph of the ellipse, hover the mousse over it and read the coordinates. Read the coordinates of F1 and F2 (right legend) or hover the mousse and read the coordinates and show that the sum of the distances $\overline{MF_1} + \overline{MF_2}$ is close to $2a$.
You may do the above activity for different values of $a, b, h$ and $k$ and as many points as needed to better understand the definition of an ellipse.
2 - Use the values of $a$ and $b$ to find $c$ using the relationship between $a, b$ and $c$ given by $a^2 = b^2 + c^2$.
Use the above results to find the coordinates of $F_1, F_2, V_1$ and $V_2$ and check the results graphically.
3 - Set $a, b, h$ and $k$ to some values. Find the x and y intercepts and check your results graphically.
4 -
Exercise:
Show by algebraic calculations that the following equation $\dfrac{(x + 2)^2}{5} + 5(y-3)^2 = 5$ is that of an ellipse and find the center, foci and vertices of the ellipse given by the equation then use the app to graph it and check your answers.
If needed, Free graph paper is available.
## More References and Links to Topics Related to the Equation of the Ellipse
Ellipse
Similar tutorials on
circle , Parabola and the hyperbola can be found in this site.
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# Louisiana - Grade 1 - Math - Measurement and Data - Length of an Object - 1.MD.2
### Description
Express the length of an object as a whole number of length units, by laying multiple copies of a shorter object (the length unit) end to end; understand that the length measurement of an object is the number of same-size length units that span it with no gaps or overlaps. Limit to contexts where the object being measured is spanned by a whole number of length units with no gaps or overlaps.
• State - Louisiana
• Standard ID - 1.MD.2
• Subjects - Math Common Core
### Keywords
• Math
• Measurement and Data
## More Louisiana Topics
Apply properties of operations to add and subtract.2 Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)
Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares.
1.OA.4 Understand subtraction as an unknown-addend problem. For example, subtract 10 – 8 by finding the number that makes 10 when added to 8. Add and subtract within 20.
1.OA.5 Relate counting to addition and subtraction (e.g., by counting on 2 to add 2).
1.OA.6 Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 – 4 = 13 – 3 – 1 = 10 – 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 – 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
1.OA.7 Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 – 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2.
1.OA.8 Determine the unknown whole number in an addition or subtraction equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 + ? = 11, 5 = _ – 3, 6 + 6 = _.
Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes.
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# Question 2
(i) We have:
T(x,y,z,t) = \begin{pmatrix}x - y + z + t \\ x + 2y - z + t \\ 3y - 2z\end{pmatrix}
So T(x,y,z,t) = 0 iff x - y + z + t = 0, x + 2y - z + t = 0 and 3y - 2z = 0. Note that subtracting the first two equations gives the third, so the third is in fact redundant. The solution set can therefore be determined only by x - y + z + t = 0 and x + 2y - z + t = 0. This is undetermined and we are free to set two variables arbitrary, say x = \lambda and y = \mu. Then \displaystyle z = \frac 3 2 \mu and \displaystyle t = y - x - z = \mu - \lambda - \frac 3 2 \mu = -\frac 1 2 \mu - \lambda. So each vector in X can be written:
\displaystyle \begin{pmatrix}\lambda \\ \mu \\ \frac 3 2 \mu \\ -\frac 1 2 \mu - \lambda\end{pmatrix} = \lambda \begin{pmatrix}1 \\ 0 \\ 0 \\ -1\end{pmatrix} + \mu \begin{pmatrix}0 \\ 1 \\ \frac 3 2 \\ -\frac 1 2\end{pmatrix}
So:
\mathrm{ker}(T) = \mathrm{span} \left\{\begin{pmatrix}1 \\ 0 \\ 0 \\ -1\end{pmatrix}, \begin{pmatrix}0 \\ 1 \\ \frac 3 2 \\ -\frac 1 2\end{pmatrix}\right\}
(note that these are linearly independent so it in fact forms a basis)
For the image of T, we are interested in:
\mathrm{span} \left\{\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix}, \begin{pmatrix}-1 \\ 2 \\ 3\end{pmatrix}, \begin{pmatrix}1 \\ -1 \\ -2\end{pmatrix}\right\}
Note that \begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix} is not a multiple of \begin{pmatrix}-1 \\ 2 \\ 3\end{pmatrix}. It remains to investigate whether \begin{pmatrix}1 \\ -1 \\ -2\end{pmatrix} is a linear combination of these two. We investigate solutions to:
\begin{pmatrix}1 \\ -1 \\ -2\end{pmatrix} = \lambda \begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix} + \mu \begin{pmatrix}-1 \\ 2 \\ 3\end{pmatrix}
The last component requires \mu = -\dfrac 2 3 and the first requires \lambda = \dfrac 1 3. This is consistent with the second component, so \begin{pmatrix}1 \\ -1 \\ -2\end{pmatrix} can be written as a linear combination of the other two vectors. So:
\mathrm{im}(T) = \mathrm{span} \left\{\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix}, \begin{pmatrix}-1 \\ 2 \\ 3\end{pmatrix}, \begin{pmatrix}1 \\ -1 \\ -2\end{pmatrix}\right\} = \mathrm{span} \left\{\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix}, \begin{pmatrix}-1 \\ 2 \\ 3\end{pmatrix}\right\}
So we have \mathrm{rank}(T) = 2 and \mathrm{Nullity}(T) = 2.
(ii)
Note that we are free to determine x_{ii} for 1 \le i \le n - 1 arbitrarily, then x_{nn} is fixed and dependent on the other x_{ii} s. The other elements of the matrix can be decided arbitrarily. Then a basis for \mathrm{ker}(T) is:
\left\{D_i \mid 1 \le i \le n - 1\right\} \cup \left\{T_{ij} \mid 1 \le i, j \le n, \, i \ne j\right\}
where:
• D_i has all elements zero except a 1 in the (i, i) position and -1 in the (n,n) position.
• T_{ij} has all elements zero except a 1 in the (i, j) position.
(noting linear independence here)
Note that since these sets are disjoint the size of the union is:
\mathrm{Nullity}(T) = |\left\{D_i \mid 1 \le i \le n - 1\right\}| + |\left\{T_{ij} \mid 1 \le i, j \le n, \, i \ne j\right\}|
The former set clearly has size n - 1. For the latter set, note that there are n^2 possible ways to pair the i, j s ignoring the requirement that i \ne j. There are n pairs with i = j, so the second set has size n^2 - n.
Hence, we have \mathrm{Nullity}(T) = n^2 - 1.
We can show T to be surjective. Let t \in \mathbb R. Then let X = (x_{ij}) be the matrix with x_{11} = t and x_{ij} = 0 for (i,j) \ne (1,1). Then \mathrm{tr}(X) = t. So T is surjective, and \mathrm{im}(T) = \mathbb R. This has dimension 1, so \mathrm{rank}(T) = 1.
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# COMPARING DIFFERENCES IN CENTERS TO VARIABILITY
## About "Comparing differences in centers to variability"
Comparing differences in centers to variability :
Recall that to find the mean absolute deviation (MAD) of a data set, first find the mean of the data. Next, take the absolute value of the difference between the mean and each data point. Finally, find the mean of those absolute values.
## Comparing differences in centers to variability - Example
The tables show the number of minutes per day students in a class spend exercising and playing video games. What is the difference of the means as a multiple of the mean absolute deviations ?
Solution :
Step 1 :
Calculate the mean number of minutes per day exercising.
0 + 7 + 7 + 18 + 20 + 38 + 33 + 24 + 22 + 18 + 11 + 6 = 204
Divide the sum by the number of students.
204 ÷ 12 = 17
Step 2 :
Calculate the mean absolute deviation for the number of minutes exercising.
|0-17| = 17|7-17| = 10|7-17| = 10|18-17| = 1|20-17| = 3|38-17| = 21 |33-17| = 16|24-17| = 7|22-17| = 5|18-17| = 1|11-17| = 6|6-17| = 11
Find the mean of the absolute values.
17 + 10 + 10 + 1 + 3 + 21+ 16 + 7 + 5 + 1 + 6 + 11 = 108
Divide the sum by the number of students.
108 ÷ 12 = 9
Step 3 :
Calculate the mean number of minutes per day playing video games. Round to the nearest tenth.
13+18+19+30+32+46+50+34+36+30+23+19 = 350
Divide the sum by the number of students.
350 ÷ 12 ≈ 29.2
Step 4 :
Calculate the mean absolute deviation for the numbers of minutes playing video games.
|13-29.2| = 16.2 |18-29.2| = 11.2 |19-29.2| = 10.2|30-29.2| = 0.8 |32-29.2| = 2.8 |46-29.2| = 16.8 |50-29.2| = 20.8 |34-29.2| = 4.8 |36-29.2| = 6.8|30-29.2| = 0.8 |23-29.2| = 6.2 |19-29.2| = 10.2
Find the mean of the absolute values. Round to the nearest tenth.
16.2 + 11.2 + 10.2 + 0.8 + 2.8 + 16.8 + 20.8 + 4.8 + 6.8 + 0.8 + 6.2 + 10.2 = 107.6
Divide the sum by the number of students.
107.6 ÷ 12 ≈ 9
Step 5 :
Find the difference in the means.
Subtract the lesser mean from the greater mean.
29.2 - 17 = 12.2
Step 6 :
Write the difference of the means as a multiple of the mean absolute deviations, which are similar but not identical.
Divide the difference of the means by the MAD.
12.2 ÷ 9 ≈ 1.36
The means of the two data sets differ by about 1.4 times the variability of the two data sets.
## Comparing differences in centers to variability - Practice question
Question :
The high jumps in inches of the students on two intramural track and field teams are shown below. What is the difference of the means as a multiple of the mean absolute deviations ?
Solution :
After having gone through the stuff given above, we hope that the students would have understood "Comparing differences in centers to variability".
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# Analyzing Functions Graphically
Sep 15, 2022
## Key Concepts
• Identify the common features of a function when given an equation or graph.
• Analyze domain and range of the function.
• Analyze maximum and minimum values of the function.
• Understand axes of symmetry of the function.
• Analyze end behaviors of the graphs.
### Graph of the functions and its features
Q 1: Plot the points of each function on a graph. Explain the features of the function that represents the graph:
Solution:
Graph A:
Common features of the graph A:
Domain: [-2, 2]
Range: [1, 5]
There is no x-intercept.
There is no y-intercept.
Graph B:
Common features of the graph B:
Domain: [-2, 2]
Range: [-1, 11]
x-intercept: 1.7
y-intercept: 5
Graph C:
Common features of the graph C:
Domain: [-2, 2]
Range: [3, 21]
There is no x-intercept.
y-intercept: 5
### Analyze Domain and Range
The domain of a function is the set of all values for which the function is defined.
The range of the function is the set of all values that the function takes.
Example 1:
The graph of the function: p(x)=|x|−1.
Domain: all real numbers
To find the range of p(x):
|x| ≥ 0
|x|−1≥−1
p(x)≥−1
Range:
y≥−1
Example 2:
The graph of the function: q(x)=−x2+3
.
Domain: all real numbers
To find the range of q(x) :
x2≥0
−x2≤0
−x2+3≤3
q(x)≤3
Range: y≤3.
### Analyze Maximum and Minimum Values
Example 3:
The graph of the function: f(x)=−2x+5.
Solution:
The graph of the linear function f(x)=−2x+5 decreases at a constant rate. So, there is no maximum or minimum value.
Example 4:
The graph of the function:
g(x)=−2x+3.
Solution:
The graph of the function g(x)=−2x+3 is a translation of an exponential function.
It is bounded above the asymptote y=3 which means that g(x)<3.
It has no maximum because it is approaching 3 but never reaches 3.
The function g also has no minimum. As x increases, g(x) decreases.
Example 5:
The graph of the function: h(x)=|x−2|−1.
Solution:
The graph of the function h(x)=|x−2|−1 is a translation of an absolute value function.
It opens upward so the function has a minimum value of – 1 at the vertex (2,−1).
### Understand Axes of Symmetry
Example 6:
The graph of the function: p(x)=5−|x+1|.
Solution:
Translations of the absolute value function always have an axis of symmetry passing through the vertex.
Here the function
p(x)=5−|x+1| has an axis of symmetry x=−1 passing through the vertex (−1, 5).
Example 7:
The graph of the function: q(x)=(x+3).
Solution:
Quadratic functions always have a vertical axis of symmetry.
Here the quadratic function q(x)=(x+3)2 has an axis of symmetry x=−3.
Example 8:
The graph of the function:
r(x)=√x+2
Solution:
The function r(x)=√x+2 does not have an axis of symmetry.
There is no way to fold the graph so that one side aligns with the other.
### Analyze End Behaviors of Graphs
Example 9:
The graph of the function:
h(x)=x2−2x+1.
Solution:
As x→∞ , the values of h(x) increases without bound.
So, h(x)→∞.
As x→−∞ , the values of h(x) also increases without bound.
So, h(x)→∞ .
Example 10:
The graph of the function:
g(x)=∛x−2.
Solution:
As x→∞ , the values of g(x) grow less and less steeply, but they do not approach to any asymptote.
So, g(x)→∞.
As x→−∞ , the values of g(x) also decreases.
So, g(x)→−∞.
## Exercise
1. Sketch the graph of the function fx=√x-4 and identify its domain and range.
2. Use the graph of the function fx=5|x|-8 to identify its maximum and minimum value if they exist.
3. Describe the end behavior of the function fx=-7x.
### What we have learned
• Identify the common features of a function when given an equation or graph.
• Analyze domain and range of the function.
• Analyze maximum and minimum values of the function.
• Understand axes of symmetry of the function.
• Analyze end behaviors of the graphs.
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […]
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# Question Video: Finding the Limit of a Quotient of a Difference of Powers at a Point Mathematics • Higher Education
Determine lim_(π₯ β 2) (π₯β΅ β 32)/(π₯ β 2).
02:53
### Video Transcript
Determine the limit as π₯ tends to two of π₯ to the power of five minus 32 over π₯ minus two.
The first thing we can do to try to evaluate this limit is just to directly substitute the limit point two into the expression. The expression is π₯ to the power of five minus 32 over π₯ minus two. Substituting two into this, we get two to the power of five minus 32 over two minus two. Two to the power of five is 32, so we get 32 minus 32 over two minus two. And this is zero divided by zero. This is an indeterminate form, which we canβt evaluate. And so direct substitution hasnβt given us the value of this limit.
Clearly, weβre not going to have any success. Well, π₯ minus two is the denominator of this fraction. So weβre going to have to rewrite this into another form, where thatβs not the case. If you know about the factor theorem, youβll know that we have good reason to expect that the numerator π₯ to the power of five minus 32 has a factor of π₯ minus two. And so if we perform a polynomial long division to this rational expression, we should expect to get a polynomial.
An alternative to performing the polynomial long division is to rewrite the 32 as two to the power of five. And then we can apply the identity π₯ to the power of π minus π to the power of π over π₯ minus π equals π₯ to the power of π minus one plus π times π₯ to the power of π minus two plus π squared times π₯ to the power of π minus three, and so on. So the last two terms are π to the power of π minus two times π₯ plus π to the power of π minus one.
So setting π equal to two and π equal to five, we get that this quotient is π₯ to the power of four plus two times π₯ to the power of three plus two squared times π₯ squared plus two to the power of three times π₯ plus two to the power of four. And we can simplify this by evaluating the powers of two. The left-hand side and right-hand side are equal. Apart from that, π₯ equals two and the left-hand side is undefined. So the limit of the left-hand side is equal to the limit of the right-hand side. And when we substitute it directly into the limits on the right-hand side, we donβt get an indeterminate form.
Replacing π₯ by two, we get two to the power of four plus two times two to the power of three plus four times two to the power of two plus eight times two plus 16. And evaluating this, we get our final answer 80. So to recap what happened, substituting directly into π₯ to the power of five minus 32 over π₯ minus two gave us an indeterminate form. But by applying an identity or using polynomial long division, we could rewrite this expression as a polynomial. And substituting directly into that polynomial gave us the value of the limit 80.
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Architecture
# What Is The Greatest Common Factor Of 18 And 12, What Is The Greatest Common Factor Of 12 And 18
GCF of 12 and 18 is the largest possible number that divides 12 and 18 exactly without any remainder. The factors of 12 and 18 are 1, 2, 3, 4, 6, 12 and 1, 2, 3, 6, 9, 18 respectively. There are 3 commonly used methods to find the GCF of 12 and 18 – prime factorization, long division, and Euclidean algorithm.
You are watching: What is the greatest common factor of 18 and 12
1 GCF of 12 and 18 2 List of Methods 3 Solved Examples 4 FAQs
Answer: GCF of 12 and 18 is 6.
Explanation:
The GCF of two non-zero integers, x(12) and y(18), is the greatest positive integer m(6) that divides both x(12) and y(18) without any remainder.
Let's look at the different methods for finding the GCF of 12 and 18.
Long Division MethodPrime Factorization MethodListing Common Factors
### GCF of 12 and 18 by Long Division
GCF of 12 and 18 is the divisor that we get when the remainder becomes 0 after doing long division repeatedly.
Step 2: Since the remainder ≠ 0, we will divide the divisor of step 1 (12) by the remainder (6).Step 3: Repeat this process until the remainder = 0.
The corresponding divisor (6) is the GCF of 12 and 18.
### GCF of 12 and 18 by Prime Factorization
Prime factorization of 12 and 18 is (2 × 2 × 3) and (2 × 3 × 3) respectively. As visible, 12 and 18 have common prime factors. Hence, the GCF of 12 and 18 is 2 × 3 = 6.
### GCF of 12 and 18 by Listing Common Factors
Factors of 12: 1, 2, 3, 4, 6, 12Factors of 18: 1, 2, 3, 6, 9, 18
There are 4 common factors of 12 and 18, that are 1, 2, 3, and 6. Therefore, the greatest common factor of 12 and 18 is 6.
☛ Also Check:
## GCF of 12 and 18 Examples
Example 1: Find the GCF of 12 and 18, if their LCM is 36.
Solution:
∵ LCM × GCF = 12 × 18⇒ GCF(12, 18) = (12 × 18)/36 = 6Therefore, the greatest common factor of 12 and 18 is 6.
Example 2: Find the greatest number that divides 12 and 18 exactly.
Solution:
The greatest number that divides 12 and 18 exactly is their greatest common factor, i.e. GCF of 12 and 18.⇒ Factors of 12 and 18:
Factors of 12 = 1, 2, 3, 4, 6, 12Factors of 18 = 1, 2, 3, 6, 9, 18
Therefore, the GCF of 12 and 18 is 6.
Example 3: For two numbers, GCF = 6 and LCM = 36. If one number is 12, find the other number.
Solution:
Given: GCF (x, 12) = 6 and LCM (x, 12) = 36∵ GCF × LCM = 12 × (x)⇒ x = (GCF × LCM)/12⇒ x = (6 × 36)/12⇒ x = 18Therefore, the other number is 18.
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## FAQs on GCF of 12 and 18
### What is the GCF of 12 and 18?
The GCF of 12 and 18 is 6. To calculate the GCF (Greatest Common Factor) of 12 and 18, we need to factor each number (factors of 12 = 1, 2, 3, 4, 6, 12; factors of 18 = 1, 2, 3, 6, 9, 18) and choose the greatest factor that exactly divides both 12 and 18, i.e., 6.
### What is the Relation Between LCM and GCF of 12, 18?
The following equation can be used to express the relation between LCM and GCF of 12 and 18, i.e. GCF × LCM = 12 × 18.
### How to Find the GCF of 12 and 18 by Long Division Method?
To find the GCF of 12, 18 using long division method, 18 is divided by 12. The corresponding divisor (6) when remainder equals 0 is taken as GCF.
### What are the Methods to Find GCF of 12 and 18?
There are three commonly used methods to find the GCF of 12 and 18.
By Prime FactorizationBy Euclidean AlgorithmBy Long Division
### How to Find the GCF of 12 and 18 by Prime Factorization?
To find the GCF of 12 and 18, we will find the prime factorization of the given numbers, i.e. 12 = 2 × 2 × 3; 18 = 2 × 3 × 3.⇒ Since 2, 3 are common terms in the prime factorization of 12 and 18. Hence, GCF(12, 18) = 2 × 3 = 6☛ What are Prime Numbers?
### If the GCF of 18 and 12 is 6, Find its LCM.
GCF(18, 12) × LCM(18, 12) = 18 × 12Since the GCF of 18 and 12 = 6⇒ 6 × LCM(18, 12) = 216Therefore, LCM = 36☛ GCF Calculator
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# Multiplication and Division of Radicals
## Rationalize the denominator
Estimated9 minsto complete
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Practice Multiplication and Division of Radicals
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Dividing Square Roots
The area of a rectangle is \begin{align*}\sqrt{30}\end{align*} and the length of the rectangle is \begin{align*}\sqrt{20}\end{align*}. What is the width of the rectangle?
### Dividing Square Roots
Dividing radicals can be a bit more difficult that the other operations. The main complication is that you cannot leave any radicals in the denominator of a fraction. For this reason we have to do something called rationalizing the denominator, where you multiply the top and bottom of a fraction by the same radical that is in the denominator. This will cancel out the radicals and leave a whole number.
4. \begin{align*}\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}\end{align*}
5. \begin{align*}\frac{\sqrt{a}}{\sqrt{b}} \cdot \frac{\sqrt{b}}{\sqrt{b}} = \frac{\sqrt{ab}}{b}\end{align*}
Let's simplify the following problems by rationalizing the denominator.
1. Simplify \begin{align*}\sqrt{\frac{1}{4}}\end{align*}.
Break apart the radical by using Rule #4.
\begin{align*}\sqrt{\frac{1}{4}}=\frac{\sqrt{1}}{\sqrt{4}}=\frac{1}{2}\end{align*}
1. Simplify \begin{align*}\frac{2}{\sqrt{3}}\end{align*}.
This might look simplified, but radicals cannot be in the denominator of a fraction. This means we need to apply Rule #5 to get rid of the radical in the denominator, or rationalize the denominator. Multiply the top and bottom of the fraction by \begin{align*}\sqrt{3}\end{align*}.
\begin{align*}\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}\end{align*}
1. Simplify \begin{align*}\sqrt{\frac{32}{40}}\end{align*}.
Reduce the fraction, and then apply the rules above.
\begin{align*}\sqrt{\frac{32}{40}}= \sqrt{\frac{4}{5}}= \frac{\sqrt{4}}{\sqrt{5}}= \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}=\frac{2\sqrt{5}}{5}\end{align*}
### Examples
#### Example 1
Earlier, you were asked to find the width of the rectangle.
Recall that the area of a rectangle equals the length times the width, so to find the width, we must divide the area by the length.
\begin{align*}\sqrt{\frac{30}{20}}\end{align*} = \begin{align*}\sqrt{\frac{3}{2}}\end{align*}.
Now we need to rationalize the denominator. Multiply the top and bottom of the fraction by \begin{align*}\sqrt{2}\end{align*}.
\begin{align*}\frac{\sqrt{3}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6}}{2}\end{align*}
Therefore, the width of the rectangle is \begin{align*}\frac{\sqrt{6}}{2}\end{align*}.
Simplify the following expressions using the Radical Rules you have learned.
#### Example 2
\begin{align*}\sqrt{\frac{1}{2}}\end{align*}
\begin{align*}\sqrt{\frac{1}{2}}=\frac{\sqrt{1}}{\sqrt{2}}=\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}= \frac{\sqrt{2}}{2}\end{align*}
#### Example 3
\begin{align*}\sqrt{\frac{64}{50}}\end{align*}
\begin{align*}\sqrt{\frac{64}{50}}=\sqrt{\frac{32}{25}}=\frac{\sqrt{16 \cdot 2}}{5}= \frac{4\sqrt{2}}{5}\end{align*}
#### Example 4
\begin{align*}\frac{4\sqrt{3}}{\sqrt{6}}\end{align*}
The only thing we can do is rationalize the denominator by multiplying the numerator and denominator by \begin{align*}\sqrt{6}\end{align*} and then simplify the fraction.
\begin{align*}\frac{4\sqrt{3}}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}}= \frac{4\sqrt{18}}{6}=\frac{4\sqrt{9 \cdot 2}}{6}= \frac{12\sqrt{2}}{6}=2\sqrt{2}\end{align*}
### Review
Simplify the following fractions.
1. \begin{align*}\sqrt{\frac{4}{25}}\end{align*}
2. \begin{align*}-\sqrt{\frac{16}{49}}\end{align*}
3. \begin{align*}\sqrt{\frac{96}{121}}\end{align*}
4. \begin{align*}\frac{5\sqrt{2}}{\sqrt{10}}\end{align*}
5. \begin{align*}\frac{6}{\sqrt{15}}\end{align*}
6. \begin{align*}\sqrt{\frac{60}{35}}\end{align*}
7. \begin{align*}8\frac{\sqrt{18}}{\sqrt{30}}\end{align*}
8. \begin{align*}\frac{12}{\sqrt{6}}\end{align*}
9. \begin{align*}\sqrt{\frac{208}{143}}\end{align*}
10. \begin{align*}\frac{21\sqrt{3}}{2\sqrt{14}}\end{align*}
Challenge Use all the Radical Rules you have learned to simplify the expressions.
1. \begin{align*}\sqrt{\frac{8}{12}} \cdot \sqrt{15}\end{align*}
2. \begin{align*}\sqrt{\frac{32}{45}} \cdot \frac{6\sqrt{20}}{\sqrt{5}}\end{align*}
3. \begin{align*}\frac{\sqrt{24}}{\sqrt{2}}+\frac{8\sqrt{26}}{\sqrt{8}}\end{align*}
4. \begin{align*}\frac{\sqrt{2}}{\sqrt{3}}+\frac{4\sqrt{6}}{\sqrt{3}}\end{align*}
5. \begin{align*}\frac{5\sqrt{5}}{\sqrt{12}}-\frac{2\sqrt{15}}{\sqrt{10}}\end{align*}
To see the Review answers, open this PDF file and look for section 5.6.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Rationalize the denominator
To rationalize the denominator means to rewrite the fraction so that the denominator no longer contains a radical.
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# Express the following numbers in standard form:$3908.78$
Last updated date: 11th Aug 2024
Total views: 363.5k
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363.5k+ views
Hint: Keep a decimal point into one digit and convert others into terms of $10$.
Now in question it is asked to convert the number in standard form,
It is difficult to read numbers like$1202000004240$ or $0.0000042852512$. To make it easy
to read very large and small numbers, we write them in standard form.
So first of all what is standard form,
Standard form is a way of writing down very large or very small numbers
easily.${{10}^{3}}=1000$, so $4\times {{10}^{3}}=4000$ . So $4000$ can be written as$4\times {{10}^{3}}$ . This idea can be used to write even larger numbers down easily in
standard form. Any number that we can write as a decimal number, between $1.0$ and
$10.0$, multiplied by a power of $10$, is said to be in standard form.
Standard form in Math is mentioned for basically decimal numbers, equations, polynomials,
linear equations, etc. Its correct definition could be explained better in terms of decimal
numbers and following certain rules.
As we know, the decimal numbers are the simplified form of fractions. Some fractions give
decimal numbers which have numbers after decimal in thousandths, hundredths or tenths
place. But there are some fractions, which gives a big decimal number.
To represent such big numbers, we use such simpler forms, which is also stated as Scientific
notation.
Let us take an example $43333.21$
So we can write $43333.21=4.333321\times {{10}^{4}}$……..(1)
In the (1) we can see it, So $4.333321\times {{10}^{4}}$is standard form of $43333.21$.
So in this way we can write the digits in standard form.
So generally Standard form is a way of writing down very large or very small numbers easily.
Now for us given number is $3908.78$ ,
So we have to convert $3908.78$ in standard form,
Multiplying and dividing by 100 to $3908.78$, We get
$=\dfrac{3908.78\times 100}{100}=\dfrac{390878}{100}$
Now simplifying it We get
$=\dfrac{3.90878\times {{10}^{5}}}{100}=\dfrac{3.90878\times {{10}^{5}}}{{{10}^{2}}}$…… (2)
Now we know the property that $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$,
So using the above property in (2) We get,
\begin{align} & =\dfrac{3.90878\times {{10}^{5}}}{{{10}^{2}}}=3.90878\times \dfrac{{{10}^{5}}}{{{10}^{2}}}=3.90878\times {{10}^{5-2}} \\ & =3.90878\times {{10}^{3}} \\ \end{align}
So $3.90878\times {{10}^{3}}$ is standard form of $3908.78$.
So we have got the standard form of $3908.78$ which is $3.90878\times {{10}^{3}}$.
Hence proved.
$3.90878\times {{10}^{3}}$ is standard form of $3908.78$ is the answer.
Note: So basically don’t jumble while converting the number. Such as $3908.78$ to $3.90878\times {{10}^{3}}$. Use the property accordingly Such as I had used this property
$\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$. Remember that the properties are used according
to the problem.
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# Mathematical Induction
Mathematical induction is a powerful, yet straight-forward method of proving statements whose "domain" is a subset of the set of integers. Usually, a statement that is proven by induction is based on the set of natural numbers. This statement can often be thought of as a function of a number n, where n = 1,2,3...
Proof by induction involves three main steps: proving the base of induction, forming the induction hypothesis, and finally proving that the induction hypothesis holds true for all numbers in the domain.
Proving the base of induction involves showing that the claim holds true for some base value (usually 0, 1, or 2). There are sometimes many ways to do this, and it can require some ingenuity. We will outline this with a simple example.
This is the base step. We simply prove that the statement holds true for at least one value of n. In the induction hypothesis step, we say that since the statement holds true for at least one value, we can assume that it will hold true for some arbitrary, fixed value of n, usually k. We can make this assumption since we know that it will at least be true for our base value (2 in our example)
This is the simplest, yet most powerful part of proof by induction. By forming the induction hypothesis, we can now use it to finish the proof.
In the final step, we prove that the induction hypothesis holds true for all values of n. To do this, we use the fact that the statement is true for n = k, and then check to see if it holds true for n = k + 1.
Using this fact, we can now state that n²≥2n for all n = 2, 3... At first it may seem too simple, but if you examine this proof, it is easy to see the logic behind it. We know that the statement is true for n = 2. So we can now assume that it is true up to some fixed number k, which is at least 2. By proving that it is true for k + 1, we now know that it is true for at least n = 3. So now k = 3, but since the statement is true for k + 1, n is at least 4. In this manner, we can repeat this pattern indefinitely. Therefore, the statement is proven true for all n.
While different statements can require different techniques to prove that the statement is true (in the base step and k + 1 step), the reasoning behind a proof by induction is always the same. You must always follow the three steps:
1) Prove the statement true for some small base value
(usually 0, 1, or 2)
2) Form the induction hypothesis by assuming the statement is true
up to some fixed value n = k
3) Prove the induction hypothesis holds true for n = k + 1
There is one very important thing to remember about using proof by induction. This technique can only be used to prove statements that have real valued inputs. If you think of the statement as a relation of functions (for example, prove f(x) ≤ g(x) for all x), the domain of these functions must be a subset of the integers. It cannot be used to prove statements true for non-integer values. For example, the proof that we did in the above example does not prove that (2.5)² ≥ 2(2.5). It only proves the statement true for integer values of k greater than 2 (k = 2, 3, ...) However, for many proofs involving statements based on subsets of the integers (usually natural numbers), proof by induction is the easiest method to use.
# Examples
1 | Prove the formula for the sum of the first n natural numbers
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## Functional Equation Problem from SMO, 2013 – Senior Section
Try this beautiful Functional Equation Problem from SMO, Singapore Mathematics Olympiad, 2013.
## Problem – Functional Equation (SMO Test)
Let M be a positive integer .It is known that whenever $|ax^2 + bx +c|\leq 1$ for all
$|x|\leq 1$ then $|2ax + b |\leq M$ for all $|x|\leq 1$. Find the smallest possible value of M.
• 4
• 5
• 6
• 10
### Key Concepts
Functional Equation
Function
Challenges and Thrills – Pre – College Mathematics
## Try with Hints
We cant this sum by assuming a,b,c as fixed quantity.
Let $f(x) = ax^2 + bx + c$.
Then $f(-1) = a – b + c$ ; $f(0) = c$ ; $f(1) = a + b + c$ ;
Try to do the rest of the sum …………………………..
Suppose $|f(x)|\leq 1$ for all $|x|\leq 1$ . Then
$|2ax + b| = | (x – \frac {1}{2} ) f(-1) – 2 f(0) x + (x+\frac {1}{2} f(1) |$
$\leq |x – \frac {1}{2}| + 2 |x| + |x + \frac {1}{2}|$
$\leq |x – \frac {1}{2} | + |x+\frac {1}{2}| + 2$
$\leq 4$
Now I guess you have already got the answer but if not ………….
From the last step we can conclude ,
$|2 x^2 – 1|\leq 1$ whenever $|x|\leq 4$ and $|2x| = 4$
is achieved at $x = \pm 1$.
Categories
# Understand the problem
[/et_pb_text][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway||||||||” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″ _i=”1″ _address=”0.0.0.1″]Find all the real Polynomials P(x) such that it satisfies the functional equation: $P(2P(x)) = 2P(P(x)) + P(x)^{2} \forall real x$.
[/et_pb_text][et_pb_tabs active_tab_background_color=”#0c71c3″ inactive_tab_background_color=”#000000″ _builder_version=”3.29.2″ tab_text_color=”#ffffff” tab_font=”||||||||” background_color=”#ffffff” hover_enabled=”0″ _i=”2″ _address=”0.1.0.2″][et_pb_tab title=”Hint 0″ _builder_version=”3.22.4″ _i=”0″ _address=”0.1.0.2.0″]Do you really need a hint? Try it first!
[/et_pb_tab][et_pb_tab title=”Hint 1″ _builder_version=”3.29.2″ _i=”1″ _address=”0.1.0.2.1″]Well, it is really good that the information polynomial is given! You should use that. What is the first thing that you check in a Polynomial Identity? Degree! Yes, check whether the degree of the Polynomial on both the LHS and RHS are the same or not. Yes, they are both the same $n^2$. But did you observe something fishy? [/et_pb_tab][et_pb_tab title=”Hint 2″ _builder_version=”3.29.2″ _i=”2″ _address=”0.1.0.2.2″]Now rewrite the equation as $P(2P(x)) - 2P(P(x)) = P(x)^{2}$. Do the Degree trick now… You see it right? Yes, on the left it is $n^2$ and on the RHS it is $2n$. So, there are two cases now… Figure them out!
[/et_pb_tab][et_pb_tab title=”Hint 3″ _builder_version=”3.29.2″ _i=”3″ _address=”0.1.0.2.3″]
Case 1: $2n = n^2$… i.e. P(x) is either a quadratic or a constant function. Case 2: $P(2P(x)) - 2P(P(x))$ has coefficient zero till $x^2n$. We will study case 1 now. Case 1: $2n = n^2$… i.e. P(x) is either a quadratic or a constant function. $P(2P(x)) - 2P(P(x)) = P(x)^{2}$ = $P(2y) - 2P(y) = y^{2}$ where $y = P(x)$. Now, expand using $P(x) = ax^2 + bx +c$, it gives $2ay^2 -c = y^2$… Now find out all such polynomials satisfying this property. For e.g. $\frac{x^2}{2}$ is a solution. If P(x) is constant, prove that $P(x) = 0 / \frac{-1}{2}$. [/et_pb_tab][et_pb_tab title=”Hint 4″ _builder_version=”3.29.2″ _i=”4″ _address=”0.1.0.2.4″]Case 2: $P(2y) - 2P(y) = y^{2}$. Assume a general form of P(x) = $latex$and show that P(x) must be quadratic or lesser degree by comparing coefficients as you have a quadratic on RHS and n degree polynomial of the LHS. Now, we have already solved it for quadratic or less degree. [/et_pb_tab][et_pb_tab title=”Techniques Revisited” _builder_version=”3.29.2″ _i=”5″ _address=”0.1.0.2.5″]
• Always Compare the Degree of Polynomials in identities like this. It provides a lot of information.
• Compare the coefficients of Polynomials on both sides to equalize the coefficient on both sides.
[/et_pb_tab][et_pb_tab title=”Food for Thought” _builder_version=”3.29.2″ hover_enabled=”0″ _i=”6″ _address=”0.1.0.2.6″]
• Find all polynomials $P(2P(x)) - 8P(P(x)) = P(x)^{2}$.
• Find all polynomials $P(cP(x)) – d P(P(x)) = P(x)^{2}$ depending on the values of c and d.
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Multiples
# What are the Multiples of 49 | All Multiples of 49 up to 1000
Written by Prerit Jain
Updated on: 12 Aug 2023
### What are the Multiples of 49 | All Multiples of 49 up to 1000
The first ten multiples of 49 are listed as follows: 49, 98, 147, 196, 245, 294, 343, 392, 441, and 490.
The multiples of 49 are a sequence of numbers that can be obtained by multiplying the number 49 with a sequence of natural numbers.
The difference between any two consecutive numbers in the sequence of multiples of 49 is always 49.
To find the multiples of 49, you need to perform a repeated addition of the number 49 or multiply the number 49 with a sequence of natural numbers.
Alternatively, the multiples of 49 are the numbers that when divided by 49 do not leave any remainder.
## What are the first five Multiples of 49
The multiples of 49 are a sequence of numbers that can be obtained by multiplying the number 49 with a sequence of natural numbers.
The difference between any two consecutive numbers in the sequence of multiples of 49 is always 49.
Alternatively, the multiples of 49 are the numbers that when divided by 49 do not leave any remainder.
To find the multiples of 49, you need to perform a repeated addition of the number 49 or multiply the number 49 with a sequence of natural numbers.
Step 1: To find the first multiple of 49, multiply 49 by 1. 49 x 1 = 49.
Step 2: To find the second multiple of 49, multiply 49 by 2. 49 x 2 = 98. Alternatively, you can add 49 to the first multiple to get the second multiple of 49. 49 + 49 = 98.
Step 3: Multiply 49 by 3 to get the third multiple of 49. 49 x 3 = 147. Alternatively, you can add 49 two times to the second multiple to get the third multiple of 49. 98 + 49 = 147.
Step 4: The fourth multiple of 49 can be found by multiplying 49 by 4. 49 x 4 = 196. Alternatively, you can add 49 three times to the third multiple to get the fourth multiple of 49. 147 + 49 = 196.
Step 5: To find the fifth multiple of 49, you need to multiply 49 by 5. 49 x 5 = 245. Alternatively, you can add 49 four times to the fourth multiple to get the fifth multiple of 49. 196 + 49 = 245.
Therefore, the first 5 multiples of 49 are listed as follows 49, 98, 147, 196, and 245.
## How to find the Multiples of 49
To find the multiples of 49 two methods can be used. The first few multiples of 49 are listed as follows 49, 98, 147, 196, 245, and so on. The multiples of 49 are a sequence of numbers that can be obtained by multiplying the number 49 with a sequence of natural numbers. The difference between any two consecutive numbers in the sequence of multiples of 49 is always 49.
### Finding multiples of 49 using the repeated addition method
The first multiple of 49 is 49 itself. To find the second multiple of 49, we add 49 to the first multiple, which gives 98. Continuing this process, we can obtain the third, fourth, and so on multiples of 49. An example of the first five multiples of 49 using the repeated addition method is given below:
49
49 + 49 = 98
98 + 49 = 147
147 + 49 = 196
196 + 49 = 245
### Finding the multiples of 49 using the multiplication method
We can find the multiples of 49 by multiplying 49 with a sequence of natural numbers. For example, the first five multiples of 49 obtained by multiplying 49 with a sequence of natural numbers are:
## Finding the first 20 multiples of 49
To find the first 20 multiples of 49 using the multiplication method, we can follow similar steps as above. First, we create the sequence of natural numbers from 1 to 20. Then, we multiply each number in the sequence by 49 to find the corresponding multiples. The first 20 multiples of 49 are given below:
## Difference between Multiples and Factors of 49
In mathematics multiples and factors are two very different concepts.
In summary, multiples of 49 are the products of 49 and any natural number, while factors of 49 are the numbers that divide 49 exactly without leaving a remainder.
## Solved Examples for Multiples of 49
1. Find the first 5 multiples of 49.
Solution: To find the first 5 multiples of 49, we can simply multiply 49 by each of the first 5 positive integers:
49 x 1 = 49
49 x 2 = 98
49 x 3 = 147
49 x 4 = 196
49 x 5 = 245
Therefore, the first 5 multiples of 49 are 49, 98, 147, 196, and 245.
1. What is the smallest multiple of 49 that is greater than 500?
Solution: To find the smallest multiple of 49 that is greater than 500, we can divide 500 by 49 and take the next integer value.
500 ÷ 49 = 10 remainder 10
The next integer value is 11. So, we can multiply 49 by 11 to get the smallest multiple of 49 that is greater than 500.
49 x 11 = 539
Therefore, the smallest multiple of 49 that is greater than 500 is 539.
1. Find the common factors of 49 and 98.
Solution: To find the common factors of 49 and 98, we can factor both numbers and find their common factors.
The prime factorization of 49 is 7 x 7, and the prime factorization of 98 is 2 x 7 x 7.
Therefore, the common factor of 49 and 98 is 7, which means their greatest common factor (GCF) is 7.
1. Is 1837 a multiple of 49?
Solution: To check if 1837 is a multiple of 49, we need to divide 1837 by 49 and see if the result is a whole number.
1837 ÷ 49 = 37 remainder 30
Since the remainder is not 0, 1837 is not a multiple of 49.
1. Express 980 as a product of its prime factors in terms of multiples of 49.
Solution: To express 980 as a product of its prime factors, we can factor it into its prime factors and use the prime factors of 49 to represent those factors in terms of multiples of 49.
980 = 2 x 2 x 5 x 7 x 7
The prime factorization of 49 is 7 x 7, so we can write:
980 = 2 x 2 x 5 x 7 x 7 = 2 x 2 x 5 x 49
Therefore, 980 can be expressed as a product of its prime factors in terms of multiples of 49 as 2 x 2 x 5 x 49.
## FAQs on Multiples of 49
### What is the significance of the number 49 in US history?
The number 49 is significant in US history because it was the year in which California was admitted as the 31st state in the Union, in 1849, during the Gold Rush era.
### What is the largest US bill that is a multiple of 49?
There is no US bill denomination that is a multiple of 49. The US Treasury only produces bills in denominations of \$1, \$2, \$5, \$10, \$20, \$50, and \$100.
### Are there any notable events or celebrations that occur on multiples of 49 in the US?
There are no major events or celebrations that occur on multiples of 49 in the US, although some individuals may choose to celebrate their 49th wedding anniversary, which is traditionally known as the “jade” anniversary.
### What is the significance of the number 49 in mathematics and science?
In mathematics, the number 49 is a perfect square, as it is the square of 7. It is also a composite number, which means it can be divided by factors other than 1 and itself. In science, the number 49 has no specific significance or use.
### What are some common multiples of 49 in the US measurement system?
In the US measurement system, common multiples of 49 include 49 feet, 98 feet, 147 feet, and so on. However, these are not commonly used measurements.
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# Writing Down a Pair of Integers Whose Sum is -5
## What is an Integer?
An integer is a whole number, meaning it has no fractional components. Integers can be positive, negative, or zero. Examples of integers include -2, -1, 0, 1, and 2.
## Finding a Pair of Integers Whose Sum is -5
To find a pair of integers whose sum is -5, you must use basic algebra. Let’s say you are looking for two numbers, x and y. You know that their sum is -5, so you can put that into an equation as follows: x + y = -5.
Now that you have an equation, you can solve for either x or y. Let’s say you want to solve for x. You can subtract y from both sides of the equation, so you get x = -5 – y. If you are looking for two numbers whose sum is -5, you can choose any value for y and then calculate the value of x.
For example, if you choose y = -3, then x = -5 – (-3), or x = -2. This means that the pair of integers (x, y) whose sum is -5 is (-2, -3).
## Other Pairs of Integers Whose Sum is -5
As mentioned above, you can choose any value for y and then calculate the value of x. Here are some other pairs of integers whose sum is -5:
• (-4, -1)
• (-3, -2)
• (-2, -3)
• (-1, -4)
• (0, -5)
## Conclusion
As you can see, it is possible to find a pair of integers whose sum is -5. You can use basic algebra to solve for either x or y, and then calculate the value of the other. There are several pairs of integers whose sum is -5, and you can find them using the same method.
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## Transcription
2 What You Should Learn Sketch graphs of equations. Find x- and y-intercepts of graphs of equations. Use symmetry to sketch graphs of equations. Find equations of and sketch graphs of circles. Use graphs of equations in solving real-life problems. 2
3 The Graph of an Equation 3
4 The Graph of an Equation We have used a coordinate system to represent graphically the relationship between two quantities. There, the graphical picture consisted of a collection of points in a coordinate plane. Frequently, a relationship between two quantities is expressed as an equation in two variables. For instance, y = 7 3x is an equation in x and y. An ordered pair (a, b) is a solution or solution point of an equation in x and y if the equation is true when a is substituted for x and b is substituted for y. 4
5 The Graph of an Equation For instance, (1, 4) is a solution of y = 7 3x because 4 = 7 3(1) is a true statement. In this section you will review some basic procedures for sketching the graph of an equation in two variables. The graph of an equation is the set of all points that are solutions of the equation. 5
6 The Graph of an Equation The basic technique used for sketching the graph of an equation is the point-plotting method. 6
7 Example 2 Sketching the Graph of an Equation Sketch the graph of y = 7 3x. Solution: Because the equation is already solved for y, construct a table of values that consists of several solution points of the equation. For instance, when x = 1, y = 7 3( 1) = 10 which implies that ( 1, 10) is a solution point of the graph. 7
8 Example 2 Solution cont d From the table, it follows that ( 1, 10), (0, 7), (1, 4), (2, 1), (3, 2), and (4, 5) are solution points of the equation. 8
9 Example 2 Solution cont d After plotting these points, you can see that they appear to lie on a line, as shown in Figure Figure 1.14 The graph of the equation is the line that passes through the six plotted points. 9
10 Intercepts of a Graph 10
11 Intercepts of a Graph It is often easy to determine the solution points that have zero as either the x coordinate or the y coordinate. These points are called intercepts because they are the points at which the graph intersects or touches the x- or y-axis. 11
12 Intercepts of a Graph It is possible for a graph to have no intercepts, one intercept, or several intercepts, as shown in Figure No x-intercepts; one y-intercept Three x-intercepts; one y-intercept One x-intercept; two y-intercepts No intercepts Figure 1.18 Note that an x-intercept can be written as the ordered pair (x, 0) and a y-intercept can be written as the ordered pair (0, y). 12
13 Intercepts of a Graph Some texts denote the x-intercept as the x-coordinate of the point (a, 0) [and the y-intercept as the y-coordinate of the point (0, b)] rather than the point itself. Unless it is necessary to make a distinction, we will use the term intercept to mean either the point or the coordinate. 13
14 Example 4 Finding x- and y-intercepts Find the x- and y-intercepts of the graph of y = x 3 4x. Solution: Let y = 0. Then 0 = x 3 4x = x(x 2 4) has solutions x = 0 and x = ±2. x-intercepts: (0, 0), (2, 0), ( 2, 0) 14
15 Example 4 Solution cont d Let x = 0. Then y = (0) 3 4(0) has one solution, y = 0. y-intercept: (0, 0) See Figure Figure
16 Symmetry 16
17 Symmetry Graphs of equations can have symmetry with respect to one of the coordinate axes or with respect to the origin. Symmetry with respect to the x-axis means that if the Cartesian plane were folded along the x-axis, the portion of the graph above the x-axis would coincide with the portion below the x-axis. 17
18 Symmetry Symmetry with respect to the y-axis or the origin can be described in a similar manner, as shown in Figure x-axis symmetry y-axis symmetry Origin symmetry Figure 1.20 Knowing the symmetry of a graph before attempting to sketch it is helpful, because then you need only half as many solution points to sketch the graph. 18
19 Symmetry There are three basic types of symmetry, described as follows. 19
20 Symmetry You can conclude that the graph of y = x 2 2 is symmetric with respect to the y-axis because the point ( x, y) is also on the graph of y = x 2 2. (See the table below and Figure 1.21.) y-axis symmetry Figure
21 Symmetry 21
22 Example 5 Testing for Symmetry Test y = 2x 3 for symmetry with respect to both axes and the origin. Solution: x-axis: y = 2x 3 y = 2x 3 y-axis: y = 2x 3 y = 2( x) 3 y = 2x 3 Write original equation. Replace y with y. Result is not an equivalent equation. Write original equation. Replace x with x. Simplify. Result is not an equivalent equation. 22
23 Example 5 Solution cont d Origin: y = 2x 3 y = 2( x) 3 y = 2x 3 y = 2x 3 Write original equation. Replace y with y and x with x. Simplify. Equivalent equation Of the three tests for symmetry, the only one that is satisfied is the test for origin symmetry (see Figure 1.22). Figure
24 Symmetry Throughout this course, you will learn to recognize several types of graphs from their equations. For instance, you will learn to recognize that the graph of a second degree equation of the form y = ax 2 + bx + c is a parabola. The graph of a circle is also easy to recognize. 24
25 Circles 25
26 Circles Consider the circle shown in Figure A point (x, y) is on the circle if and only if its distance from the center (h, k) is r. By the Distance Formula, By squaring each side of this equation, you obtain the standard form of the equation of a circle. Figure
27 Circles From this result, you can see that the standard form of the equation of a circle with its center at the origin, (h, k) = (0, 0), is simply x 2 + y 2 = r 2. Circle with center at origin 27
28 Example 8 Finding the Equation of a Circle The point (3, 4) lies on a circle whose center is at ( 1, 2), as shown in Figure Write the standard form of the equation of this circle. Figure
29 Example 8 Solution The radius of the circle is the distance between ( 1, 2) and (3, 4). Distance Formula Substitute for x, y, h, and k. Simplify. Simplify. Radius 29
30 Example 8 Solution cont d Using (h, k) = ( 1, 2) and r = circle is (x h) 2 + (y k) 2 = r 2 [x ( 1)] 2 + (y 2) 2 = ( ) 2 (x + 1) 2 + (y 2) 2 = 20., the equation of the Equation of circle Substitute for h, k, and r. Standard form 30
31 Application 31
32 Application In this course, you will learn that there are many ways to approach a problem. Three common approaches are illustrated in Example 9. A Numerical Approach: Construct and use a table. A Graphical Approach: Draw and use a graph. An Algebraic Approach: Use the rules of algebra. 32
33 Example 9 Recommended Weight The median recommended weight y (in pounds) for men of medium frame who are 25 to 59 years old can be approximated by the mathematical model y = 0.073x x , 62 x 76 where x is the man s height (in inches). (Source: Metropolitan Life Insurance Company) a. Construct a table of values that shows the median recommended weights for men with heights of 62, 64, 66, 68, 70, 72, 74, and 76 inches. 33
34 Example 9 Recommended Weight cont d b. Use the table of values to sketch a graph of the model. Then use the graph to estimate graphically the median recommended weight for a man whose height is 71 inches. c. Use the model to confirm algebraically the estimate you found in part (b). 34
35 Example 9(a) Solution You can use a calculator to complete the table, as shown below. 35
36 Example 9(b) Solution cont d The table of values can be used to sketch the graph of the equation, as shown in Figure From the graph, you can estimate that a height of 71 inches corresponds to a weight of about 161 pounds. Figure
37 Example 9(c) Solution cont d To confirm algebraically the estimate found in part (b), you can substitute 71 for x in the model. y = 0.073(71) (71) So, the graphical estimate of 161 pounds is fairly good. 37
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# Find the moment of inertia about axis 1 (diameter of hemisphere) of solid hemisphere $(m,R)$.(A) $\dfrac{{m{R^2}}}{5}$(B) $\dfrac{2}{5}m{R^2}$(C) $\dfrac{2}{3}m{R^2}$(D) $\dfrac{{173}}{{320}}m{R^2}$
Last updated date: 20th Jun 2024
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Hint First of all get the mass $dm$ of the disc by using the expression:
$dm = \left( {\dfrac{m}{{{V_H}}}} \right) \times {V_D}$
where, $m$ is the mass of hemisphere
${V_H}$ is the volume of hemisphere
${V_D}$ is the volume of disc
Now, use the formula
${I_{yy'}} = {I_{cm}} + m{r^2}$
Then, integrate it with the limit from $0$ to $R$
After solving we will get the value of ${I_{yy'}}$ then, use the value of $\rho = \dfrac{{3m}}{{2\pi {R^3}}}$
Then, if $I$ displace them parallel to the axis and again join them to complete a sphere
$2I = \dfrac{2}{5}m{R^2}$
Complete Step by Step Solution
Suppose that hemisphere is constructed by small discs
Mass $dm$of disc= $dm = \left( {\dfrac{m}{{{V_H}}}} \right) \times {V_D}$
$\therefore dm = \left( {\dfrac{m}{{\dfrac{2}{3}\pi {R^3}}}} \right)\left( {\pi {y^2}dx} \right) \\ dm = \rho \left( {\pi {y^2}dx} \right) \\$
Because, $\rho = \dfrac{m}{{\dfrac{2}{3}\pi {R^3}}}$
According to parallel and perpendicular axis theorem we get
${I_{yy'}} = {I_{cm}} + m{r^2}$
$\therefore d{I_{yy'}} = (dm)\dfrac{{{y^2}}}{4} + (dm){x^2}$
Now, integrating both sides
${I_{yy'}} = \int\limits_0^R {(dm)\dfrac{{{y^2}}}{4} + } \int\limits_0^R {dm} {x^2} \\ \Rightarrow \int\limits_0^R {(\rho \pi {y^2}dx)\dfrac{{{y^2}}}{4} + } \int\limits_0^R {(\rho \pi {y^2}dx){x^2}} \\$
Now, put the limits
${I_{yy'}} = \dfrac{{\rho \pi }}{4}\left[ {\dfrac{{{R^5}}}{5} + {R^5} - 2{R^2}\dfrac{{{R^3}}}{3}} \right] + \left[ {\rho \pi {R^5}\left( {\dfrac{{{R^3}}}{3}} \right) - \rho \pi \left( {\dfrac{{{R^5}}}{5}} \right)} \right] \\ = \rho \pi {R^5}\left[ {\dfrac{1}{4} + \dfrac{1}{{20}} - \dfrac{1}{4} \times \dfrac{2}{3}} \right] + \rho \pi {R^5}\left[ {\dfrac{1}{3} - \dfrac{1}{5}} \right] \\ = \rho \pi {R^5}\left( {\dfrac{4}{{15}}} \right) \\$
Now, putting the value of $\rho$ in the above equation
${I_{yy'}} = \dfrac{{3m}}{{2\pi {R^3}}}\pi {R^5}\dfrac{4}{{15}} = \dfrac{2}{5}m{R^2} \\ \therefore I = \dfrac{2}{5}m{R^2} \\$
But if $I$displaces the parallel to axis and join them to complete the sphere then,
$2I = \dfrac{2}{5}m{R^2}$
After cancelling 2 on both sides we get
$I = \dfrac{{m{R^2}}}{5}$
Hence, option (A) is the correct answer
Note Moment of Inertia of continuous bodies is $\dfrac{2}{5}m{R^2}$
Moment of Inertia of disc about centre of mass is $\dfrac{{m{R^2}}}{2}$
Moment of inertia of sphere about centre of mass is $\dfrac{2}{5}m{R^2}$
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# Problem of the Week
## Updated at Jan 24, 2022 8:49 AM
To get more practice in equation, we brought you this problem of the week:
How would you solve the equation $$4\times \frac{5}{{(2+t)}^{2}}=\frac{20}{49}$$?
Check out the solution below!
$4\times \frac{5}{{(2+t)}^{2}}=\frac{20}{49}$
1 Simplify $$4\times \frac{5}{{(2+t)}^{2}}$$ to $$\frac{20}{{(2+t)}^{2}}$$.$\frac{20}{{(2+t)}^{2}}=\frac{20}{49}$2 Multiply both sides by $${(2+t)}^{2}$$.$20=\frac{20}{49}{(2+t)}^{2}$3 Simplify $$\frac{20}{49}{(2+t)}^{2}$$ to $$\frac{20{(2+t)}^{2}}{49}$$.$20=\frac{20{(2+t)}^{2}}{49}$4 Multiply both sides by $$49$$.$20\times 49=20{(2+t)}^{2}$5 Simplify $$20\times 49$$ to $$980$$.$980=20{(2+t)}^{2}$6 Divide both sides by $$20$$.$\frac{980}{20}={(2+t)}^{2}$7 Simplify $$\frac{980}{20}$$ to $$49$$.$49={(2+t)}^{2}$8 Take the square root of both sides.$\pm \sqrt{49}=2+t$9 Since $$7\times 7=49$$, the square root of $$49$$ is $$7$$.$\pm 7=2+t$10 Switch sides.$2+t=\pm 7$11 Break down the problem into these 2 equations.$2+t=7$$2+t=-7$12 Solve the 1st equation: $$2+t=7$$.1 Subtract $$2$$ from both sides.$t=7-2$2 Simplify $$7-2$$ to $$5$$.$t=5$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$t=5$13 Solve the 2nd equation: $$2+t=-7$$.1 Subtract $$2$$ from both sides.$t=-7-2$2 Simplify $$-7-2$$ to $$-9$$.$t=-9$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$t=-9$14 Collect all solutions.$t=5,-9$Donet=5,-9
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# Graphs Using Slope-Intercept Form
## Use the y-intercept and the 'rise over run' to graph a line
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Practice Graphs Using Slope-Intercept Form
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Graphs of Lines from Equations
Can you graph the linear function on a Cartesian grid?
### Watch This
Khan Academy Slope and y-Intercept Intuition
### Guidance
The graph of any linear function can be plotted using the slope-intercept form of the equation.
• Step 1: Solve the equation for if it is not already in the form .
• Step 2: To graph the function, start by plotting the -intercept.
• Step 3: Use the slope to find another point on the line. From the -intercept, move to the right the number of units equal to the denominator of the slope and then up or down the number of units equal to the numerator of the slope. Plot the point.
• Step 4: Connect these two points to form a line and extend the line.
Note: You can repeat Step 3 multiple times in order to find more points on the line if you wish.
Because the equations of horizontal and vertical lines are special, these types of lines can be graphed differently:
• The graph of a horizontal line will have an equation of the form where is the y-intercept of the line. You can simply draw a horizontal line through the y-intercept to sketch the graph.
• The graph of a vertical line will have an equation of the form , where is the x-intercept of the line. You can simply draw a vertical line through the x-intercept to sketch the graph.
#### Example A
For the following linear function, state the -intercept and the slope: .
Solution:
The first step is to rewrite the equation in the form . To do this, solve the equation for ‘’.
The -intercept is (0, –3) and the slope is .
#### Example B
Graph the linear function on a Cartesian grid.
Solution:
The -intercept is (0, 7) and the slope is . Begin by plotting the -intercept on the grid.
From the -intercept, move to the right (run) 5 units and then move downward (rise) 3 units. Plot a point here.
Join the points with a straight line. Use a straight edge to draw the line.
#### Example C
Plot the following linear equations on a Cartesian grid.
i)
ii)
Solution:
i) A line that has as its equation passes through all points that have –3 as the coordinate. The line also has a slope that is undefined. This line is parallel to the -axis.
ii) A line that has as its equation passes through all points that have 5 as the coordinate. The line also has a slope of zero. This line is parallel to the -axis.
#### Concept Problem Revisited
Plot the linear function on a Cartesian grid.
The first step is to rewrite the function in slope-intercept form.
The slope of the line is and the -intercept is (0, 4)
Plot the -intercept at (0, 4). From the -intercept, move to the right 4 units and then move upward 5 units. Plot the point. Using a straight edge, join the points.
### Vocabulary
Slope-Intercept Form
The slope-intercept form is one method for writing the equation of a line. The slope-intercept form is where refers to the slope and identifies the -intercept. This form is used to plot the graph of a linear function.
### Guided Practice
1. Using the slope-intercept method, graph the linear function
2. Using the slope-intercept method, graph the linear function
3. Graph the following lines on the same Cartesian grid. What shape is formed by the graphs?
(a)
(b)
(c)
(d)
1. The slope of the line is and the -intercept is (0, –1). Plot the -intercept. Apply the slope to the -intercept. Use a straight edge to join the two points.
2. Write the equation in slope-intercept form.
The slope is and the -intercept is (0, –5). Plot the -intercept. Apply the slope to the -intercept. Use a straight edge to join the two points.
3. There are four lines to be graphed. The lines and are lines with a slope of zero and are parallel to the -axis. The lines and are lines that have a slope that is undefined and are parallel to the -axis. The shape formed by the intersections of the lines is a rectangle.
### Practice
For each of the following linear functions, state the slope and the -intercept:
Using the slope-intercept method, graph the following linear functions:
Graph the following linear equations and state the slope of the line:
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 4.4.
### Vocabulary Language: English
Cartesian Plane
Cartesian Plane
The Cartesian plane is a grid formed by a horizontal number line and a vertical number line that cross at the (0, 0) point, called the origin.
linear equation
linear equation
A linear equation is an equation between two variables that produces a straight line when graphed.
Linear Function
Linear Function
A linear function is a relation between two variables that produces a straight line when graphed.
Slope-Intercept Form
Slope-Intercept Form
The slope-intercept form of a line is $y = mx + b,$ where $m$ is the slope and $b$ is the $y-$intercept.
1. [1]^ License: CC BY-NC 3.0
2. [2]^ License: CC BY-NC 3.0
3. [3]^ License: CC BY-NC 3.0
4. [4]^ License: CC BY-NC 3.0
5. [5]^ License: CC BY-NC 3.0
6. [6]^ License: CC BY-NC 3.0
7. [7]^ License: CC BY-NC 3.0
8. [8]^ License: CC BY-NC 3.0
9. [9]^ License: CC BY-NC 3.0
### Explore More
Sign in to explore more, including practice questions and solutions for Graphs Using Slope-Intercept Form.
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##### Geometry – Triangles and Trapezoids. All Triangles are related to rectangles or parallelograms : Each rectangle or parallelogram is made up of two triangles!
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the new team logo. All submitted entries had to use two parallelograms and one triangle with the dimensions given below. What would the total area be for the winning logo? Area of triangle= b x h ÷ 2 A = 6 x 6 ÷ 2 A = 18 Area of parallelogram = b x h A = 8 x 8 A = 64 Total area of new logo = 1 triangle + 2 parallelograms A = 18 + (2 x 64) A = 146 The total/
##### AREA OF SHAPES. MEMORY GAME Have a look at these Formulas below and try and memorise them SHAPEPICTUREFORMULA Square Rectangle Triangle Parallelogram.
AREA OF SHAPES MEMORY GAME Have a look at these Formulas below and try and memorise them SHAPEPICTUREFORMULA Square Rectangle Triangle Parallelogram Trapezium Kite a a a ab ba h MEMORY GAME Have a look at these Formulas below and try and memorise them SHAPEPICTUREFORMULA Squarea x a = a 2 Rectanglea x b = ab Triangle/2 ab a a a ab ba h Which Formula’s is missing? SHAPEPICTUREFORMULA Square Rectanglea x b = ab Triangle 1 / 2 x a x b = 1 / 2 ab Parallelogram Trapezium 1 / 2 x (a + b) x h Kite 1 / 2 x a x b = 1/
##### Areas of Triangles, Parallelograms, & Trapezoids.
Areas of Triangles, Parallelograms, & Trapezoids Rectangle Area = number of square units contained in the figure. Calculated A = lw l w Rectangle Sometimes referred to as base and height A = bh b h Parallelogram Especially, when we are looking at a parallelogram A = bh b h Parallelogram Height is always PERPENDICULAR to the base A = bh b h Triangles Now let’s consider triangles b h How would we calculate the Area of the triangle? Triangles A = ½ bh/
##### POLYGONS and AREA Classifying Polygons Angles in Polygons Area of Squares and Rectangles Area of Triangles Area of Parallelograms Area of Trapezoids Circumference.
#8 POLYGONS and AREA Area of Triangles POLYGONS and AREA Area of Triangles MENU STUDENT PROBLEMS Find the AREA of this regular octogon: 10m 12m 5 Area of 1 triangle: # of triangles: 30 16 Area: 30x16 =480m 2 POLYGONS and AREA Area of Parallelograms MENU POLYGONS and AREA Area of Parallelograms MENU To calculate the area of a parallelogram… Just Multiply base and height B H Area of a Parallelogram h b *Base and height make a right angle. POLYGONS and AREA Area of Parallelograms MENU POLYGONS and AREA Area of/
##### Geometry Geometry: Part IV Area and Volume By Dick Gill, Julia Arnold and Marcia Tharp for Elementary Algebra Math 03 online.
to you that this triangle is half of some parallelogram? If we flip this triangle and join the two triangles, we get a parallelogram whose area is defined by A = bh. b hThe area of the triangle then will be A = ½ bh. Do you think you can envision every triangle as half of a parallelogram with the same height and base as the triangle? Before you click, imagine each triangle as half of a parallelogram with the same height/
##### AREA OF TRIANGLES. What are the base and height of a triangle? The base can be any of the three sides. The height is the distance from the vertex (corner)
related to parallelograms? Every triangle is half of a parallelogram This means that two of the same triangle combine to form a parallelogram How are the area formulas for triangles and parallelograms related? Since a triangle is half of a parallelogram, the area of a triangle is half of the area of the parallelogram with the same base and height Area of a triangle = ½×base×height A = ½×b×h or A = Review: What is the area of this parallelogram? What is the area of this triangle? Area of parallelogram = base/
##### What does the word “polygon” mean? What is the smallest number of sides a polygon can have? What is the largest number of sides a polygon can have?
= 4 4 A = bh Area of a parallelogram. Substitute 4 for b and 4 for h. A composite figure is made up of basic geometric shapes such as rectangles, triangles, trapezoids, and circles. To find the area of a composite figure, find the areas of the geometric shapes and then add the areas. Additional Example 3: Finding Area and Perimeter of a Composite Figure Find the perimeter and area of the figure. The length of the side that is not/
##### Exploring Area for Sixth Grade Click the triangle to watch a short introduction.
rectangle on the left is 24 sq mm Rectangles and Parallelograms A parallelogram can be cut and reconstructed to form a rectangle. This is why rectangles and parallelograms have the same formula! Study the diagram below to see the relationship! Formula for Area of a Parallelogram: Area = base x height or A = b x h Triangles A triangle is related to parallelograms, just like parallelograms are related to rectangles. Study the diagram below to/
##### Confidential2 1.Find the area of a parallelogram whose one side is 5 m and the corresponding altitude is 3 m. Answer: 15 m² 2. Find the area of a parallelogram.
three sides. Examples of triangles: Confidential6 Area of a Triangle The area of a triangle is given by "half of base times height“. Area = where b is the length of the base h is the height of the triangle. Note: The height is the length of a line segment perpendicular to the base of the triangle. 1212 x b x h Confidential7 Finding Area of Triangle Example1: Find the area of the triangle whose base is 10 cm and height is 4 cm/
##### Areas of Parallelograms and Triangles Lesson 7-1.
Areas of Parallelograms and Triangles Lesson 7-1 Thm 7-1 Area of a Rectangle For a rectangle, A=bh. (Area = base · height) h b AREA OF A PARALLELOGRAM To do this let’s cut the left triangle and… h b slide it… h h b h h b h h b h h b …thus, changing it to a rectangle. What is the area of the rectangle? h b Thm 7-2 Area of a Parallelogram For a/
##### Copyright©amberpasillas2010. The area of a rectangle is equal to the base times the height. Also known as length times width. height base (h) (b) A =
happens when instead we use 2 isosceles triangles. Given an isosceles triangle Make a similar triangle, Given an isosceles triangle Make a similar triangle, flip it and put both triangles next to each other What polygon is this? A Parallelogram copyright©amberpasillas2010 base height h How do you find the area of the parallelogram? copyright©amberpasillas2010 9 cm 5 cm6 cm 3 cm # 4 Area of a Triangle 8 m 6 m A = 8/
##### Quadrilaterals 8 8.1Properties of Parallelograms Chapter Summary Case Study 8.2Conditions for Parallelograms 8.5Mid-point Theorem 8.3Rhombuses, Rectangles,
can be proved by using the properties of parallelograms. The straight line joining the mid-points of 2 sides of a triangle has some properties as described below. These properties are called the mid-point theorem. Mid-point Theorem The line segment joining the mid-points of 2 sides of a triangle is parallel to the third side and is half the length of the third side. In the figure/
##### Areas of Parallelograms and Triangles LESSON 11–1.
= 8 in., h = 15 in. Example 4 A.base = 56 in. and height = 10 in. B.base = 28 in. and height = 40 in. C.base = 20 in. and height = 56 in. D.base = 26 in. and height = 38 in. ALGEBRA The height of a triangle is 12 inches more than its base. The area of the triangle is 560 square inches. Find the base and the height. Areas of Parallelograms and Triangles LESSON 11–1
##### Geometry Unit. Identify the following shapes Important Definitions O Parallelogram: a four sided plane with opposite parallel sides. O Trapezoid: a quadrilateral.
diagonal line dividing the rectangle in half. What two shapes make up the rectangle?? Decomposing shapes 1. Draw a parallelogram on your paper 2. Draw a diagonal line dividing up your parallelogram What shapes make up a parallelogram?? Area of Triangles Since Triangles are half of parallelograms, squares, and rectangles use that knowledge and see if you can find the area of the triangle below: Try this one….. Area of a Triangle Click here for practice with area of triangle
##### Splash Screen. Lesson Menu Five-Minute Check (over Chapter 10) NGSSS Then/Now New Vocabulary Postulate 11.1: Area Addition Postulate Key Concept: Area.
1–6) Find perimeters and areas of parallelograms. Find perimeters and areas of triangles. Vocabulary base of a parallelogram height of a parallelogram base of a triangle height of a triangle Concept 1 Concept 2 Example 1 Perimeter and Area of a Parallelogram Find the perimeter and area of PerimeterSince opposite sides of a parallelogram are congruent, RS UT and RU ST. So UT = 32 in. and ST = 20 in. Example 1 Perimeter and Area of a Parallelogram Area Find the height of the parallelogram. The height forms a/
##### SPI Solve contextual problems that require calculating the area of triangles and parallelograms. SPI Decompose irregular shapes to find.
reason why understanding perimeter is important. You will share this with a partner. PART 2 Lesson 3 Area of Rectangles Lesson 4 Area of Parallelograms Lesson 5 Area of Triangles Lesson 6 Area of Irregular Shapes SPI Lesson 3 Area of Rectangles SPI 0506.4.2 Decompose irregular shapes to find perimeter and area. Essential Questions How is finding area different from finding perimeter? Assessment After Lesson 3 you will complete Quick Check 12-4/
##### Shape and Space 2 PGCE Seminar Dr David Bolden 0191 334 8325 1.
area of parallelograms 7 We already know how to calculate the area of a rectangle. Well, a parallelogram is simply a sheared rectangle. Area of the rectangle= base × height parallelogram Area of the rectangle base height perpendicular height 8 Area of any parallelogram = base height Perpendicular height Does this works for any parallelogram? base height YES 9 The area of trapeziums 10 cm 3 cm 4 cm 6 cm Area of trapezium = area of parallelogram + area of triangle/the middle. 12 3 14 And all the way round the /
##### Area I will find the area of triangles, special quadrilaterals and polygons in solving real-world and mathematical problems.
shape. Add at least one triangle and combine it with a parallelogram and/or rectangle. Name: Real World Design a 2,000 square foot house. The customer hates rectangular houses and wants a more unique shape. Add at least one triangle and combine it with a parallelogram and/or rectangle. Name: How do you find the area of a figure made up of parallelograms and triangles? OLS Lesson Fundamentals of Geometry and Algebra – Unit 2 Lesson 1/
##### Area Honors Geometry.
the same way for other parallelograms? Parallelogram Area Theorem: The area of a parallelogram is given by the formula A=bh, where A is the Area, b is the length of the base, and h is the height of the parallelogram. What formula does this give us for the area of a triangle? Triangles What formula does this give us for the area of a triangle? Area of a triangle? A triangle is half of a parallelogram, therefore its area is given by the formula/
##### This material is made freely available at www.njctl.org and is intended for the non-commercial use of students and teachers. These materials may not be.
Slide 9 for an example.) Setting the PowerPoint View Table of Contents Click on a topic to go to that section Area of Rectangles Area of Irregular Figures Area of Shaded Regions Common Core: 6.G.1-4 Area of Parallelograms Area of Triangles Area of Trapezoids Mixed Review: Area 3-Dimensional Solids Surface Area Volume Polygons in the Coordinate Plane Area of Rectangles Return to Table of Contents Area - The number of square units (units 2 ) it takes to cover the/
##### Perimeter Area Volume.
in • 4 in A = 24 in2 5 in 4 in Triangles We can use the formula for a parallelogram to help us find the area of any triangle. Let’s use our GeoBoards to help us determine this formula. How do you find the area of a triangle? To find the area of a triangle use the formula Area = ½ base ● height and solve using the dimensions given. REMEMBER F.S.S.L!!! A/
##### Heron Heron of Alexandria (c. 10–70 AD) was an ancient Greek mathematician and engineer. He is considered the greatest experimenter of antiquity and his.
the height. Area of a Parallelogram Theorem The area of a parallelogram is the product of a base and its corresponding height. A = bh Area of a Parallelogram Theorem The area of a parallelogram is the product of a base and its corresponding height. A = bh Example Find the area of parallelogram PQRS. Example What is the height of a parallelogram that has an area of 7.13 m 2 and a base 2.3 m long? Example Find the area of each triangle or parallelogram. 1.2/
##### Areas of Polygons & Circumference of Circles
a. A right triangle was cut from one end of the rectangle and slid to the other side to create a non-rectangular parallelogram. c. Based on your observation, write a sentence describing the area of a parallelogram. d. Write a formula for the area of a parallelogram. The area of the rectangle is equal to the area of the parallelogram. The width of the rectangle is equal to the height of the parallelogram and the length is equal/
##### Activity: 1)Draw a rectangle measuring 6cm by 3cm using a ruler in your notebook 2)Draw a triangle inside the rectangle so that: a) at least one side.
) shade the triangle d) estimate the area of the triangle 3) Cut out the rectangle then cut out the triangle. 4) Arrange the unshaded pieces to cover the triangle. 5) What is the area of the triangle and how do you know? What is the formula for the area of a parallelogram? = A = bh What is the connection between a parallelogram and a triangle? = a triangle is ½ a parallelogram What would be the formula for the area of a triangle? A/
##### Parallelograms, Trapezoids and Triangles Section 3-4-5.
+c where a, b and c are the lengths of the sides. The area of a triangle is A= ½ bh where b is the base and h is the height perpendicular to the base. To illustrate why the area formula works, make a parallelogram out of two identical triangles. The height and base of the parallelogram is the same as that of the triangle. The area of the parallelogram is base times height and is twice the desired area of the triangle.
##### Perimeters, areas and other measurements In many careers it is essential to have the ability to recognize 2-dimensional images as 3-dimensional objects.
the opposite corner. This forms two congruent triangles. Finding the area of a triangle from here is fairly simple, just take ½ of the area of the square, rectangle, or parallelogram, So, the area A of a triangle is half the product of its base b and its height h A = ½ bh Find the area of the triangle: It may be necessary, when working with an obtuse triangle, to look outside the triangle to find the height. Notice how/
##### Area of Rectangles, Squares, Parallelograms, Triangles, and Trapezoids.
7 or 28 cm. 2. Area of Parallelograms For a parallelogram, we use the terms base and height. Remember, the height is measured straight up and down (like the doctor measures your height!). Therefore, the height of this parallelogram is 4 cm., not 5 cm. 7 cm 5 cm 4 cm The area of a parallelogram can be found by multiplying the base times the height. Area of Triangles 7 cm 4 cm Starting/
##### Area of Parallelograms & Triangles Jacob Shaffer March 31, 2014.
Area of Parallelograms & Triangles Jacob Shaffer March 31, 2014 Find the Perimeter of Disneyland P = 2L + 2W Perimeter Formula of a Parallelogram P = 2 (3,500 ft) + 2 (2,500 ft) Substitute 3,500 ft for Length and 2,500 ft for width P = 7,000 + 5,000 Multiply P = 12,000 ft Add to find total perimeter Area of a Parallelogram Area Example Find the area of this parallelogram: Area Problem 1 Find the area of the parallelogram: A/
##### Area of Quadrilaterals SECTION 5.02. After completing this lesson, you will be able to say: I can use composition and decomposition to determine the area.
A right trapezoid has a side that is perpendicular to its parallel bases. Area of Quadrilaterals The area of quadrilaterals can be found by decomposing the shape into rectangles and triangles. Recall the formulas for calculating the area for both shapes. Area of Parallelogram How can we decompose this parallelogram into triangles and rectangles? Area of Parallelogram A parallelogram can be decomposed into two right triangles with a rectangle in between them. Drawing vertical lines from the corners/
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Ex 7.1
Chapter 7 Class 10 Coordinate Geometry
Serial order wise
### Transcript
Ex 7.1, 10 Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4). Let the points be A(x , y) , B(3, 6) , C(−3, 4) According to question, point A is equidistant from B & C Hence AB = AC So, we will find AB & AC using distance formula Finding AB x1 = x, y1 = y x2 = 3, y2 = 6 AB = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 3 −𝑥)2+(6 −𝑦)2) Similarly, Finding AC x1 = x, y1 = y x2 = −3, y2 = 4 AC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( −3 −𝑥)2+(4 −𝑦)2) We know AB = AC √(( 3 −𝑥)2+(6 −𝑦)2) = √(( −3 −𝑥)2+(4 −𝑦)2) Squaring both sides (√(( 3 −𝑥)2+(6 −𝑦)2) ")2 = (" √(( −3 −𝑥)2+(4 −𝑦)2))^2 (3 − x)2 + (6 − y)2 = (−3 − x) 2 + (4 − y) 2 32 + x2 – 6x + 62 + y2 – 12y = (−3) 2 + x2 + (–2) × (−3) × x + 42 + y2 – 8y 9 + x2 – 6x + 36 + y2 – 12y = 9 + x2 + 6x + 16 + y2 – 8y (x2 − x2) – 6x – 6x + y2 − y2 −12y + 8y + 9 – 9 – 16 + 36 = 0 −12x – 4y + 20 = 0 −4(3x + y – 5) = 0 3x + y – 5 = 0/5 3x + y – 5 = 0 Note that we only have to give relation between x & y and not solve it So, answer is 3x + y – 5 = 0
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# Asymptotic behavior
Deepen your knowledge in algorithm analysis. Know the importance of asymptotic behavior of the functions generated by the analysis process.
In a simplified and more objective way, the asymptotic behavior can be understood as the growth curve of the function generated by the algorithm analysis process.
It is known that the objective of algorithm analysis is to find a Closed-form expression (function) T(n) that represents the number of instructions executed by the algorithm as a function of its input n.
For example: the algorithm findMax() (in the worst case) has complexity T(n)=4n. This means that if your input n is equal to 4 (an array of size 4), the number of instructions executed by the algorithm will be equal to:
• T(4) = 4*4 = 16 instructions.
Likewise, if the input n is equal to 10 (an array of size 10), the number of instructions executed by the algorithm will be equal to:
• T(10) = 4*10 = 40 instructions.
The functions T(n) found by the analysis process are of great importance to measure the efficiency of the algorithms. Computer science uses these T(n) functions to separate algorithms into specific performance classes. And it does this by analyzing its “asymptotic behavior“.
To begin to understand the concept of asymptotic behavior, we need to graph the function and observe how the number of instructions grows as a function of the value of the input n. To this end, we will analyze the T(n) functions of three hypothetical algorithms in the next section. Are they:
• Algorithm 1. Complexity: T(n)=7
• Algorithm 2. Complexity: T(n)=3n+2
• Algorithm 3. Complexity: T(n)=5n2-n+1
## Functions growth curve
For each of the algorithms, we will create a graph: (input size n) x (quantity of instructions executed).
### Asymptotic behavior of algorithm 1: T(n)=7
This is a constant complexity algorithm. If we look at the graph of the function illustrated below, we can see how constant its behavior is.
It can be seen in the graph above that the quantity of instructions executed (vertical axis) does not grow as a function of values n (horizontal axis). The number of instructions always remains equal to 7.
### Asymptotic behavior of algorithm 2: T(n)=3n+2
This is a linear complexity algorithm. If we look at the graph of the function illustrated below, we can see that the quantity of instructions executed by the algorithm (vertical axis) grows linearly, in relation to the values of n (horizontal axis).
### Asymptotic behavior of algorithm 3: T(n)=5n²-n+1
Algorithm 3 has quadratic complexity. If we look at the graph of the function illustrated below, we can see that the quantity of instructions executed by the algorithm (vertical axis) grows quadratically in relation to the values of n (horizontal axis).
### Comparison between algorithms
When we superimpose the graphs of the complexity functions of the three algorithms (image below), we can see the growth trend of each of the functions. Notice how the quadratic function tends to present an ever-increasing quantity of instructions for each new size n of the input! This means that the quadratic algorithm is less efficient than the other two, which have a smaller growth trend.
## A mistaken asymptotic behavior
Let’s look at a very illustrative example to show the importance of the asymptotic behavior of algorithms. An analysis was performed on two algorithms (A and B) which resulted in the following complexity functions:
Algorithm A
Algorithm B
Question: What is the most efficient algorithm?
See below a table that shows the quantity of instructions executed, by each one of them, for the initial values of n:
Apparently (looking at the table) we are tempted to say that algorithm A is more efficient because it presents much smaller amounts of instructions for the same values of n.
But observe below what happens when we analyze the quentity of instructions for larger values of n:
Note that up to a value of n=209, algorithm A is more efficient than B, but from a value of n=210 things change and algorithm B becomes more efficient than A.
The multiplicative and additive constants, respectively 100 and 1000 (in the function of algorithm B), give the false impression that it is less efficient, but they only postpone the inevitable.
See below the graph plotted for both functions. Notice how, starting from the value of n=210, the quantity of instructions executed by algorithm B is surpassed by that of algorithm A:
What we have just done in this example was analyze the asymptotic behavior of the two algorithms. We analyzed the growth trend of algorithm instructions when n has a very large value, when n tends to infinity (n → ∞).
The fact that algorithm A is quadratic already gives us the clue that, for very large values of n (n → ∞), its quantity of instructions will exceed the quantity of instructions of algorithm B, which in turn is linear. For this reason, all additive and multiplicative constants of the complexity function of an algorithm have no impact on its asymptotic behavior (its tendency to grow on the graph).
In this way, it is understood that neither the denominator 2 (of algorithm A) nor the factor 100 and the portion 1000 (of algorithm B) can prevent algorithm B from being more efficient than A for large values of n.
## Exercise
So that you can practice the concept of asymptotic behavior of functions, try to solve the proposed exercise.
• Algoritmo A: T(n) = n(n+5)
• Algoritmo B: T(n) = 10n+10
Graphically analyze the time complexity functions T(n) of algorithms A and B above and determine:
• What is the “most efficient” and “less efficient” algorithm.
• At what value of n does the “less efficient” outperform the “most efficient”?
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Solve the recurrence $$T(n) = 3T(\sqrt n) + \log n$$. by making a change of variables. Your solution should be asymptotically tight. Do not worry about whether values are integral.
Note: I don’t why suddenly $$\log$$ is used in this problem statement, whereas till now everywhere $$\lg$$ was used. I’m assuming in this context, $$\log$$ still means $$\log_2 = \lg$$.
Let us assume $$n = 2^m$$, $$m = \lg n$$.
Hence, the recurrence takes the form
$$T(2^m) = 3T(2^{m/2}) + m$$.
Now, assuming $$S(m) = T(2^m)$$, the recurrence takes the form
$$S(m) = 3S(m/2) + m$$.
Let’s assume, $$S(m) \le cm^{\lg 3}$$, for all $$m \ge m_0$$, where $$c$$ and $$m_0$$ are some positive constants.
\begin {aligned} S(m) & = 3S(m/2) + m \\ & \le 3c(m/2)^{\lg 3} + m \\ & = 3c \frac {m^{\lg 3}} {2^{\lg 3}} + m \\ & = cm^{\lg 3} + m \end {aligned}
With this we cannot prove our assumption in it’s exact form.
So, let us modify our assumption by subtracting a lower-order term.
Let’s assume $$S(m) \le cm^{\lg 3} - bm$$.
\begin {aligned} S(m) & = 3S(m/2) + m \\ & \le 3(c(m/2)^{\lg 3} - b(m/2)) + m \\ & = 3c \frac {m^{\lg 3}} {2^{\lg 3}} - 3bm/2 + m \\ & = cm^{\lg 3} - bm - (b/2 - 1)m \\ & \le cm^{\lg 3} - bm \end {aligned}
The last step holds as long as $$(b/2 - 1)m$$ is positive.
If we set $$m_0 = 1$$, we need $$b \ge 2$$.
Changing our variable back to $$n$$, we have:
$T(n) = O(m^{\lg 3}) = O((\lg n)^{\lg 3}) = O(\lg^{\lg 3} n)$
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Jackson Huckaby
Circumcenter of Triangles
Our goal for this unit is to find the center point of a tetrahedron that will also be the center of a sphere that passes through all four of the tetrahedrons vertices. To begin the process of finding this point we will explore a 2d version of a similar scenario.
Lets begin with a simple triangle in a plane, with a goal of constructing a circle that passes through the three vertices of the triangle.
Using Geometers Sketchpad, here, try and construct a circle that will pass through the 3 vertices of each triangle.
Here are my estimations:
Now that we have successfully estimated our circles, we need to discover the method for finding the circumcenter of our triangles that will provide the exact center of our circumcenter!
Take a moment and look at the 3 circles above and try and come up with some construction methods that make provide us with a circumcenter.
First let us think about the geometric definition of a circle. The set of points that are equidistant from a particular point on a plane.
So if our circumcircle passes through all three vertices on our triangle, then what can be said about the center of the circumcircle?
Here's a hint:
Correct! All three of our vertices will be equidistant from the circumcenter.
Using this knowledge, how can we decide how to find this center?
Lets look at triangle ABC.
We want to find a point that is equidistant from A and B. So lets start by locating all the possible points. We will find this by first locating the midpoint between A and B, we will call this point M. This is one possible point that is equidistant between A and B. But we want all possible points. What can we do to represent them all?
Correct! We will create a perendicular bisector through point M to line AB. An interactive example of why this creates the set of all points equidistant to points A and B can be found here.
What next?
While we know our circumcircle has to be equidistant between A and B. It also has to be equidistant between A and C, as well as B and C.
So lets repeat this process for A and C.
We have located our circumcenter! I chose to find the points between A and C, but B and C would have given us the same answer. If you wish to do all three perpendicular bisectors you may.
Congrats! We can now test our center by creating a circle with this point as the center (point O) and a radius of length OA. (or OB, or OC).
Extension.
Refer back to the three guesses we made at the top of the page. The triangle on the right's circumcenter lies on the outside of the triangle. Why is this? What other special case can you make?
Using GSP or paper and compass try and create three cases where the circumcenter is distinct.
My file can be found here.
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# What is 1 divided by 3/356
If you have the whole number 1 and you want to divide it by the fraction 3/356 then you have found the perfect article. In this quick math lesson, we'll show you how you can divide any whole number by a fraction. If dividing numbers by fractions is your jam, read on my friend!
Want to quickly learn or show students how to divide a whole number by a fraction? Play this very quick and fun video now!
Now, remember kids, the number above the fraction like is called the numerator, and the number below it is called the denominator. We'll be using these terms throughout the guide. Pretty simple stuff, but it's always nice to do a quick term recap.
Let's put our whole number and fraction side by side so we can visualize the problem we're trying to solve:
1 ÷ 3 / 356
The trick to working out 1 divided by 3/356 is similar to the method we use to work out dividing a fraction by a whole number.
All we need to do here is multiply the whole number by the numerator and make that number the new numerator. The old numerator then becomes the new denominator. Let's write this down visually:
1 x 356 / 3 = 356 / 3
So, the answer to the question "what is 1 divided by 3/356?" is:
356 / 3
Sometimes, after calculating the answer we can simplify the resulting fraction down to lower terms. In this example though 356/3 is already in it's lowest possible form.
If you made it this far you must really love your fractions and dividing whole numbers by them. Hopefully this simple guide was easy for you to follow along and you can now go forth and divide more whole numbers by as many fractions as your heart desires.
## Convert 1 divided by 3/356 to Decimal
One last little calculation before you go. It's common to want to express your result as a decimal and, to do that, all you need to do is divide your numerator by your denominator:
356 / 3 = 118.6667
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
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## Whole Number Divided by Fraction
Enter a whole number, numerator, denominator
÷
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Nikoismusic.com Other How do you factor polynomials with solutions?
# How do you factor polynomials with solutions?
## How do you factor polynomials with solutions?
The following outlines a general guideline for factoring polynomials:
1. Check for common factors. If the terms have common factors, then factor out the greatest common factor (GCF).
2. Determine the number of terms in the polynomial.
3. Look for factors that can be factored further.
4. Check by multiplying.
What is factoring in high school math?
Common factoring is the process of finding numbers and/or variables that are a multiple of every term in an expression and removing them. Once these have been identified, we divide them out of the expression and determine what remains. Keeping with the above example, factoring out 2 and x would give 2x(x^2 + 4x + 6).
### What are the factoring techniques?
The following factoring methods will be used in this lesson:
• Factoring out the GCF.
• The sum-product pattern.
• The grouping method.
• The perfect square trinomial pattern.
• The difference of squares pattern.
What is an example of factoring?
Factoring is employed at every algebra level for solving polynomials, graphing functions, and simplifying complex expressions. Generally, factoring is the inverse operation of expanding an expression. For example, 3(x − 2) is a factored form of 3x − 6, and (x − 1) (x + 6) is a factored form of x2 + 5x − 6.
#### What is factoring in math grade 8?
In algebra, factoring is used to simplify an algebraic expression by finding the greatest common factors that are shared by the terms in the expression.
What are the different ways to factor polynomials?
To factor the polynomial. for example, follow these steps: Break down every term into prime factors. This expands the expression to. Look for factors that appear in every single term to determine the GCF. In this example, you can see one 2 and two x’s in every term. These are underlined in the following:
## What are the factors of polynomials?
Factor of a Polynomial Factorization of a Polynomial. A factor of polynomial P ( x ) is any polynomial which divides evenly into P ( x ). For example, x + 2 is a factor of the polynomial x 2 – 4. The factorization of a polynomial is its representation as a product its factors. For example, the factorization of x 2 – 4 is ( x – 2) ( x + 2).
How do you factor polynomial?
To factor a polynomial completely is to find the factors of least degree that, when multiplied together, make the original polynomial. Stated mathematically, to factor a polynomial P(x), is to find two or more polynomials, say Q(x) and R(x), of lesser degree such that P(x) = Q(x) · R(x).
### What is the factored form of a polynomial?
The factored form of a polynomial is a polynomial expressed as factors with the least degree. A factor of the least degree is a number, a first degree polynomial, or a second degree polynomial with no real roots.
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# 3.3: Proving Lines Parallel
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Use the converses of the Corresponding Angles Postulate, Alternate Interior Angles Theorem, Alternate Exterior Angles Theorem, and the Same Side Interior Angles Theorem to show that lines are parallel.
• Construct parallel lines using the above converses.
• Use the Parallel Lines Property.
## Review Queue
Answer the following questions.
1. Write the converse of the following statements:
1. If it is summer, then I am out of school.
2. I will go to the mall when I am done with my homework.
3. If two parallel lines are cut by a transversal, then the corresponding angles are congruent.
2. Are any of the three converses from #1 true? Why or why not? Give a counterexample.
3. Determine the value of x\begin{align*}x\end{align*} if l || m\begin{align*}l \ || \ m\end{align*}.
Know What? Here is a picture of the support beams for the Coronado Bridge in San Diego. This particular bridge, called a girder bridge, is usually used in straight, horizontal situations. The Coronado Bridge is diagonal, so the beams are subject to twisting forces (called torque). This can be fixed by building a curved bridge deck. To aid the curved bridge deck, the support beams should not be parallel. If they are, the bridge would be too fragile and susceptible to damage.
This bridge was designed so that 1=92\begin{align*}\angle 1 = 92^\circ\end{align*} and 2=88\begin{align*}\angle 2 = 88^\circ\end{align*}. Are the support beams parallel?
## Corresponding Angles Converse
Recall that the converse of a statement switches the conclusion and the hypothesis. So, if a\begin{align*}a\end{align*}, then b\begin{align*}b\end{align*} becomes if b\begin{align*}b\end{align*}, then a\begin{align*}a\end{align*}. We will find the converse of all the theorems from the last section and will determine if they are true.
The Corresponding Angles Postulate says: If two lines are parallel, then the corresponding angles are congruent. The converse is:
Converse of Corresponding Angles Postulate: If corresponding angles are congruent when two lines are cut by a transversal, then the lines are parallel.
Is this true? For example, if the corresponding angles both measured 60\begin{align*}60^\circ\end{align*}, would the lines be parallel? YES. All eight angles created by l, m\begin{align*}l, \ m\end{align*} and the transversal are either 60\begin{align*}60^\circ\end{align*} or 120\begin{align*}120^\circ\end{align*}, making the slopes of l\begin{align*}l\end{align*} and m\begin{align*}m\end{align*} the same which makes them parallel. This can also be seen by using a construction.
Investigation 3-5: Creating Parallel Lines using Corresponding Angles
1. Draw two intersecting lines. Make sure they are not perpendicular. Label them l\begin{align*}l\end{align*} and m\begin{align*}m\end{align*}, and the point of intersection, A\begin{align*}A\end{align*}, as shown.
2. Create a point, B\begin{align*}B\end{align*}, on line m\begin{align*}m\end{align*}, above A\begin{align*}A\end{align*}.
3. Copy the acute angle at A\begin{align*}A\end{align*} (the angle to the right of m\begin{align*}m\end{align*}) at point B\begin{align*}B\end{align*}. See Investigation 2-2 in Chapter 2 for the directions on how to copy an angle.
4. Draw the line from the arc intersections to point B\begin{align*}B\end{align*}.
From this construction, we can see that the lines are parallel.
Example 1: If m8=110\begin{align*}m \angle 8 = 110^\circ\end{align*} and m4=110\begin{align*}m \angle 4 = 110^\circ\end{align*}, then what do we know about lines l\begin{align*}l\end{align*} and m\begin{align*}m\end{align*}?
Solution: 8\begin{align*}\angle 8\end{align*} and 4\begin{align*}\angle 4\end{align*} are corresponding angles. Since m8=m4\begin{align*}m \angle 8 = m \angle 4\end{align*}, we can conclude that l || m\begin{align*}l \ || \ m\end{align*}.
## Alternate Interior Angles Converse
We also know, from the last lesson, that when parallel lines are cut by a transversal, the alternate interior angles are congruent. The converse of this theorem is also true:
Converse of Alternate Interior Angles Theorem: If two lines are cut by a transversal and alternate interior angles are congruent, then the lines are parallel.
Example 3: Prove the Converse of the Alternate Interior Angles Theorem.
Given: l\begin{align*}l\end{align*} and m\begin{align*}m\end{align*} and transversal t\begin{align*}t\end{align*}
36\begin{align*}\angle 3 \cong \angle 6\end{align*}
Prove: l || m\begin{align*}l \ || \ m\end{align*}
Solution:
Statement Reason
1. l\begin{align*}l\end{align*} and m\begin{align*}m\end{align*} and transversal t36\begin{align*}t \angle 3 \cong \angle 6\end{align*} Given
2. 32\begin{align*}\angle 3 \cong \angle 2\end{align*} Vertical Angles Theorem
3. 26\begin{align*}\angle 2 \cong \angle 6\end{align*} Transitive PoC
4. l || m\begin{align*}l \ || \ m\end{align*} Converse of the Corresponding Angles Postulate
Prove Move: Shorten the names of these theorems. Discuss with your teacher an appropriate abbreviations. For example, the Converse of the Corresponding Angles Theorem could be “Converse CA Thm” or “ConvCA.”
Notice that the Corresponding Angles Postulate was not used in this proof. The Transitive Property is the reason for Step 3 because we do not know if l\begin{align*}l\end{align*} is parallel to m\begin{align*}m\end{align*} until we are done with the proof. You could conclude that if we are trying to prove two lines are parallel, the converse theorems will be used. And, if we are proving two angles are congruent, we must be given that the two lines are parallel.
Example 4: Is l || m\begin{align*}l \ || \ m\end{align*}?
Solution: First, find m1\begin{align*}m \angle 1\end{align*}. We know its linear pair is 109\begin{align*}109^\circ\end{align*}. By the Linear Pair Postulate, these two angles add up to 180\begin{align*}180^\circ\end{align*}, so m1=180109=71\begin{align*}m \angle 1 = 180^\circ - 109^\circ = 71^\circ\end{align*}. This means that \begin{align*}l \ || \ m\end{align*}, by the Converse of the Corresponding Angles Postulate.
Example 5: Algebra Connection What does \begin{align*}x\end{align*} have to be to make \begin{align*}a \ || \ b\end{align*}?
Solution: Because these are alternate interior angles, they must be equal for \begin{align*}a \ || \ b\end{align*}. Set the expressions equal to each other and solve.
\begin{align*}3x+16^\circ&=5x-54^\circ\\ 70^\circ&=2x\\ 35^\circ&=x \qquad \quad \text{To make}\ a \ || \ b, \ x = 35^\circ.\end{align*}
## Converse of Alternate Exterior Angles & Consecutive Interior Angles
You have probably guessed that the converse of the Alternate Exterior Angles Theorem and the Consecutive Interior Angles Theorem areal so true.
Converse of the Alternate Exterior Angles Theorem: If two lines are cut by a transversal and the alternate exterior angles are congruent, then the lines are parallel.
Example 6: Real-World Situation The map below shows three roads in Julio’s town.
Julio used a surveying tool to measure two angles at the intersections in this picture he drew (NOT to scale). Julio wants to know if Franklin Way is parallel to Chavez Avenue.
Solution: The labeled \begin{align*}130^\circ\end{align*} angle and \begin{align*}\angle a\end{align*} are alternate exterior angles. If \begin{align*}m \angle a = 130^\circ\end{align*}, then the lines are parallel. To find \begin{align*}m \angle a\end{align*}, use the other labeled angle which is \begin{align*}40^\circ\end{align*}, and its linear pair. Therefore, \begin{align*}\angle a + 40^\circ = 180^\circ\end{align*} and \begin{align*}\angle a = 140^\circ\end{align*}. \begin{align*}140^\circ \neq 130^\circ\end{align*}, so Franklin Way and Chavez Avenue are not parallel streets.
The final converse theorem is of the Same Side Interior Angles Theorem. Remember that these angles are not congruent when lines are parallel, they are supplementary.
Converse of the Same Side Interior Angles Theorem: If two lines are cut by a transversal and the consecutive interior angles are supplementary, then the lines are parallel.
Example 7: Is \begin{align*}l \ || \ m\end{align*}? How do you know?
Solution: These are Same Side Interior Angles. So, if they add up to \begin{align*}180^\circ\end{align*}, then \begin{align*}l \ || \ m\end{align*}. \begin{align*}113^\circ + 67^\circ = 180^\circ\end{align*}, therefore \begin{align*}l \ || \ m\end{align*}.
## Parallel Lines Property
The Parallel Lines Property is a transitive property that can be applied to parallel lines. Remember the Transitive Property of Equality is: If \begin{align*}a = b\end{align*} and \begin{align*}b = c\end{align*}, then \begin{align*}a = c\end{align*}. The Parallel Lines Property changes = to \begin{align*}||\end{align*}.
Parallel Lines Property: If lines \begin{align*}l \ || \ m\end{align*} and \begin{align*}m \ || \ n\end{align*}, then \begin{align*}l \ || \ n\end{align*}.
Example 8: Are lines \begin{align*}q\end{align*} and \begin{align*}r\end{align*} parallel?
Solution: First find if \begin{align*}p \ || \ q\end{align*}, followed by \begin{align*}p \ || \ r\end{align*}. If so, then \begin{align*}q \ || \ r\end{align*}.
\begin{align*}p \ || \ q\end{align*} by the Converse of the Corresponding Angles Postulate, the corresponding angles are \begin{align*}65^\circ\end{align*}. \begin{align*}p \ || \ r\end{align*} by the Converse of the Alternate Exterior Angles Theorem, the alternate exterior angles are \begin{align*}115^\circ\end{align*}. Therefore, by the Parallel Lines Property, \begin{align*}q \ || \ r\end{align*}.
Know What? Revisited: The CoronadoBridge has \begin{align*}\angle 1\end{align*} and \begin{align*}\angle 2\end{align*}, which are corresponding angles. These angles must be equal for the beams to be parallel. \begin{align*}\angle 1 = 92^\circ\end{align*} and \begin{align*}\angle 2 = 88^\circ\end{align*} and \begin{align*}92^\circ \neq 88^\circ\end{align*}, so the beams are not parallel, therefore a sturdy and safe girder bridge.
## Review Questions
1. Construction Using Investigation 3-1 to help you, show that two lines are parallel by constructing congruent alternate interior angles. HINT: Steps 1 and 2 will be exactly the same, but at step 3, you will copy the angle in a different location.
2. Construction Using Investigation 3-1 to help you, show that two lines are parallel by constructing supplementary consecutive interior angles. HINT: Steps 1 and 2 will be exactly the same, but at step 3, you will copy a different angle.
For Questions 3-5, fill in the blanks in the proofs below.
1. Given: \begin{align*}l \ || \ m, \ p \ || \ q\end{align*} Prove: \begin{align*}\angle 1 \cong \angle 2\end{align*}
Statement Reason
1. \begin{align*}l \ || \ m\end{align*} 1.
2. 2. Corresponding Angles Postulate
3. \begin{align*}p \ || \ q\end{align*} 3.
4. 4.
5. \begin{align*}\angle 1 \cong \angle 2\end{align*} 5.
1. Given: \begin{align*}p \ || \ q, \ \angle 1 \cong \angle 2\end{align*} Prove: \begin{align*}l \ || \ m\end{align*}
Statement Reason
1. \begin{align*}p \ || \ q\end{align*} 1.
2. 2. Corresponding Angles Postulate
3. \begin{align*}\angle 1 \cong \angle 2\end{align*} 3.
4. 4. Transitive PoC
5. 5. Converse of Alternate Interior Angles Theorem
1. Given: \begin{align*}\angle 1 \cong \angle 2, \ \angle 3 \cong \angle 4\end{align*} Prove: \begin{align*}l \ || \ m\end{align*}
Statement Reason
1. \begin{align*}\angle 1 \cong \angle 2\end{align*} 1.
2. \begin{align*}l \ || \ n\end{align*} 2.
3. \begin{align*}\angle 3 \cong \angle 4\end{align*} 3.
4. 4. Converse of Alternate Interior Angles Theorem
5. \begin{align*}l \ || \ m\end{align*} 5.
For Questions 6-9, create your own two column proof.
1. Given: \begin{align*}m \perp l, \ n \perp l\end{align*} Prove: \begin{align*}m \ || \ n\end{align*}
2. Given: \begin{align*}\angle 1 \cong \angle 3 \end{align*} Prove: \begin{align*}\angle 1\end{align*} and \begin{align*}\angle 4\end{align*} are supplementary
3. Given: \begin{align*}\angle 2 \cong \angle 4 \end{align*} Prove: \begin{align*}\angle 1 \cong \angle 3\end{align*}
4. Given: \begin{align*}\angle 2 \cong \angle 3 \end{align*} Prove: \begin{align*}\angle 1\cong \angle 4\end{align*}
In 10-15, use the given information to determine which lines are parallel. If there are none, write none. Consider each question individually.
1. \begin{align*}\angle LCD \cong \angle CJI\end{align*}
2. \begin{align*}\angle BCE \ and \ \angle BAF\end{align*} are supplementary
3. \begin{align*}\angle FGH \cong \angle EIJ\end{align*}
4. \begin{align*}\angle BFH \cong \angle CEI\end{align*}
5. \begin{align*}\angle LBA \cong \angle IHK\end{align*}
6. \begin{align*}\angle ABG \cong \angle BGH\end{align*}
In 16-22, find the measure of the lettered angles below.
1. \begin{align*}m \angle 1\end{align*}
2. \begin{align*}m \angle 2\end{align*}
3. \begin{align*}m \angle 3\end{align*}
4. \begin{align*}m \angle 4\end{align*}
5. \begin{align*}m \angle 5\end{align*}
6. \begin{align*}m \angle 6\end{align*}
7. \begin{align*}m \angle 7\end{align*}
For 23-27, what does \begin{align*}x\end{align*} have to measure to make the lines parallel?
1. \begin{align*}m \angle 3 = (3x+25)^\circ\end{align*} and \begin{align*}m \angle 5 = (4x-55)^\circ\end{align*}
2. \begin{align*}m \angle 2 = (8x)^\circ\end{align*} and \begin{align*}m \angle 7 = (11x-36)^\circ\end{align*}
3. \begin{align*}m \angle 1 = (6x-5)^\circ\end{align*} and \begin{align*}m \angle 5 = (5x+7)^\circ\end{align*}
4. \begin{align*}m \angle 4 = (3x-7)^\circ\end{align*} and \begin{align*}m \angle 7 = (5x-21)^\circ\end{align*}
5. \begin{align*}m \angle 1 = (9x)^\circ\end{align*} and \begin{align*}m \angle 6 = (37x)^\circ\end{align*}
6. Construction Draw a straight line. Construct a line perpendicular to this line through a point on the line. Now, construct a perpendicular line to this new line. What can you conclude about the original line and this final line?
7. How could you prove your conjecture from problem 28?
8. What is wrong in the following diagram, given that \begin{align*}j \ || \ k\end{align*}?
## Review Queue Answers
1. If I am out of school, then it is summer.
2. If I go to the mall, then I am done with my homework.
3. If corresponding angles created by two lines cut by a transversal are congruent, then the two lines are parallel.
1. Not true, I could be out of school on any school holiday or weekend during the school year.
2. Not true, I don’t have to be done with my homework to go to the mall.
3. Yes, because if two corresponding angles are congruent, then the slopes of these two lines have to be the same, making the lines parallel.
1. The two angles are supplementary.
\begin{align*}(17x + 14)^\circ + (4x - 2)^\circ &= 180^\circ\\ 21x + 12^\circ &= 180^\circ\\ 21x &= 168^\circ\\ x &= 8^\circ\end{align*}
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# GATE | GATE-CS-2004 | Question 90
• Difficulty Level : Hard
• Last Updated : 28 Jul, 2021
The recurrence equation
```T(1) = 1
T(n) = 2T(n - 1) + n, n ≥ 2 ```
evaluates to
a. 2n + 1– n – 2
b. 2n – n
c. 2n + 1 – 2n – 2
d. 2n + n
(A) a
(B) b
(C) c
(D) d
Explanation: If draw recursion tree, we can notice that total work done is,
T(n) = n + 2(n-1) + 4(n-2) + 8(n-3) + 2n-1 * (n – n + 1)
T(n) = n + 2(n-1) + 4(n-2) + 8(n-3) + 2n-1 * 1
To solve this series, let us use our school trick, we multiply T(n) with 2 and subtract after shifting terms.
```2*T(n) = 2n + 4(n-1) + 8(n-2) + 16(n-3) + 2n
T(n) = n + 2(n-1) + 4(n-2) + 8(n-3) + 2n-1 * 1```
We get
```2T(n) - T(n) = -n + 2 + 4 + 8 + ..... 2n
T(n) = -n + 2n+1 - 2 [Applying GP sum formula for 2, 4, ...]
= 2n+1 - 2 - n```
```Alternate Way to solve is to use hit and try method.
Given T(n) = 2T(n-1) + n and T(1) = 1
For n = 2, T(2) = 2T(2-1) + 2
= 2T(1) + 2
= 2.1 + 2 = 4
Now when you will put n = 2 in all options,
only 1st option 2^(n+1) - n - 2 satisfies it. ```
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# How to Calculate and Solve for the Area, Length and Height of a Trapezium | The Calculator Encyclopedia
The image above is a Trapezium
To compute the area of a trapezium, three essential parameter is needed and they are the length of the top side (a), length of the bottom side (b) and the height of the trapezium (h).
The formula for calculating the area of a trapezium is;
A = 0.5[a + b]h
Where;
A = Area of a trapezium
a = length of the top side of the trapezium
b = length of the bottom side of the trapezium
h = height of the trapezium
Let’s solve an example;
Find the area of a trapezium where length of top side (a) is 7 cm, length of bottom side (b) is 12 cm and height of trapezium (h) is 10 cm.
This implies that;
a = Length of top side of the trapezium = 7 cm
b = Length of bottom side of the trapezium = 12 cm
h = Height of the trapezium = 10 cm
A = 0.5[a + b]h
A = 0.5[7 + 12]10
A = 0.5[19]10
A = 95
Therefore, the area of a trapezium is 95 cm2
How to Calculate the Height of a Trapezium when the Area, Length of top side and Length of bottom side of the Trapezium is given.
The formula is h = A / 0.5(a + b)
Where;
A = Area of the trapezium
a = Length of the top side of the trapezium
b = Length of the bottom side of the trapezium
Let’s solve for an example;
Given that the length of top side (a) is 10 cm, length of bottom side (b) is 14 cm and the area of the trapezium is 20 cm2 Find the height of the trapezium?
This implies that;
A = Area of the trapezium = 20 cm2
a = Length of top side of the trapezium = 10 cm
b = Length of bottom side of the trapezium = 14 cm
h = A / 0.5(a + b)
h = 20 / 0.5(10 + 14)
h = 20 / 0.5(24)
h = 20 / 12
h = 1.667
Therefore, the height of the trapezium is 1.667 cm.
How to Calculate the Length of top side of a Trapezium when the Area of the trapezium, Length of bottom side of the Trapezium and the height of the trapezium is given.
a = 2A / h – b
where;
A = Area of the trapezium
b = Length of the bottom side of the trapezium
h = Height of the trapezium
Let’s solve an example;
Find the length of top side of a trapezium when the height of the trapezium is 14 cm, length of bottom side of the trapezium is 16 cm and area of the trapezium is 240 cm2
This implies that;
h = Height of the trapezium = 14 cm
b = Length of bottom side of the trapezium = 16 cm
A = Area of the trapezium = 240 cm2
a = 2A / h – b
a = 2(240) / 14 – 16
a = 480 / 14 – 16
a = 34.26 – 16
a = 18.256
Therefore, the length of top side of the trapezium is 18.256 cm.
How to Calculate the Length of bottom side of a Trapezium when the Area of the trapezium, Length of top side of the Trapezium and the height of the trapezium is given.
b = 2A / h – a
where;
A = Area of the trapezium
a = Length of the top side of the trapezium
h = Height of the trapezium
Let’s solve an example;
Find the length of the bottom side of a trapezium when length of the top side of the trapezium is 20 cm, height of the trapezium is 25 cm and area of the trapezium is 300 cm2
This implies that;
a = Length of top side of the trapezium = 20 cm
h = height of the trapezium = 25 cm
A = Area of the trapezium = 300 cm2
b = 2A / h – a
b = 2(300) / 25 – 20
b = 600 / 25 – 20
b = 24 – 20
b = 4
Therefore, the length of bottom side of the trapezium is 4 cm.
Nickzom Calculator – The Calculator Encyclopedia is capable of calculating the area of a trapezium.
To get the answer and workings of the area of a trapezium using the Nickzom Calculator – The Calculator Encyclopedia. First, you need to obtain the app.
You can get this app via any of these means:
To get access to the professional version via web, you need to register and subscribe for NGN 1,500 per annum to have utter access to all functionalities.
You can also try the demo version via https://www.nickzom.org/calculator
Once, you have obtained the calculator encyclopedia app, proceed to the Calculator Map, then click on Mensuration under the Mathematics section
Now, click on Area of a trapezium under Mensuration
The screenshot below displays the page or activity to enter your value, to get the answer for the area of a trapezium according to the respective parameters which is the length of top side of the trapezium, length of bottom side of the trapezium and height of the trapezium
Now, enter the values appropriately and accordingly for the parameters as required by the example above where the length of the top side of the trapezium is 7 cm, length of the bottom side of the trapezium is 12 cm and height of the trapezium is 10 cm
Finally, click on Calculate
As you can see from the screenshot above, Nickzom Calculator – The Calculator Encyclopedia solves for the area of a trapezium and presents the formula, workings and steps too.
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## Monday, August 17, 2015
### Prove that $\frac{cot(x)}{1-tan(x)} + \frac{tan(x)}{1 - cot(x)}=1 + tan(x) + cot(x)$
Q. Prove: $\frac{cot(x)}{1-tan(x)} + \frac{tan(x)}{1 - cot(x)}=1 + tan(x) + cot(x)$}
A. $$\begin{array}{lll}\\ \frac{cot(x)}{1-tan(x)} + \frac{tan(x)}{1 - cot(x)} &= \frac{\frac{1}{tan(x)}}{1 - tan(x)} + \frac{tan(x)}{1 - \frac{1}{tan(x)}}&\mbox{Expressing } cot(x)=1/tan(x)\\ &= \frac{1}{tan(x)(1-tan(x))} + \frac{\tan^2(x)}{tan(x) - 1}& \mbox{Simplifying fractions.}\\ &= \frac{1}{tan(x)(1-tan(x))} - \frac{\tan^2(x)}{1- tan(x)}& \mbox{Last term as a subtraction }\\ &= \frac{1 - tan^3(x)}{tan(x)(1-tan(x)}& \mbox{Rewriting as a single fraction}\\ &= \frac{(1 - tan(x)){(1 + tan(x)} + \tan^2(x)}{tan(x)(1-tan(x))}& \mbox{Factoring a difference of cubes}\\ &= \frac{(1 + tan(x) +\tan^2(x)}{tan(x)}&\mbox{Cancellation by division}\\ &= cot(x) + 1 + tan(x)& \mbox{Division}\\ &= 1 + tan(x) + cot(x)& \mbox{Rearranging terms.}\\ \end{array}$$
## Derivation of the length of a chord
Let an arc on the unit circle with starting point at $P_0 = (1, 0)$ subtend an angle of $\beta$ radians. Recall that its endpoint will be
$P_1 = (\cos(\beta), \sin(\beta))$. The length of the chord connecting the point (1,0) to the terminal point can be determined using the
distance formula $$d = \sqrt{(\Delta x)^2 + (\Delta y)^2}=\sqrt{(\cos(\beta) - 1)^2 + (\sin(\beta) - 0) ^2}$$. Simplifying we will have
$d = \sqrt{\cos^2(\beta) + \sin^2 (\beta) - 2\cos(\beta) + 1}= \sqrt{2- 2\cos(\beta)}.$
## The fundamental formula for the cosine of the difference in angles: $\cos(\alpha - \beta)$
Now let $\alpha$ be an angle greater than $\beta$. The new endpoint will be at $P_2 =(\cos(\alpha), sin(\alpha))$. Consider the chord $P_1P_2$.
By the distance formula, the length of this chord will be given by $\sqrt{(\cos (\alpha) - \cos(\beta))^2 +(\sin(\alpha)-\sin(\beta))^2}$
and expanding we obtain
$\sqrt{(\cos^2(\alpha) - 2 cos(\alpha)cos(\beta) + cos^2(\beta) + (\sin^2(\alpha) - 2 sin(\beta)\sin(\alpha)+\sin^2(\beta))}$
Further simplifying we obtain $\sqrt{(\cos^2(\alpha) + \sin^2(\alpha) + \cos^2(\beta) + \sin^2(\beta) - 2\cos(\alpha) cos\beta - \sin(\alpha) \sin(\beta)}$.
and finally we have $d = \sqrt{2 -2 (\cos(\alpha) cos\beta + \sin(\alpha) \sin(\beta)}$
But this chord lengthis also given by $\sqrt{ 2 - 2 cos(\alpha - \beta)}$.
Equating the two expressions , we obtain the formula
$$\cos(\alpha - \beta) = \cos(\alpha) cos(\beta) + \sin(\alpha) \sin(\beta)$$
This is an important result. Let us try to derive further formulas from this.
Substiture $-\beta$ for $\beta$. We sould then have $\cos(\alpha - (-\beta)) = \cos(\alpha) cos(-\beta) + \sin(\alpha) \sin(-\beta)$.
or $$\cos(\alpha + \beta)) = \cos(\alpha) cos(\beta) - \sin(\alpha) \sin(\beta)$$ using the facts that the cosine function is even $\cos(-alpha) = \cos(alpha)$
that the sine function is odd $sin(-\beta) = -sin(beta)$.We can combine the two identities as $$\cos(\alpha \pm \beta) = cos(\alpha) cos(\beta) \mp \sin(\beta) sin(\alpha)$$.
## Wednesday, August 12, 2015
### The angle between the hour and minute hands of a clock.
Q. The time shown in the clock is 7:35 what is the angle between the hour and minute hands of a clock?
A. Consider the simpler problem if the time is 7:00. As there are 12 hours in a complete revolution of the hour hand, one hour subtends an angle
of 360/12 = 30 degrees. Thus, if H is the number of hours, the angle made by the hour hand from the noon positioon would be 30H.
Now let us consider the minute hand. if M is the number of minutes, consider that if M is 15 minutes , the angle which the minute hand makes is
90 degrees. This means that each minute ssubends an angle of 90/15 = 6 degrees or 6M.
But each revolution of the minute hand imparts 1/12 revolution of the hour hand. In other words, in the time covered by the minute hand, the hour hand will cover an
The angle between the hour hand and minute hand for a time reading of H:M will then be |30H + M/2 - 6M|. If this angle is greater than 180 degrees,
we take the difference from 360 degrees.
Getting back to our original problem, for a time of 7:35, the angle beween the hour and minute hands is |30(7) + 35/2 - 6(35)| = 17.5 degrees.
## Monday, August 20, 2012
### Statistics Problem Set Aug-21-2012
1. Which of the following formulas measure symmetry of a sample data distribution?
(a)$(1/n) \sum (x-\overline{x})^2$ (b) $(1/n) \sum (x-\overline{x})^3$ (c)$(1/n) \sum (x-\overline{x})^4$ (d.) Not listed
2. The following were determined for a sample data: n = 10, min=-2, max= 10, sd = 3,
$\overline{x}=5$. The data is invalid since
$max > \overline{x} + (n-1) sd.$ (b) $min < \overline{x} - (n-1) sd$ (c) median is not specified (d) None of the above.
3. A sample has current mean mean 4.0 with sample size 10. If 5 is added to all sample values, the recomputed mean will :
stay the same! (b) will increase by 5.0 (c) will decrease by 5 (d)will increase by 0.5 (e) None of the above
4.A sample has current variance of 2.0 with sample size 10. If 5 is added to all sample values, the recomputed variance will :
stay the same! (b) will increase by 5.0 (c) will decrease by 5 (d)will increase by 0.5 (d) None of the above.
5. The formula $e^{[(1/n)\sum_{i=1}^{i = n} \ln(x_i)}$ is another way of solving for the
(a) HM (b) GM (c) AM (d) RMS.
6. A new data value 2.5 is added to a sample with mean 3 and sample size 10. The recomputed mean will
increase (b) decrease (c) stay the same (d) None of the above
7. A sample has a sample size 10 and median 50, with all sample values unique. If the minimum value in the sample is removed,
the new median will
stay the same! (b) increase (c) decrease (d) None of the above.
8. Express (-1) + 2 + (-3) + 4 + (-5) + 6 in summation form with index k $of summation varying from 1 to 6. You cannot use a variable X as these are not stored in an a vector or array!. 9. If X = c( 4,5,6,7,8), what is the value of$\sum_{i=1}^4 (x_i)(x_{i-1})$?. 10.If X is the same above and Y= c(2,3,1,0,4) what is the value of$\sum_{i=1}^5 {(x_i -\overline{x}) (y_i-\overline{y})}\$?
Solutions next week! Enjoy first solving.
## Tuesday, July 31, 2012
### Descriptive Statistics
Modified True or False. If the statement is true, write T and if the statement if false , write F and explain why!
Q1. The values computed for a sample, arithmetic mean =-4.5, and root mean square= -4 are valid since the root mean square is always greater than or equal to the arithmetic mean.
Q2. Variances computed from a sample were sample variance = 0.4629, and population variance = 0.5141 are valid.
Q3. The variances computed with a sample size n = 10, population variance = 0.4629, and sample variance = 0.60, are valid.
Q4. Various means computed for a sample: harmonic mean, HM = 0.2, geometric mean, GM = 0.2, arithmetic mean, AM = .3,
root mean square, RMS = .4 are valid since they satisfy $HM\le GM\le AM\le RMS$.
Q5. If the variance of a sample X is 5.3 then the variance of $Y = -2 (X + 5)$ is $2 \cdot 5.3 = 10.6$.
1. F. The reasoning may sound valid but the root mean square is never negative!
2. F. Sample variance is always greater than population variance!
3. F. Sample varince may be obtained using the formula: $$\sigma_{n-1}^2= \frac{n}{n-1} \sigma_n^2$$. But $$\frac{10}{9} 0.4629 = 0.5141$$ which is not equal to 0.60.
4. Although the given values does satisfy the inequality for various means, whenever any two of the means are equal, then all means should be equal!
5. If X has the variance $V_X$, then $Y = a(X + b)$ will have variance $$a^2 V_X$$. Therefore, the varice of Y will be 4 times the variance of X or 4 (5.3)= 21.2.
## Monday, May 7, 2012
### Problems with complex numbers.
Q1. Convert the complex number z = 6 + 8j to polar form.
A1. The magnitude $|z|= \sqrt{6^2 + 8^2}= \sqrt{36+ 64} = \sqrt{100} = 10.$
Since both x and y components are positive, the complex number is in the first quadrant.
The argument $\theta = atan(\frac{8}{6}) = 0.9273$, in radians. Multiply by $180/\pi$ to obtain the argument degrees. The value would be $53.3^\circ$.
Q2, Convert to rectangular form the complex number in polar form $12 \angle 210^\circ$.
A2. The complex number is in the third quadrant. This means that both components are negative.
$x = 12 \cos(210^\circ) = -10.392, y = 12 \sin(210^\circ) = -6.0$,from which $z = -10.392 - 6j.$
## Saturday, May 5, 2012
### Fomula mass of chemical compound
Q. Calculate the formula mass of the compounds KBr and $NH_4CN$
A. The atomic masses of K and Br are respectively 39.102 and 79.904 respectively. Thus the formula mass is 39.102+79.904= 119.006 amu.
On the other hand, NH4CN has a formula mass calculated in the following table:
Component AmuTotal
N 14.0067114.0067
H41.008(4)4.032
C 1212
N 14.006714.0067
Total44.107
The formula mass is then 44.107.
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# Maths Wiz Solutions Class 8 Chapter 2
## Maths Wiz Class 8 Solutions Chapter 2 Powers (Exponents)
Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Maths Wiz Class 8 Math Book, Chapter 2, Powers (Exponents). Here students can easily find step by step solutions of all the problems for Powers (Exponents), Exercise 2A and 2B Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 2 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.
Powers (Exponents) Exercise 2A Solution :
Question no – (1)
Solution :
(a) 90
= 1
(b) (1/3)0
= 1
(c) (- 3 1/21)0
= 1
(d) (395.008)0
= 1
(e) (3xy/ab – z)0
= 1
(f) 1/a0 + b0 + c0
= 1/1 + 1 + 1
= 1/3
(g) (80 – 20) × 250
= (1 – 1) × 1
= 0
(h) (350 + 20 + 30) ÷ 3
= 3 ÷ 3
= 1
Question no – (2)
Solution :
(a) 6-1
= 1/6
(b) 10-1
= 1/10
(c) 4-3
= 1/43
= 1/64
(e) 3-3
= 1/33
= 1/27
(f) p-1
= 1/p
(g) (x-2)2
= 1/x4
(h) (10a)-2
= 1/(10a)2
(i) (x + y)-2
= 1/(x + y)2
(j) (m5)-3
= 1/m15
Question no – (3)
Solution :
(a) (3)-2
= 1/32
= 1/9
(b) (-2)-5
= 1/(-2)5
= – 1/32
(c) (2/5)-3
= 1/(2/5)3
= 25/4
(d) (2/3)-4
= (2/3)-4
= (3/4)4
Question no – (4)
Solution :
(a) (- 2)-4 × (- 2)-6
= 1/24 × 1/26
= 1/210
(b) 37 × 3-9 × 36
= 37 × 36/39
= 34
(c) 3× (3/5)-7
= 37 × (5/3)7
= 57
(d) 2-7 × [(2)-3]-4
= 27 × 212
= 219
Question no – (5)
Solution :
(a) (80 + 5-1) × (1/5)-2
= (1 + 1/5) × 52
= 6/5 × 52
= 6 × 5
= 30
(b) (300 + 150) × (1/2)-3
= (1 + 1) × 23
= 21 × 23
= 24
(c) (7-1 + 8-1 + 9-1)0 ÷ (- 1/9)-2
= (1/7 + 1/8 + 1/9) ÷ 92
= 1 × 9-2
= 1/81
(d) [(1/3)-2 + (1/4)-2 + (1/6)-2]
= [3+ 42 + 62]
= 9 + 16 + 36
= 51
(e) {(-4/5)-2}-2
= {(5/4)2}-2
= (5/4)-4
= (4/5)4
(f) 9-1 × 43/3-4
= 43 × 34/9
= 43 × 9 × 32/9
= 43 × 32
(g) (5-1 × 4-1) × (3/2)-2
= (1/5 × 1/4) (2/3)2
= 1/20 × 4/9
= 1/45
Question no – (6)
Solution :
(a) [(1/4)-1 – (1/5)-1] + 1
= [4 + 5] + 1
= 9 + 1
= 10
(b) [(11/12)-10 ÷ (11/12)-12] × 1 23/121
= [(12/11)10 ÷ (12/11)12] × 144/121
= [(12/11)10 × (11)12/(12)12 × 122/112
= 112/122 × 122/112
= 1
Question no – (7)
Solution :
Given, 72m/7 – 4 = 7-6
= 72m × 74 = 7-6
= 72m + 4 = 7-6
= 72m+4 = 7–6
So, 2m + 4 = – 6
= 2m = – 10
= m = – 5
Therefore, the value of m will be -5
Powers (Exponents) Exercise 2B Solution :
Question no – (1)
Solution :
(a) 10-4
= 1/104
= 1/10000
(b) 10-7
= 1/17
= 1/10000000
(c) 10-8
= 1/108
= 1/100000000
(d) 10-10
= 1/1010
= 1/10000000000
Question no – (2)
Solution :
(a) 5,700,000 in scientific notation,
= 57 × 101000
(b) 8,19,000 in scientific notation,
= 81.90 × 1000
(c) 4,000,000,000 in scientific notation,
= 2020000
(d) 56,000 in scientific notation,
= 560 × 10100
Question no – (3)
Solution :
(a) 0.000053 in scientific notation,
= 53/16
(b) 0.000000084 in scientific notation,
= 84/109
(c) 0.0009 in scientific notation,
= 9/104
(d) 0.0603 in scientific notation,
= 603/104
Question no – (6)
Solution :
(a) Distance form earth to sun is 149600000000
In Standard form,
= 1.496 × 10^11 m
(b) The average diameter of a red blood cell is 0.000007 mm.
In Standard form,
= 7 × 10-6 mm
(c) Diameter of a wire in a computer chip is 0.000003 m.
In Standard form,
= 3 × 10-6 m
(d) Size of a plant cell is 0.00001275 m.
In Standard form,
= 1275 × 10-7 m
Next Chapter Solution :
Updated: June 19, 2023 — 7:36 am
|
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# ZingPath: Using Probability
Searching for
## Using Probability
Learn in a way your textbook can't show you.
Explore the full path to learning Using Probability
Pre-Algebra
### Learning Made Easy
Design a probability experiment to test predictions about fairness related to the probabilities for the sum of two dice.
### Now You Know
After completing this tutorial, you will be able to complete the following:
• Determine if an event is more likely or equally likely to occur.
• Design a probability experiment to test predictions about fairness related to the probabilities for the sum of two dice.
### Everything You'll Have Covered
The likelihood line demonstrates the continuum of probability that an event will occur.
Since probability is a ratio, the probability of any event can be plotted on a number line from 0 to 1.
The event, which falls on the far left side of the likelihood line is the least likely to occur because it has a probability closest to zero. If an event falls on the far right side of the likelihood line, it has a probability of very close to one whole. Events falling in the center of the line have about a fifty percent chance of occurring.
Likelihood lines are typically used in early probability instruction to provide a quick overview of how likely an event can occur. Students can generate a list of real-life events and plot them on the likelihood line.
In this Activity Object, students will create a game using the outcomes of rolling two dice in such a way that one pawn is more likely to win or both are equally likely to win.
Theoretical probability is calculated by dividing the number of outcomes for a specific event by the number of total events possible.
For example, when flipping a penny, the penny could land on either heads or tails resulting in two possible outcomes. To determine the probability of the penny landing on heads:
In this Activity Object, the term theoretical is not used directly; however, since the probability of the sum of two dice is not tested in an actual experiment, it is important for students to know that all of the probabilities discussed in this Activity Object are theoretical. At the end of the Activity Object, students are given the opportunity to play the game. The results of the game represent the experimental probability, the probability of the event actually occurring.
Theoretical probability for the sum of two dice.
A simple chart can be used to list the outcomes of the sum of two dice.
Using this chart, you can easily see that two dice with the sum of 7 will occur most often, and two dice with a sum of either 2 or 12 will occur least often.
A game is fair when all players are equally likely to win.
In this Activity Object, students determine if the game they create is fair. In a fair game, the probability of each player winning must match. For example, we would determine the fairness of the game below by finding the probability of each player winning.
Player 1 has one out of two chances of landing on red. The theoretical probability of landing on red when spinning ten times is 5 out of 10. Player 2 has one out of five chances of landing on red. The theoretical probability of landing on red when spinning ten times is 2 out of 10. Therefore, this game is unfair because player one and player two do not have an equally likely chance of winning the game.
### Tutorial Details
Approximate Time 15 Minutes Pre-requisite Concepts concept of probability, likelihood of an event Course Pre-Algebra Type of Tutorial Concept Development Key Vocabulary equally likely, fairness, probability
|
This IIT-JEE maths examination video is based on the topic probability. This topic is also interesting and you will enjoy while solving problems based on probability. It is simple because you will not have to remember many formulas in this maths video lecture. Also, you should have thorough knowledge of permutation and combinations for understanding probability better.
Students need to understand the various principles of probability to solve the problem easily. If you understand this topic well then there you can easily score good marks in probability. Many a times student do not get good tutorials for understanding this topic and they end up getting confused. But this video lecture has simplified the understanding of the topic and will help you to solve the problems correctly. Many questions are asked in JEE examination and you will enjoy while watching these video lectures.
Random experiment is an experiment whose outcome cannot be predicted or cannot be said anything certainly. Example: Tossing affair coin. Now when we toss a coin, you may get a head or a tail, but you cannot predict the result. Example: Drawing a card from a well shuffled pack of cards. When we play cards, and take out a card from a well shuffled pack, then which card will come out is a random process. Example: Throwing an unbiased dice. When you throw a die, you don’t know what number you will get, it can be anything between one to six. Hence, any experiment the outcome of which cannot be predicted is called a random experiment in probability.
These video lectures on IIT JEE maths probability examination are based on several examples on probability which will help you understand the topic better.
Sample space: if you perform a trial, then a set in this experiment, say a set of all possible outcomes/ results of a random experiment are called a sample space. And each element of the sample space is called sample point. Example: If you toss a coin, then there are only two possible results. Sample space will be given by either head or tail. Hence sample space will contain two elements, head or tail = S {H, T}. Similarly, throwing a die, sample space will contain 6 digits because out of 6 digits you can have any outcome such as S = { 1,2,3,4,5,6}. Similarly, you will be able to understand several other concepts in this video lecture.
|
# 60 Rate Of Change Worksheet
## Introduction
Welcome to our comprehensive guide on rate of change worksheets. In this article, we will explore the concept of rate of change, its importance in various disciplines, and provide you with a collection of worksheets to help you practice and master this fundamental mathematical concept. Whether you are a student looking to improve your skills or a teacher searching for resources to enhance your lessons, you've come to the right place.
## Understanding Rate of Change
### Defining Rate of Change
Rate of change, also known as slope, is a measure of how one variable changes with respect to another variable. It quantifies the steepness or direction of a relationship between two quantities. In mathematical terms, it is calculated by dividing the change in the dependent variable by the change in the independent variable.
### Real-World Applications
Rate of change is a fundamental concept in various fields, including physics, economics, and engineering. For example, in physics, rate of change is used to determine the velocity of an object, while in economics, it helps analyze trends in prices or demand. Understanding rate of change is essential for making predictions, analyzing data, and solving real-world problems.
## Types of Rate of Change
### Constant Rate of Change
A constant rate of change occurs when the dependent variable changes by the same amount for every unit change in the independent variable. This results in a straight line on a graph, indicating a consistent relationship between the variables.
### Variable Rate of Change
A variable rate of change occurs when the dependent variable changes by different amounts for different units of change in the independent variable. This results in a curved line on a graph, indicating a changing relationship between the variables.
## Worksheet 1: Calculating Rate of Change
### Worksheet Overview
This worksheet is designed to help you practice calculating the rate of change using given data. It consists of a series of questions with different scenarios, allowing you to apply the formula and determine the rate of change in each case.
### Example Question
Question: The temperature in a city decreases by 2 degrees Celsius every hour. What is the rate of temperature change?
Solution: In this scenario, the temperature change is the dependent variable, and time is the independent variable. Since the temperature decreases by 2 degrees Celsius for every hour, the rate of change is -2 degrees Celsius per hour.
## Worksheet 2: Interpreting Rate of Change
### Worksheet Overview
This worksheet focuses on interpreting the rate of change in various contexts. It presents you with different scenarios and asks you to analyze the rate of change, its meaning, and its implications.
### Example Question
Question: The population of a town is increasing by 500 people per year. What does the rate of change signify?
Solution: In this case, the rate of change of 500 people per year indicates that the population is growing at a steady pace. It means that every year, the town's population increases by 500 individuals.
## Worksheet 3: Graphing Rate of Change
### Worksheet Overview
This worksheet focuses on graphing the rate of change using given data. It provides you with sets of data points and asks you to plot them on a graph, identify the rate of change, and interpret the results.
### Example Question
Question: The distance traveled by a car over time is given by the following data:
Time (hours): [0, 1, 2, 3, 4, 5]
Distance (miles): [0, 50, 100, 150, 200, 250]
Plot the data points on a graph and determine the rate of change.
Solution: By plotting the data points on a graph, we can observe a straight line, indicating a constant rate of change. The rate of change is 50 miles per hour, as the distance increases by 50 miles for every hour of travel.
## Worksheet 4: Applications of Rate of Change
### Worksheet Overview
This worksheet explores the practical applications of rate of change in real-world scenarios. It presents you with problems from various fields and challenges you to apply the concept to solve them.
### Example Question
Question: A car travels at a speed of 60 miles per hour. How long will it take to cover a distance of 180 miles?
Solution: To solve this problem, we can use the rate of change formula. Since the distance is the dependent variable and time is the independent variable, we divide the distance (180 miles) by the rate of change (60 miles per hour). The car will take 3 hours to cover a distance of 180 miles.
## Conclusion
Rate of change is a fundamental mathematical concept with numerous applications in various disciplines. By mastering the calculations, interpretation, and graphing of rate of change, you can enhance your problem-solving skills and gain a deeper understanding of how different variables relate to each other. We hope these worksheets provide you with valuable practice and help you excel in your studies or teaching endeavors.
|
# Graph solver
Apps can be a great way to help students with their algebra. Let's try the best Graph solver. Our website can help me with math work.
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Here, we will show you how to work with Graph solver. How to solve an equation by elimination. The first step is to understand what an equation is. An equation is a mathematical sentence that shows that two things are equal. In order to solve an equation, you need to find the value of the variable that makes the two sides of the equation equal. There are many different methods of solving equations, but one of the simplest is called "elimination." Elimination involves adding or subtracting terms from both sides of the equation in order to cancel out one or more of the variables.
Algebra is a math discipline that studies mathematical symbols and the rules for manipulating these symbols. In elementary algebra, students are introduced to solving linear equations and graphing linear equations. In intermediate algebra, students learn how to solve quadratic equations. Advanced algebra includes the study of polynomial equations, rational equations, and logarithmic equations. Algebraic methods can be used to solve problems in physics and engineering. Algebra is also used in computer science and in economics. Algebra is a critical tool for solving problems in many different fields.
An equation is a mathematical statement that two things are equal. For example, the equation 2+2=4 states that two plus two equals four. In order to solve for x, one must first identify what x represents in the equation. In the equation 2x+4=8, x represents the unknown quantity. In order to solve for x, one must use algebraic methods to determine what value x must be in order to make the equation true. There are many different methods that can be used to solve for x, but the most common method is to use algebraic equations. Once the value of x has been determined, it can be plugged into the original equation to check if the equation is still true. For example, in the equation 2x+4=8, if x=2 then 2(2)+4=8 which is true. Therefore, plugging in the value of x allows one to check if their solution is correct. While solving for x may seem like a daunting task at first, with a little practice it can be easily mastered. With a little perseverance and patience anyone can learn how to solve for x.
Math can be a difficult subject for many students, but it doesn't have to be. There are now many online math solvers that can help you work through problems and even show you the work that was done to get the answer. This can be a great way to check your work or to see how to do a problem if you are stuck. Some math solvers will even give you step-by-step instructions on how to solve a problem. All you need is a internet connection and you can get help with math anytime, anywhere. So if you are struggling with math, don't hesitate to use one of these online solvers. You might just find it makes math a lot easier.
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Solution to mathematical problems Math help websites for algebra Implicit differentiation solver Systems by substitution solver Free algebra solver
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# How do you write an equation in slope intercept form given (5, 7) and (8, 22)?
Apr 8, 2018
$y = 5 x - 18$
#### Explanation:
$\text{the equation of a line in "color(blue)"slope-intercept form}$ is.
•color(white)(x)y=mx+b
$\text{where m is the slope and b the y-intercept}$
$\text{to calculate m use the "color(blue)"gradient formula}$
•color(white)(x)m=(y_2-y_1)/(x_2-x_1)
$\text{let "(x_1,y_1)=(5,7)" and } \left({x}_{2} , {y}_{2}\right) = \left(8 , 22\right)$
$\Rightarrow m = \frac{22 - 7}{8 - 5} = \frac{15}{3} = 5$
$\Rightarrow y = 5 x + b \leftarrow \textcolor{b l u e}{\text{is the partial equation}}$
$\text{to find b substitute either of the 2 given points into}$
$\text{the partial equation}$
$\text{using "(5,7)" then}$
$7 = 25 + b \Rightarrow b = 7 - 25 = - 18$
$\Rightarrow y = 5 x - 18 \leftarrow \textcolor{red}{\text{in slope-intercept form}}$
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Scientific Notation & Significant Figures
# Scientific Notation & Significant Figures | Physics Class 11 - NEET PDF Download
Table of contents Introduction Scientific Notation What are Significant Figures? Addition and Subtraction Multiplication and Division Order of Magnitude
## Introduction
Many times in the study of chemistry, one has to deal with experimental data as well as theoretical calculations.
There are meaningful ways to handle the numbers conveniently and present the data realistically with certainty to the extent possible like:
• Scientific Notation
• Significant Figures
• Dimensional Analysis
## Scientific Notation
• Scientific Notation is a way of expressing numbers that are too big or too small to be conveniently written in decimal form.
• In which any number can be represented in form N × 10(where n is an exponent having positive or negative values and N can vary between 1 to 10).
Example: We can write 232.508 as 2.32508 × 102 in scientific notation. Similarly, 0.00016 can be written as 1.6 × 10–4.
Thus, we can write 232.508 as 2.32508 × 102 in scientific notation. Note that while writing it, the decimal had to be moved to the left by two places and the same is the exponent (2) of 10 in the scientific notation.
Similarly, 0.00016 can be written as 1.6 × 10–4. Here the decimal has to be moved four places to the right and (– 4) is the exponent in the scientific notation.
### Multiplication and Division for Exponential Numbers
• These two operations follow the same rules which are there for exponential numbers, i.e.
### Addition and Subtraction for Exponential Numbers
• For these two operations, first, the numbers are written in such a way that they have the same exponent. After that, the coefficients (digit terms) are added or subtracted as the case may be.
• Thus, for adding 6.65 × 104 and 8.95 × 103, the exponent is made the same for both numbers. Thus, we get (6.65 × 104) + (0.895 × 104).
### Examples:
Q.1. Which of the following options is not correct?
(a) 8008 = 8.008 x 103
(b) 208 = 3
(c) 5000 = 5.0 x 103
(d) 2.0034 = 4
Ans: (d)
Solution:
2.0034 = 4
Q.2. Exponential notation in which any number can be represented in the form, Nx 10here N is termed as
(a) non –digit term
(b) digit term
(c) numeral
(d) base term
Ans: (b)
Solution:
In exponential notation N × 10n, N is a number called digit term which varies between 1.000 and 9.000….
## What are Significant Figures?
It is important to understand that the way a measurement is taken affects its accuracy. For example, you could measure the length of a leaf with a ruler that had markings every centimeter (cm). In this example, illustrated below, the leaf is longer than 3 cm and shorter than 4 cm, so you might estimate that the leaf is 3.5 cm long.
On the other hand, if you measured the same leaf with a ruler that had markings every millimeter (mm), as drawn below, you can see that the end of the leaf actually falls between the markings for 3.5 and 3.6 cm (or 35 and 36 mm). Because it’s closer to the 3.5 marking, you might estimate that the leaf is 3.52 cm (or 35.2 mm) long.
Using the second ruler, it’s possible to estimate that the leaf is 3.52 cm long, but it is not possible to measure that accurately with the first ruler. In this way, the number of digits in the measured value gives us an idea of the maximum accuracy of the measurement. These are called significant digits or significant figures.
RULES FOR SIGNIFICANT FIGURES
• All non-zero numbers ARE significant. The number 33.2 has THREE significant figures because all of the digits present are non-zero.
• Zeros between two non-zero digits ARE significant. 2051 has FOUR significant figures. The zero is between a 2 and a 5.
• Leading zeros are NOT significant. They're nothing more than "place holders." The number 0.54 has only TWO significant figures. 0.0032 also has TWO significant figures. All of the zeros are leading.
• Trailing zeros to the right of the decimal ARE significant. There are FOUR significant figures in 92.00.
92.00 is different from 92: a scientist who measures 92.00 milliliters knows his value to the nearest 1/100th milliliter; meanwhile his colleague who measured 92 milliliters only knows his value to the nearest 1 milliliter. It's important to understand that "zero" does not mean "nothing." Zero denotes actual information, just like any other number. You cannot tag on zeros that aren't certain to belong there.
• Trailing zeros in a whole number with the decimal shown ARE significant. Placing a decimal at the end of a number is usually not done. By convention, however, this decimal indicates a significant zero.
For example, "540." indicates that the trailing zero IS significant; there are THREE significant figures in this value.
• Trailing zeros in a whole number with no decimal shown are NOT significant. Writing just "540" indicates that the zero is NOT significant, and there are only TWO significant figures in this value.
• Exact numbers have an INFINITE number of significant figures. This rule applies to numbers that are definitions.
For example, 1 meter = 1.00 meters
= 1.0000 meters
= 1.0000000000000000000 meters, etc.
So now back to the example posed in the Rounding Tutorial: Round 1000.3 to four significant figures. 1000.3 has five significant figures (the zeros are between non-zero digits 1 and 3, so by rule 2 above, they are significant.) We need to drop the final 3, and since 3 < 5, we leave the last zero alone. so 1000. is our four-significant-figure answer. (from rules 5 and 6, we see that in order for the trailing zeros to "count" as significant, they must be followed by a decimal. Writing just "1000" would give us only one significant figure.)
• For a number in scientific notation: N × 10x, all digits comprising N ARE significant by the first 6 rules; "10" and "x" are NOT significant. 5.02 × 104 has THREE significant figures: "5.02." "10 and "4" are not significant.
Rule 8 provides the opportunity to change the number of significant figures in a value by manipulating its form.
For example, let's try writing 1100 with THREE significant figures. By rule 6, 1100 has TWO significant figures; its two trailing zeros are not significant. If we add a decimal to the end, we have 1100., with FOUR significant figures (by rule 5.) But by writing it in scientific notation: 1.10 × 103, we create a THREE-significant-figure value.
### Addition and Subtraction
How are significant figures handled in calculations? It depends on what type of calculation is being performed. If the calculation is an addition or a subtraction, the rule is as follows: limit the reported answer to the rightmost column that all numbers have significant figures in common. For example, if you were to add 1.2 and 4.71, we note that the first number stops its significant figures in the tenths column, while the second number stops its significant figures in the hundredths column. We therefore limit our answer to the tenths column.
We drop the last digit—the 1—because it is not significant to the final answer.
The dropping of positions in sums and differences brings up the topic of rounding. Although there are several conventions, in this text we will adopt the following rule: the final answer should be rounded up if the first dropped digit is 5 or greater, and rounded down if the first dropped digit is less than 5.
## Multiplication and Division
For multiplication or division, the rule is to count the number of significant figures in each number being multiplied or divided and then limit the significant figures in the answer to the lowest count. An example is as follows:
The final answer, limited to four significant figures, is 4,094. The first digit dropped is 1, so we do not round up.
Scientific notation provides a way of communicating significant figures without ambiguity. You simply include all the significant figures in the leading number. For example, the number 450 has two significant figures and would be written in scientific notation as 4.5 × 102, whereas 450.0 has four significant figures and would be written as 4.500 × 102. In scientific notation, all significant figures are listed explicitly.
## Order of Magnitude
• Order of magnitude of a quantity is the power of 10 required to represent that quantity. This power is determined after rounding off the value of the quantity properly. For rounding off, the last digit is simply ignored if it is less than 5 and, is increased by one if it is 5 or more than 5.
• When a number is divided by 10(where x is the order of the number) the result will always lie between 0.5 and 5 i.e. 0.5 < N/10< 5
Example. Order of magnitude of the following values can be determined as follows:
(a) 49 = 4.9 × 101 » 101 \ Order of magnitude = 1
(b) 51 = 5.1 × 101 » 102 \ Order of magnitude = 2
(c) 0.049 = 4.9 × 10–2 » 10–2 \ Order of magnitude = –2
(d) 0.050 = 5.0 × 10–2 » 10–1 \ Order of magnitude = –1
(e) 0.051 = 5.1 × 10–2 » 10–1 \ Order of magnitude = –1
### ROUNDING
When doing calculations using significant figures, you will find it necessary to round your answer to the nearest significant digit. There are therefore a few rules of rounding that help retain as much accuracy as possible in the final answer.
• Complete all sequential calculations BEFORE doing any rounding, since rounding early reduces the number of significant figures available for subsequent calculations.
• If the last significant digit is followed by a 6, 7, 8, or 9, round up.
Example: 5.677, rounded to three significant digits, is 5.68
• If the last significant digit is followed by a 0, 1, 2, 3, or 4, simply drop the trailing digits.
Example: 561200, rounded to three significant digits, is 561000
• If the last significant digit is EVEN, and is followed by a 5, drop the trailing digits.
Example: 45850, rounded to three significant digits, is 45800
• If the last significant digit is ODD, and is followed by a 5, round up.
Example: 3.47588, rounded to three significant digits, is 3.48
AMBIGUOUS ZEROS
So what happens if your calculation or measurement ends in a zero? For example, what if you measured a branch that was 200 cm (not 199 or 201 cm) long? The zeros in a measured value of 200 cm in this case appear ambiguous, since it could suggest that there is only one significant digit.
One way to reduce this ambiguity is to use significant figures with scientific notation.
The document Scientific Notation & Significant Figures | Physics Class 11 - NEET is a part of the NEET Course Physics Class 11.
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## Physics Class 11
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## FAQs on Scientific Notation & Significant Figures - Physics Class 11 - NEET
1. What is scientific notation?
Scientific notation is a way to express numbers that are very large or very small in a more concise and manageable format. It is written in the form of a number between 1 and 10, multiplied by a power of 10. For example, the number 500,000,000 can be written as 5 × 10^8 in scientific notation.
2. What are significant figures?
Significant figures are the digits in a number that carry meaning or contribute to its precision. They include all the certain digits plus one uncertain or estimated digit. Significant figures are important because they indicate the precision of a measurement or calculation. For example, the number 34.56 has four significant figures.
3. How do you perform addition and subtraction with significant figures?
When adding or subtracting numbers with significant figures, the result should be rounded to the same decimal place as the least precise number in the calculation. For example, if you are adding 3.45 and 2.1, the result should be rounded to the tenths place, giving you 5.6.
4. How do you perform multiplication and division with significant figures?
When multiplying or dividing numbers with significant figures, the result should be rounded to the same number of significant figures as the least precise number in the calculation. For example, if you are multiplying 2.34 and 1.2, the result should be rounded to two significant figures, giving you 2.8.
5. What is the order of magnitude in scientific notation?
The order of magnitude in scientific notation refers to the power of 10 that is multiplied by the number. It represents the approximate size of the number in terms of magnitude. For example, in the scientific notation 4.5 × 10^3, the order of magnitude is 3, indicating that the number is in the thousands range.
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# Section 11.4 Question 1
## What is a derivative function?
Let’s examine the tangent line on the function pictured in Figure 1.
Figure 1 – A quadratic function g(x) (blue) with a tangent line (red) at x = 1.
We can approximate the value of the derivative at x = 1 or g′(1) by calculating the slope of the tangent line at x = 1. Any two points on the tangent line can be used to make an estimate of the slope.
Figure 2 – The points (0, 4) and (2, 0) can be used to estimate the slope of the tangent line at x = 1.
From Figure 2, we see that the slope of the tangent line or g′(1) is approximately -4/2 or -2. Any other two points on the line could be used to calculate this slope. For instance, the points (1, 2) and (2, 0) are on the tangent line. The slope between these points is
$$g’\left( 1 \right) \approx \frac{{2 – 0}}{{1 – 2}} = – 2$$
Different points don’t always lead to exactly the same slope, but they should be close. If we look at the tangent line for g(x) at x = 2, we see that it is horizontal and passes through the point (2, 1).
Figure 3 – The tangent line to g(x) at x = 2 is a horizontal tangent line.
Since the tangent line is horizontal, its slope is zero. This tells us that g′(2) = 0.
Let’s estimate the slope of one more tangent line at x = 4.
Figure 4 – The tangent line to g(x) at x = 4 passes through the points (3,1) and (5,9).
The slope is calculated as
$$g’\left( 4 \right) \approx \frac{{9 – 1}}{{5 – 3}} = \frac{8}{2} = 4$$
With these three tangent lines, we can organize the slopes in a table.
If we graph these values with the x values as the independent variable and the slopes as the dependent variable, we see a pattern emerge.
Figure 5 – In this graph, the slopes of the quadratic function are graphed at corresponding x values.
The slopes appear to lie along a straight line. If we draw this line through the points we can use it to find the slope of the tangent line at other x values.
Figure 6 – Each of the slopes of the quadratic function lie along a straight line.
For instance, at x = 3 the line passes though y = 2 . This means the slope of the tangent line at x = 3 on the quadratic function is 2. The function corresponding to this line is called the derivative of the quadratic function g(x) and is denoted by g′(x).
The derivative function g′(x) is a function whose output is the slope of the tangent line to the function y = g(x) at an input x.
All of these concepts can be a bit confusing. So let’s recap these different ideas using a function g(x). Keep in mind that we could use any name in place of g and any variable in place of x.
### Example 1 Find the Derivative Function
The graph of the function f (x) is shown below.
Find the graph of the derivative function f ′(x) using the tangent lines to the function f (x).
Solution To find the derivative of the function, we’ll draw tangent lines along the function, record the slopes in a table and graph the corresponding ordered pairs.
Using the grid on the graph, we see that the tangent line passes through (-3, 0) and (-2, 1.8). The slope of the tangent line is approximately
$$f'( – 3) = \frac{{1.8 – 0}}{{ – 2 – ( – 3)}} = 1.8$$
We’ll continue to fill out the table of tangent line slopes:
Our ability to calculate the slope is constrained by our ability to locate points on the tangent line. For this tangent line, the points (-2, 1) and (1, 1.9) are on the line. The slope of the tangent line is approximately
$$f'( – 2) = \frac{{1.9 – 1}}{{1 – ( – 2)}} = 0.3$$
At the next row in the table, we find the slope of the tangent line at x = -1:
Any two points on the tangent line can be used to find the slope. For this tangent line, the points (-3, 2) and (2, -1) are on the line. The slope of the tangent line is approximately
$$f'( – 1) = \frac{{ – 1 – 2}}{{2 – ( – 3)}} = – 0.6$$
If we continue to find the slopes of tangent lines at each of the x values in the table, we can graph the ordered pairs in the form (x, f ′(x)). This means that we graph the x values horizontally and the f ′(x) values vertically.
These ordered pairs lie on the derivative function f ′(x).
If we were to graph more points on the derivative function, we would see a graph like the one below.
Usually it is not necessary to graph a large number of points to graph the derivative function. Typically a few points on the graph are enough to display the overall pattern. Then the rest of the graph can be sketched to approximate the pattern.
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# Angle Between Two Lines
Let y = m1x + c1 and y = m2x + c2 be the equations of two lines in a plane where,
```m1 = slope of line 1
c1 = y-intercept made by line 1
m2 = slope of line 2
c2 = y-intercept made by line 2```
```<BAX = θ1
<DCX = θ2
∴ m1 = tan θ1 and
m2 = tan θ2 ```
Let the angle between the lines AB and CD be Ø (<APC) then,
Also if we consider <APD as the angle between lines,
Hence, the angle between two lines is,
Condition for perpendicularity
The two lines are perpendicular means. Ø = 90°
Thus, the lines are perpendicular if the product of their slope is -1.
Condition for parallelism
The two lines are perpendicular means, Ø = 0°
Thus, the lines are parallel if their slopes are equal.
### Angle Between Two Lines Examples
1. Find the angle between the lines 2x-3y+7 = 0 and 7x+4y-9 = 0.
Solution:
Comparing the equation with equation of straight line, y = mx + c,
Slope of line 2x-3y+7=0 is (m1) = 2/3
Slope of line 7x+4y-9=0 is (m2) = -7/4
Let, Ø be the angle between two lines, then
2. Find the equation of line through point (3,2) and making angle 45° with the line x-2y = 3.
Solution:
Let m be the slope of the required line passing through (3,2). So, using slope point form, its equation is
y-2 = m(x-3) ——— (i)
Slope of line x-2y = 3 is 1/2.
Since, these lines make an angle of 45° so,
Substituting values of m in equation (i), we get
y-2 = 3(x-3) and y-2 = -(x-3)/3
or, 3x-y-7 = 0 and x+3y-9 = 0 are the required equations of line.
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## Write a system of two equations in two variables to solve the problem. On a mother’s 23-mile commute to work, she drops her daughter off at
Question
Write a system of two equations in two variables to solve the problem. On a mother’s 23-mile commute to work, she drops her daughter off at a child care center. The first part of the trip is 5 miles less than the second part. How long is each part of her morning commute? first part mi second part m
in progress 0
1 week 2021-09-09T02:29:04+00:00 1 Answer 0
1. Answer: the first part is 9 miles.
The second part is 14 miles.
Step-by-step explanation:
Let x represent the distance of the first part of her morning commute.
Let y represent the distance of the second part of her morning commute.
On a mother’s 23-mile commute to work, she drops her daughter off at a child care center. This means that
x + y = 23- – – – – – – – – – – – 1
The first part of the trip is 5 miles less than the second part. It means that
x = y – 5
Substituting x = y – 5 into equation 1, it becomes
y – 5 + y = 23
y + y = 23 + 5
2y = 28
y = 28/2 = 14
x = y – 5 = 14 – 5
x = 9
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# The Cali Garmo
does Math
## Ferrers Diagram
By Cali G , Published on Tue 07 January 2020
Category: math / symmetric functions
## Background
Both Young diagrams and Ferrers diagrams represent the same thing: a partition. The main difference between the two is what they use to represent the diagram:
Definitions: A Ferrers diagram is a way to represent a partition using dots. A Young diagram is a way to represent a partition using square boxes. If $\lambda = (\lambda_1, \lambda_2, \ldots)$ is a partition, then its Ferrers diagram has $\lambda_1$ dots in the first row, $\lambda_2$ dots in the second row, etc. Likewise, its Young diagram has $\lambda_1$ boxes in the first row, $\lambda_2$ boxes in the second row, etc.
As an example, let $\lambda = (4,2,1)$ be a partition of $7$. Then its Ferrers diagram is given by and its Young diagram is given by
There are 3 types of ways we can draw a Young diagram. The way given is known as the "French notation". There are two other notations: English and Russian.
The English notation flips the Young diagram upside down so that the rows go from top to bottom. The Russian notation puts everything on a diagonal so that the boxes are going towards the north-east.
In our example for $(4,2,1)$ the English notation is given by: and the Russian notation is given by:
## Distribution of Young diagram
We next consider how the Young diagrams are distributed. For this we'll need the following definition.
Definition 1: Let $\lambda$ be a partition of $n$. The order $\abs{\lambda}$ of $\lambda$ is $n$. Alternatively, it is the number of boxes in its Young diagram.
Recall that $n^k = (n,n, \ldots, n)$ (with $k$ number of $n$s) is a partition of $n \cdot k$. Recall also that for two partitions $\lambda$ and $\mu$ then $\lambda \subseteq \mu$ if $\lambda_i \leq \mu_i$ for every $i$. An alternative way to see this is through Young diagrams: $\lambda \subseteq \mu$ if the Young diagram of $\lambda$ is contained in the Young diagram of $\mu$.
To see the distribution, we will also recall some $q$ notations.
Theorem:
We show this in an example. Let $n = 3$ and $k = 2$. Then $n^k = (3,3)$ and is represented by the Young diagram:
As our sum is over every $\lambda$ which is contained in this Young diagram, we list every possible Young diagram that is contained in the above one:
Partitions of $5$:
Partitions of $4$:
Partitions of $3$:
Partitions of $2$:
Partitions of $1$:
Partitions of $0$: $\emptyset$
Therefore we have Note that the coefficient $c_i$ is just the number of Young diagram of a partition of $i$ that fits inside $n^k$.
Alternatively:
So we see that the two are equal in our example.
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CHAPTER 10 FACTORING POLYNOMIALS A factor of a quantity N, as defined in chapter 2 of this course, is any expression which can be divided into N without producing a remainder. Thus 2 and 3 are factors of 6, and the factors of 5x are 5 and x. Conversely, when all of the factors of N are multiplied together, the product is N. This definition is extended to include polynomials. The factors of a polynomial are two or more expressions which, when multiplied together, give the polynomial as a product. For example, 3, x, and x2 - 4 are factors of 3x3 - 12x, as the following equation shows: (3)(x)(x2 - 4) = 3x3 - 12x The factors 3 and x, which are common to both terms of the polynomial 3x3 - 12x, are called COMMON FACTORS. The distributive principle, mentioned in chapters 3 and 9 of this course, is an important part of the concept of factoring. It may be stated as follows: If the sum of two or more quantities is multiplied by a third quantity, the. product is found by applying the multiplier to each of the original quantities separately and summing the resulting expressions. It is this principle which allows us to separate common factors from the terms of a polynomial. Just as with numbers, an algebraic expression is a prime factor if it has no other factors except itself and 1. The factor x2 - 4 is not prime, since it can be separated into x - 2 and x + 2. The factors x -2 and x+2 are both prime factors, rime they cannot be separated into other factors. The process of finding the factors of a polynomial is called FACTORING. An expression is said to be factored completely when it has been separated into its prime factors. The polynomial 3x3 - 12x is factored completely as follows: 3x3 - 12x = 3x(x - 2)(x + 2) COMMON FACTORS Factoring any polynomial begins with the removal of common factors. Notice that "removal" of a factor does not mean discarding it. To remove a factor is to insert parentheses and move the factor outside the parentheses as a common multiplier. The removal of common factors proceeds as follows: 1. Inspect the polynomial and find the factors which are common to all terms. These common factors, multiplied together, comprise the ‘largest common factor ." 2. Mentally divide each term of the polynomial by the largest common factor and write the quotients within a set of parentheses. 3. Write the largest common factor outside the parentheses as a common multiplier. For example, the expression x2y - xy2 contains xy as a factor of each term. Therefore, it is factored as follows: x2y - xy2 = xy(x - Y) Other examples of factoring by the removal of common factors are found in the following expressions: In selecting common factors, always remove as many factors as possible from each term in order to factor completely. For example, x is a factor of 3ax2 - 3ax, so that 3ax2 - 3ax is equal to x(3ax -3a). However, 3 and a are also factors. Thus the largest common factor is 3ax. When factored completely, the expression is as follows: 3ax2 - 3ax = 3ax(x - 1) Practice problems: Remove the common factors: LITERAL EXPONENTS It is frequently necessary to remove common factors involving literal exponents; that is, exponents composed of letters rather than numbers. A typical expression involving literal exponents is x2a + xa , ’ in which xa is a common factor. The factored form is xa(xa + 1). Another example of this type is am+n, + 2am. Remember that (am·n) is equivalent to (am)(an). Thus the factored form is as follows:
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# A mirror produces an image that is located 32.0 cm behind the mirror when the object is located...
## Question:
A mirror produces an image that is located 32.0 cm behind the mirror when the object is located 8.00 cm in front of the mirror. What is the focal length of the mirror?
## Spherical Mirror Formula:
First we have to understand the spherical mirror formula to solve this problem:
The spherical mirror formula is given by: {eq}\displaystyle \frac{1}{f} = \frac{1}{v} + \frac{1}{u} {/eq} where:
• {eq}u {/eq} is the object distance from the mirror.
• {eq}v {/eq} is the image distance from the mirror.
• {eq}f {/eq} is the focal length of the mirror. It is also defined as: {eq}\displaystyle f = \frac{R}{2} {/eq} where {eq}R {/eq} is radius of sphere.
## Answer and Explanation:
Given:
• The image distance from the spherical mirror is: {eq}v = 32 \ \rm cm {/eq}.
• The object distance from the spherical mirror is: {eq}u = - 8 \ \rm cm {/eq}. Here the object distance is negative because the object is placed left of side of the image.
We will compute the object distance from the mirror.
Apply the spherical mirror formula:
\begin{align*} \displaystyle \frac{1}{f} &= \frac{1}{v} + \frac{1}{u} &\text{(Where } f = 34 \ \rm cm \text{ and } v = 13 \text{)}\\ \frac{1}{f} &= \frac{1}{32} + \frac{1}{- 8} &\text{(Plugging in all given values)}\\ \frac{1}{f} &= \frac{1}{32} - \frac{1}{8} \\ \frac{1}{f} &= \frac{1 - 4}{32} \\ \frac{1}{f} &= \frac{- 3}{32} \\ - 3 \ f &= 32 &\text{(Doing cross multiplication)}\\ f &= - \frac{32}{3} &\text{(Dividing both sides by 3)}\\ f \ &\boxed{= - 10.667 \ \rm cm } &\text{(Negative sign is showing, the mirror is concave mirror)}\\ \end{align*}
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# 4. solving inequalities
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### 4. solving inequalities
1. 1. 1.4 Solving Inequalities OBJECTIVES a Determine whether a given number is a solution of an inequality. b Graph an inequality on the number line. c Solve inequalities using the addition principle. d Solve inequalities using the multiplication principle. e Solve inequalities using the addition principle and the multiplication principle together. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 1
2. 2. 1.4 Solving Inequalities a Determine whether a given number is a solution of an inequality. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 2
3. 3. 1.4 Solving Inequalities SOLUTION A replacement that makes an inequality true is called a solution. The set of all solutions is called the solution set. When we have found the set of all solutions of an inequality, we say that we have solved the inequality. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 3
4. 4. 1.4 a Solving Inequalities Determine whether a given number is a solution of an inequality. EXAMPLE Determine whether 2 is a solution of x < 2. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 4
5. 5. 1.4 Solving Inequalities Determine whether a given number is a solution of an a inequality. EXAMPLE Determine whether 6 is a solution of Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 5
6. 6. 1.4 b Solving Inequalities Graph an inequality on the number line. A graph of an inequality is a drawing that represents its solutions. An inequality in one variable can be graphed on the number line. An inequality in two variables can be graphed on the coordinate plane. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 6
7. 7. 1.4 Solving Inequalities b Graph an inequality on the number line. EXAMPLE The solutions are all those numbers less than 2. They are shown on the number line by shading all points to the left of 2. The open circle at 2 indicates that 2 is not part of the graph. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 7
8. 8. 1.4 Solving Inequalities b Graph an inequality on the number line. EXAMPLE The solutions are shown on the number line by shading the point for –3 and all points to the right of –3. The closed circle at –3 indicates that –3 is part of the graph. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 8
9. 9. 1.4 Solving Inequalities b Graph an inequality on the number line. EXAMPLE The inequality is read “–3 is less than or equal to x and x is less than 2,” or “x is greater than or equal to –3 and x is less than 2.” In order to be a solution of this inequality, a number must be a solution of both and x < 2. We can see from the graphs that the solution set consists of the numbers that overlap in the two solution sets in Examples 5 and 6. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 9
10. 10. 1.4 Solving Inequalities b Graph an inequality on the number line. EXAMPLE The open circle at 2 means that 2 is not part of the graph. The closed circle at –3 means that is part of the graph. The other solutions are shaded. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 10
11. 11. 1.4 Solving Inequalities c Solve inequalities using the addition principle. Any solution of one inequality is a solution of the other—they are equivalent. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 11
12. 12. 1.4 Solving Inequalities THE ADDITION PRINCIPLE FOR INEQUALITIES For any real numbers a, b, and c: In other words, when we add or subtract the same number on both sides of an inequality, the direction of the inequality symbol is not changed. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 12
13. 13. 1.4 Solving Inequalities c Solve inequalities using the addition principle. As with equation solving, when solving inequalities, our goal is to isolate the variable on one side. Then it is easier to determine the solution set. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 13
14. 14. 1.4 c Solving Inequalities Solve inequalities using the addition principle. EXAMPLE Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 14
15. 15. 1.4 c Solving Inequalities Solve inequalities using the addition principle. A shorter notation for sets is called set-builder notation. is read Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 15
16. 16. 1.4 Solving Inequalities THE MULTIPLICATION PRINCIPLE FOR INEQUALITIES For any real numbers a and b, and any positive number c: For any real numbers a and b, and any negative number c: Similar statements hold for Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 16
17. 17. 1.4 Solving Inequalities THE MULTIPLICATION PRINCIPLE FOR INEQUALITIES In other words, when we multiply or divide by a positive number on both sides of an inequality, the direction of the inequality symbol stays the same. When we multiply or divide by a negative number on both sides of an inequality, the direction of the inequality symbol is reversed. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 17
18. 18. 1.4 Solving Inequalities d Solve inequalities using the multiplication principle. EXAMPLE Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 18
19. 19. 1.4 Solving Inequalities d Solve inequalities using the multiplication principle. EXAMPLE Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 19
20. 20. 1.4 Solving Inequalities e Solve inequalities using the addition principle and the multiplication principle together. Remember to reverse the inequality symbol when multiplying or dividing on both sides by a negative number. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 20
21. 21. 1.4 e Solving Inequalities Solve inequalities using the addition principle and the multiplication principle together. EXAMPLE Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 21
22. 22. 1.4 Solving Inequalities Solve inequalities using the addition principle and the e multiplication principle together. EXAMPLE Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 22
23. 23. 1.4 e Solving Inequalities Solve inequalities using the addition principle and the multiplication principle together. EXAMPLE First, we use the distributive law to remove parentheses. Next, we collect like terms and then use the addition and multiplication principles for inequalities to get an equivalent inequality with x alone on one side. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 23
24. 24. 1.4 Solving Inequalities Solve inequalities using the addition principle and the e multiplication principle together. EXAMPLE Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 24
25. 25. 1.4 Solving Inequalities Solve inequalities using the addition principle and the e multiplication principle together. EXAMPLE The greatest number of decimal places in any one number is two. Multiplying by 100, which has two 0’s, will clear decimals. Then we proceed as before. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 25
26. 26. 1.4 Solving Inequalities Solve inequalities using the addition principle and the e multiplication principle together. EXAMPLE Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 26
27. 27. 1.4 Solving Inequalities Solve inequalities using the addition principle and the e multiplication principle together. EXAMPLE The number 6 is the least common multiple of all the denominators. Thus we first multiply by 6 on both sides to clear the fractions. Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 27
28. 28. 1.4 Solving Inequalities Solve inequalities using the addition principle and the e multiplication principle together. EXAMPLE Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 28
29. 29. 1.4 Solving Inequalities Solve inequalities using the addition principle and the e multiplication principle together. EXAMPLE Copyright 2012, 2008, 2004, 2000 Pearson Education, Inc. Slide 29
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# Playing with Numbers
## Playing with Numbers
### General Form of Numbers
Various types of numbers such as Natural numbers, Whole numbers, Integers, Rational numbers and the various properties such as closure, associative, commutative and distributive.
2-digit number
The number in the general form can be written as for example 26 = 2 x 10 + 6. General Form of a 2-digit Number 10 × a + 1 × b
1. The sum of a 2-digit number and the number obtained by interchanging its digits is always divisible by 11.
2. The difference between a 2-digit number and the number obtained by interchanging its digits is always divisible by 9.
Assume ab is a 2-digit number.
• a is the tens digit
• b is the ones digit
ab = 10 × a + 1 × b.
3-digit number
General form of a 3-digit number is 100 × a + 10 × b + 1 × c.
1. The difference between a 3-digit number and a number obtained by reversing its digits is always divisible by 99.
Assume abc is a 3-digit number, where:
• a is the hundreds digit
• b is the tens digit
• c is the ones digit
abc = 100 × a + 10 × b + 1 × c
Letters for digits
Assume that each letter in a puzzle stands for just one digit and each digit is represented by a one letter, so it is like cracking a code. Here problems of addition and multiplication to solving puzzles.
For example, eighty one will be written as 81 not as 081 or 0081.
### Tests of Divisibility
The tests of divisibility with 10, 5, 2, 3, 6, 4, 8, 9 and 11. In this, learn why the numbers are divisible by 10, 5, 2, 3, 6, 4, 8, 9 and 11.
A number is said to be divisible by another number, when the remainder is zero.
• A number is divisible by 10, if its ones digit is 0.
• A number is divisible by 5, if its ones digit is 0 or 5.
• A number is divisible by 2, if its ones digit is 0, 2, 4, 6 or 8.
• If a number is divisible by 10, then the number is also divisible by 2 and 5.
• A number is divisible by 9, if the sum of its digits is divisible by 9.
• A number is divisible by 3, if the sum of its digits is divisible by 3.
• If a number is divisible by 9, then the number is also divisible by 3.
• If a number is divisible by 3, then it may not necessarily be divisible by 9.
• If a number is divisible by 6, then the number is divisible by 2 as well as 3.
• If a number is divisible by 11, then the difference of its digits in odd places and the sum of its digits in
even places is either 0 or a multiple of 11.
• If a number is divisible by 4, then the number formed by its digits in units and tens places is divisible by 4.
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# 5.1: Linear Equations in Slope-Intercept Form
Difficulty Level: At Grade Created by: CK-12
Previously, you learned how to graph solutions to two-variable equations in slope-intercept form. This lesson focuses on how to write an equation for a graphed line. There are two things you will need from the graph to write the equation in slope-intercept form:
1. The y\begin{align*}y-\end{align*}intercept of the graph
2. The slope of the line
Having these two pieces of information will allow you to make the appropriate substitutions in the slope-intercept formula. Recall from the last chapter,
Slope-intercept form: y=(slope)x+(yintercept)\begin{align*}y=(slope)x+ (y-intercept)\end{align*} or y=mx+b\begin{align*}y=mx+b\end{align*}
Example 1: Write the equation for a line with a slope of 4 and a y\begin{align*}y-\end{align*}intercept (0, –3).
Solution: Slope-intercept form requires two things: the slope and y\begin{align*}y-\end{align*}intercept. To write the equation, you substitute the values into the formula.
yyy=(slope)x+(yintercept)=4x+(3)=4x3\begin{align*}y& =(slope)x+ (y-intercept)\\ y& =4x+(-3)\\ y& =4x-3\end{align*}
You can also use a graphed line to determine the slope and y\begin{align*}y-\end{align*}intercept.
Example 2: Use the graph below to write its equation in slope-intercept form.
Solution: The y\begin{align*}y-\end{align*}intercept is (0, 2). Using the slope triangle, you can determine the slope is riserun=31=31\begin{align*}\frac{rise}{run}=\frac{-3}{-1}=\frac{3}{1}\end{align*}. Substituting the value 2 for b\begin{align*}b\end{align*} and the value 3 for m\begin{align*}m\end{align*}, the equation for this line is y=3x+2\begin{align*}y=3x+2\end{align*}.
## Writing an Equation Given the Slope and a Point
Sometimes it may be difficult to determine the y\begin{align*}y-\end{align*}intercept. Perhaps the y\begin{align*}y-\end{align*}intercept is rational instead of an integer. Maybe you don’t know the y\begin{align*}y-\end{align*}intercept. All you have is the slope and an ordered pair. You can use this information to write the equation in slope-intercept form. To do so, you will need to follow several steps.
Step 1: Begin by writing the formula for slope-intercept form y=mx+b\begin{align*}y=mx+b\end{align*}.
Step 2: Substitute the given slope for m\begin{align*}m\end{align*}.
Step 3: Use the ordered pair you are given (x,y)\begin{align*}(x, y)\end{align*} and substitute these values for the variables x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} in the equation.
Step 4: Solve for b\begin{align*}b\end{align*} (the y\begin{align*}y-\end{align*}intercept of the graph).
Step 5: Rewrite the original equation in Step 1, substituting the slope for m\begin{align*}m\end{align*} and the y\begin{align*}y-\end{align*}intercept for b\begin{align*}b\end{align*}.
Example 3: Write an equation for a line with slope of 4 that contains the ordered pair (–1, 5).
Solution:
Step 1: Begin by writing the formula for slope-intercept form.
y=mx+b\begin{align*}y=mx+b\end{align*}
Step 2: Substitute the given slope for m\begin{align*}m\end{align*}.
y=4x+b\begin{align*}y=4x+b\end{align*}
Step 3: Use the ordered pair you are given (–1, 5) and substitute these values for the variables x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} in the equation.
5=(4)(1)+b\begin{align*}5=(4)(-1)+b\end{align*}
Step 4: Solve for b\begin{align*}b\end{align*} (the y\begin{align*}y-\end{align*}intercept of the graph).
55+49=4+b=4+4+b=b\begin{align*}5& =-4+b\\ 5+4& =-4+4+b\\ 9& =b\end{align*}
Step 5: Rewrite y=mx+b\begin{align*}y=mx+b\end{align*}, substituting the slope for m\begin{align*}m\end{align*} and the y\begin{align*}y-\end{align*}intercept for b\begin{align*}b\end{align*}.
y=4x+9\begin{align*}y=4x+9\end{align*}
Example 4: Write the equation for a line with a slope of –3 containing the point (3, –5).
Solution: Using the five-steps from above:
yy554y=(slope)x+(yintercept)=3x+b=3(3)+b=9+b=b=3x+4\begin{align*}y& =(slope)x+(y-intercept)\\ y& =-3x+b\\ -5& =-3(3)+b\\ -5& =-9+b\\ 4& =b\\ y& =-3x+4\end{align*}
## Writing an Equation Given Two Points
In many cases, especially real-world situations, you are given neither the slope nor the y\begin{align*}y-\end{align*}intercept. You might have only two points to use to determine the equation of the line.
To find an equation for a line between two points, you need two things:
1. The y\begin{align*}y-\end{align*}intercept of the graph
2. The slope of the line
Previously, you learned how to determine the slope between two points. Let’s repeat the formula here.
The slope between any two points (x1,y1)\begin{align*}(x_1, y_1 )\end{align*} and (x2,y2)\begin{align*}(x_2, y_2)\end{align*} is: slope=y2y1x2x1\begin{align*}slope=\frac{y_2-y_1}{x_2-x_1}\end{align*}.
The procedure for determining a line given two points is the same five-step process as writing an equation given the slope and a point.
Example 5: Write the equation for the line containing the points (3, 2) and (–2, 4).
Solution: You need the slope of the line. Find the line's slope by using the formula. Choose one ordered pair to represent (x1,y1)\begin{align*}(x_1,y_1)\end{align*} and the other ordered pair to represent (x2,y2)\begin{align*}(x_2,y_2)\end{align*}.
slope=y2y1x2x1=4223=25\begin{align*}slope=\frac{y_2-y_1}{x_2-x_1}=\frac{4-2}{-2-3}=-\frac{2}{5}\end{align*}
Now use the five-step process to find the equation for this line.
Step 1: Begin by writing the formula for slope-intercept form.
y=mx+b\begin{align*}y=mx+b\end{align*}
Step 2: Substitute the given slope for m\begin{align*}m\end{align*}.
\begin{align*}y=\frac{-2}{5} x+b\end{align*}
Step 3: Use one of the ordered pairs you are given (–2, 4) and substitute these values for the variables \begin{align*}x\end{align*} and \begin{align*}y\end{align*} in the equation.
\begin{align*}4=\left (\frac{-2}{5}\right )(-2)+b\end{align*}
Step 4: Solve for \begin{align*}b\end{align*} (the \begin{align*}y-\end{align*}intercept of the graph).
\begin{align*}4& =\frac{4}{5}+b\\ 4-\frac{4}{5}& =\frac{4}{5}-\frac{4}{5}+b\\ \frac{16}{5}& =b\end{align*}
Step 5: Rewrite \begin{align*}y=mx+b\end{align*}, substituting the slope for \begin{align*}m\end{align*} and the \begin{align*}y-\end{align*}intercept for \begin{align*}b\end{align*}.
\begin{align*}y=\frac{-2}{5} x+\frac{16}{5}\end{align*}
Example 6: Write the equation for a line containing the points (–4, 1) and (–2, 3).
Solution:
1. Start with the slope–intercept form of the line \begin{align*}y=mx+b\end{align*}.
2. Find the slope of the line: \begin{align*}m=\frac{y_2-y_1}{x_2-x_1}=\frac{3-1}{-2-(-4)}=\frac{2}{2}=1\end{align*}.
3. Substitute the value of slope for \begin{align*}m: y=(1)x+b\end{align*}.
4. Substitute the coordinate (–2, 3) into the equation for the variables \begin{align*}x\end{align*} and \begin{align*}y : 3=-2+b \Rightarrow b=5\end{align*}.
5. Rewrite the equation, substituting the slope for \begin{align*}m\end{align*} and the \begin{align*}y-\end{align*}intercept for \begin{align*}b\end{align*}: \begin{align*}y=x+5\end{align*}.
## Writing a Function in Slope-Intercept Form
Remember that a linear function has the form \begin{align*}f(x)=mx+b\end{align*}. Here \begin{align*}f(x)\end{align*} represents the \begin{align*}y\end{align*} values of the equation or the graph. So \begin{align*}y=f(x)\end{align*} and they are often used interchangeably. Using the functional notation in an equation often provides you with more information.
For instance, the expression \begin{align*}f(x)=mx+b\end{align*} shows clearly that \begin{align*}x\end{align*} is the independent variable because you substitute values of \begin{align*}x\end{align*} into the function and perform a series of operations on the value of \begin{align*}x\end{align*} in order to calculate the values of the dependent variable, \begin{align*}y\end{align*}.
In this case when you substitute \begin{align*}x\end{align*} into the function, the function tells you to multiply it by \begin{align*}m\end{align*} and then add \begin{align*}b\end{align*} to the result. This process generates all the values of \begin{align*}y\end{align*} you need.
Example 7: Consider the function \begin{align*}f(x)=3x-4.\end{align*} Find \begin{align*}f(2), f(0),\end{align*} and \begin{align*}f(-1)\end{align*}.
Solution: Each number in parentheses is a value of \begin{align*}x\end{align*} that you need to substitute into the equation of the function.
\begin{align*}f(2)=2; f(0)=-4; \ and \ f(-1)=-7\end{align*}
Function notation tells you much more than the value of the independent variable. It also indicates a point on the graph. For example, in the above example, \begin{align*}f(-1)=-7\end{align*}. This means the ordered pair (–1, –7) is a solution to \begin{align*}f(x)=3x-4\end{align*} and appears on the graphed line. You can use this information to write an equation for a function.
Example 8: Write an equation for a line with \begin{align*}m=3.5\end{align*} and \begin{align*}f(-2)=1\end{align*}.
Solution: You know the slope and you know a point on the graph (–2, 1). Using the methods presented in this lesson, write the equation for the line.
Begin with slope-intercept form.
\begin{align*}&& y& =mx+b\\ \text{Substitute the value for the slope.} && y& =3.5x+b\\ \text{Use the ordered pair to solve for} \ b. && 1& =3.5(-2)+b\\ && b& =8\\ \text{Rewrite the equation.} && y& =3.5x+8 \\ \text{or} && f(x)& =3.5x+8\end{align*}
## Solve Real-World Problems Using Linear Models
Let’s apply the methods we just learned to a few application problems that can be modeled using a linear relationship.
Example 9: Nadia has $200 in her savings account. She gets a job that pays$7.50 per hour and she deposits all her earnings in her savings account. Write the equation describing this problem in slope–intercept form. How many hours would Nadia need to work to have 500 in her account? Solution: Begin by defining the variables: \begin{align*}y=\end{align*} amount of money in Nadia’s savings account \begin{align*}x=\end{align*} number of hours The problem gives the \begin{align*}y-\end{align*}intercept and the slope of the equation. We are told that Nadia has200 in her savings account, so \begin{align*}b=200\end{align*}.
We are told that Nadia has a job that pays 7.50 per hour, so \begin{align*}m=7.50\end{align*}. By substituting these values in slope–intercept form \begin{align*}y=mx+b\end{align*}, we obtain \begin{align*}y=7.5x+200\end{align*}. To answer the question, substitute500 for the value of \begin{align*}y\end{align*} and solve.
\begin{align*}500 = 7.5x+200 \Rightarrow 7.5x=300 \Rightarrow x=40\end{align*}
Nadia must work 40 hours if she is to have 500 in her account. Example 10: A stalk of bamboo of the family Phyllostachys nigra grows at steady rate of 12 inches per day and achieves its full height of 720 inches in 60 days. Write the equation describing this problem in slope–intercept form. How tall is the bamboo 12 days after it started growing? Solution: Define the variables. \begin{align*}y=\end{align*} the height of the bamboo plant in inches \begin{align*}x=\end{align*} number of days The problem gives the slope of the equation and a point on the line. The bamboo grows at a rate of 12 inches per day, so \begin{align*}m=12\end{align*}. We are told that the plant grows to 720 inches in 60 days, so we have the point (60, 720). \begin{align*}\text{Start with the slope-intercept form of the line.} && y& =mx+b\\ \text{Substitute 12 for the slope.} && y& =12x+b\\ \text{Substitute the point} \ (60,720). && 720& =12(60)+b \Rightarrow b=0\\ \text{Substitute the value of} \ b \ \text{back into the equation.} && y& =12x\end{align*} To answer the question, substitute the value \begin{align*}x=12\end{align*} to obtain \begin{align*}y=12(12)=144\end{align*} inches. ## Practice Set Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. 1. What is the formula for slope-intercept form? What do the variables \begin{align*}m\end{align*} and \begin{align*}b\end{align*} represent? 2. What are the five steps needed to determine the equation of a line given the slope and a point on the graph (not the \begin{align*}y-\end{align*}intercept)? 3. What is the first step in finding the equation of a line given two points? In 4 – 20, find the equation of the line in slope–intercept form. 1. The line has slope of 7 and \begin{align*}y-\end{align*}intercept of –2. 2. The line has slope of –5 and \begin{align*}y-\end{align*}intercept of 6. 3. The line has \begin{align*}\text{slope}=-2\end{align*} and a \begin{align*}y-\end{align*}intercept = 7. 4. The line has \begin{align*}\text{slope}=\frac{2}{3}\end{align*} and a \begin{align*}y-\end{align*}intercept = \begin{align*}\frac{4}{5}\end{align*}. 5. The line has slope of \begin{align*}-\frac{1}{4}\end{align*} and contains point (4, –1). 6. The line has slope of \begin{align*}\frac{2}{3}\end{align*} and contains point \begin{align*}\left(\frac{1}{2},1\right )\end{align*}. 7. The line has slope of –1 and contains point \begin{align*}\left (\frac{4}{5},0\right )\end{align*}. 8. The line contains points (2, 6) and (5, 0). 9. The line contains points (5, –2) and (8, 4). 10. The line contains points (3, 5) and (–3, 0). 11. The slope of the line is \begin{align*}-\frac{2}{3}\end{align*} and the line contains point (2, –2). 12. The slope of the line is –3 and the line contains point (3, –5). 13. The line contains points (10, 15) and (12, 20). In 21 – 28, find the equation of the linear function in slope–intercept form. 1. \begin{align*}m=5, f(0)=-3\end{align*} 2. \begin{align*}m=-2\end{align*} and \begin{align*}f(0)=5\end{align*} 3. \begin{align*}m=-7, f(2)=-1\end{align*} 4. \begin{align*}m=\frac{1}{3}, f(-1)=\frac{2}{3}\end{align*} 5. \begin{align*}m=4.2, f(-3)=7.1\end{align*} 6. \begin{align*}f\left (\frac{1}{4}\right )=\frac{3}{4}, f(0)=\frac{5}{4}\end{align*} 7. \begin{align*}f(1.5)=-3, f(-1)=2\end{align*} 8. \begin{align*}f(-1)=1\end{align*} and \begin{align*}f(1)=-1\end{align*} 9. To buy a car, Andrew puts in a down payment of1500 and pays \$350 per month in installments. Write an equation describing this problem in slope-intercept form. How much money has Andrew paid at the end of one year?
10. Anne transplants a rose seedling in her garden. She wants to track the growth of the rose, so she measures its height every week. In the third week, she finds that the rose is 10 inches tall and in the eleventh week she finds that the rose is 14 inches tall. Assuming the rose grows linearly with time, write an equation describing this problem in slope-intercept form. What was the height of the rose when Anne planted it?
11. Ravi hangs from a giant exercise spring whose length is 5 m. When his child Nimi hangs from the spring, its length is 2 m. Ravi weighs 160 lbs. and Nimi weighs 40 lbs. Write the equation for this problem in slope-intercept form. What should we expect the length of the spring to be when his wife Amardeep, who weighs 140 lbs., hangs from it?
12. Petra is testing a bungee cord. She ties one end of the bungee cord to the top of a bridge and to the other end she ties different weights. She then measures how far the bungee stretches. She finds that for a weight of 100 lbs., the bungee stretches to 265 feet and for a weight of 120 lbs., the bungee stretches to 275 feet. Physics tells us that in a certain range of values, including the ones given here, the amount of stretch is a linear function of the weight. Write the equation describing this problem in slope–intercept form. What should we expect the stretched length of the cord to be for a weight of 150 lbs?
Mixed Review
1. Translate into an algebraic sentence: One-third of a number is seven less than that number.
2. The perimeter of a square is 67 cm. What is the length of its side?
3. A hockey team played 17 games. They won two more than they lost. They lost 3 more than they tied. How many games did they win, lose, and tie?
4. Simplify \begin{align*}\frac{(30-4+4 \div 2) \div (21 \div 3)}{2}\end{align*}.
5. What is the opposite of 16.76?
6. Graph the following on a number line: \begin{align*}\left \{6,\frac{11}{3},-5.65,\frac{21}{7}\right \}\end{align*}.
7. Simplify: \begin{align*}[(-4+4.5)+(18-|-13|)+(-3.3)]\end{align*}.
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# Equation of Motion by Integration Method
## Equation of Motion by Integration Method
Relation among velocity, distance, time and acceleration is called equations of motion. There are three equations of motion.
The final velocity (v) of a moving object with uniform acceleration (a) after time (t).
Let, Initial velocity = v₀,
Final velocity = v,
Time = t,
Acceleration = a
First Equation of Motion: Acceleration is the first derivative of velocity with respect to time.
Acceleration (a) = dv/dt
⇒ dv = a x dt
v₀v dv = ₀∫t a dt
⇒ (v – v₀) = a (t – 0)
⇒ v – v₀ = at
∴ v = v₀ + at
Second Equation of Motion: Velocity is the first derivative of position with respect to time.
⇒ Velocity (v) = ds/ dt
⇒ ds = v dt
⇒ ds = (v₀ + at) dt
s₀s ds = ₀∫t (v₀ + at) dt
⇒ $$s-{{s}_{0}}=\left[ {{v}_{0}}t+\frac{a{{t}^{2}}}{2} \right]_{0}^{t}$$,
⇒ s – s₀ = v₀t + (at²/2)
∴ s = s₀ + v₀t + (at²/2)
Third Equation of Motion:
Acceleration (a) = dv/ dt
= (dv/ ds) x (ds/ dt)
= (dv/ ds) x v
∴ Acceleration (a) = v dv/ ds
⇒ v dv/ ds = a
v₀v v dv = s₀s a x ds
⇒ $$\left[ \frac{{{v}^{2}}}{2} \right]_{{{v}_{0}}}^{v}=a\left[ s \right]_{{{s}_{0}}}^{s}$$,
⇒ ½ [v² – v²₀] = a [s – s₀]
⇒ v² – v²₀ = 2a [s – s₀]
∴ v² = v²₀ + 2a [s – s₀].
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# Line Graphs
#### Chapter 37
5 Steps - 3 Clicks
# Line Graphs
### Introduction
Line Graphs are represented by specific points joined together by straight lines. The points are plotted on a two-dimensional plane taking one parameter on the horizontal axis and the other on vertical axis.
### Methods
Line graphs are categorised into two types. They are:
1. One dependent variable historigram,
2. Multi dependent variable historigram.
• Historigram is a graph of time series where the changes in a given variable is illustrated s a function of time.
• From one period of time to another, the graph shows changes in the value of one or more variables.
• These graphs can be either on a natural scale or on a ratio scale.
• Natural scale is used, if the absolute values of a variable are to be represented. In natural scale, equal distances on an axis represent equal values/amounts.
• One dependent variable historigram is the simplest type. The values of only one variable is to be plotted along the Y-axis and as the time factor being independent variable, it is taken on X-axis.
• In more than one dependent variable historigram, distribution of two variables is represented distinctly. For all values, the vertical scale is common.
Example 1:
Directions for questions 1 to 5: Given below are line graph examples with questions, which show the annual food grain production from 1992 to 1997. Refer to the graph & answer the questions based on line graph as given below.
1. What is an approximate percentage decrease in production from 1993 to 1994?
A. 87.5%
B. 37.5%
C. 9.09%
D. None of these
Solution:
Here we look up the values of the production for the 2 years first. Locate 1993 on the X axis, which denotes years. Move vertically up along the Y axis direction in 1993, and we get value of production in 1993 as 110. In the same way we get the value of production in 1994 as 100.
In calculating % increases & decreases: In this case 1993, it is very important to remember that the original year is the one that is used as the reference year.
Which is 110 – 100 = 10.
Now we have to express 10 as a percentage of the production in 1993, which is 110.
So the required answer is 100 × 10/110 = 9.09%. Hence answer is option C.
2. The average production of 1994 and 1996 was approximately equal to production of which year?
A. 1996
B. 1992
C.1995
D. 1994
Solution:
The production in 1994 was 100, that in 1996 was 62.5.
The average production of 1994 and 1996 = (100 + 62.5)/2 = 81.25.
This was approximately equal to production in 1992. Hence Answer is option B.
3. The average production of the given years was more than to the production in how many years?
A. 1
B. 2
C. 3
D. 4
Solution:
Average production for 6 years = 80 + 110 + 130 + 100 +120 +65 / 6 = 101
Clearly in 1992, 1994 and 1996, the production was less than average. Hence answer is option C.
4. If the production in 1992 is estimated to be 60% more than that in 1991, the estimate of 1991 production is?
A. 120 units
B. 100 units
C.60 units
D. 50 units
Solution:
The production in 1992 is 80 units. The Required production in 1991 =
80 / 160 x 100 = 50 Hence answer is option D.
5. The maximum increase in food grain production has been in
A. 1997
B. 1996
C.1995
D. 1994
Solution:
Here remember that absolute increase has been asked. The difference in Y coordinates of the adjacent years will give us the increase. Since the distance between consecutive years is the same on the X axis, we can check the slopes of various segments. By visual inspection, maximum increase in food grain production has been in the year 1997 (55 units by calculation).Hence, answer is option A
Example 2:
Directions for question 1-5: Study the following graph carefully and answer accordingly. Following graph shows the percent profit earned by two companies A and B on their investments. (Revenue = Investment + Profit)
1. Revenue of company B in 2000 was Rs.1239 lakhs. What was the investment in that year (in Rs lakhs) of company B?
A. 700
B. 800
C. 650
D. 193.03
Solution:
Investment of company A in 2000 = 1239 x 100 / 177 = Rs.700 lakh
2. Investment of company B in 1998 was 20% more than that in the previous year. Profit in 1998 of company B was what per cent of its profit in 1997?
A. 10%
B. 102(2/3) %
C. 106(2/3) %
D. None of these
Solution:
Let us assume the investment of Company B in 1997 is Rs. 100. Investment of company B in 1998 was 20% more than that in the previous year.
So the investment of Company B in 1998 is Rs. 120.
Profit in 1997 of company B= 90 % of 100 = Rs. 90
Profit in 1998 of company B= 85% of 120 = Rs. 102.
Required % = 102 / 90 x 100 = 113 (1/3) %
3. In which of the following years is the ratio of investment and profit maximum for company A?
A. 2001
B. 1995
C. 1998
D. 2000
Solution:
Quicker Method: The ratio of investment and profit will be maximum when percentage profit is minimum. Therefore, the ratio of investment and profit will be maximum in 2001.
4. If the revenue of company B in 1996 was same as the revenue of company B in 1999, what would be the ratio of investment of company B in 1999 to the investment of company B in 1996?
A. 9 : 10
B. 10 : 9
C. 13 : 15
D. 15 : 13
Solution:
Required ratio = (100 / 185) x (166.5 / 100) = (333 / 370) = (37*9 / 37*10) = 9:10
5. In which of the following years was the investment minimum for company B?
A. 1995
B. 1999
C. 2000
D. Can’t be determined
Solution:
We don’t have the data for investment. So, we can’t determine the answer.
### Samples
1. Study the following graph carefully and answer the given questions:
%profit = $$\frac{Income – Expenditure}{Expenditure}$$ x 100
1. If the income of company A in 1998 was equal to its expenditure in 2000, what was the ratio between company’s expenditure in the years 1998 and 2000 respectively?
(A) 29 : 20
(B) 20 : 29
(C) 19 : 20
(D) cannot be determined
(E) None of these.
2. If the total expenditure of the two companies in 2001 was Rs. 18 lakhs and expenditures of companies A and B in that year were in the ratio of 4 : 5 respectively, then what was the income of the company B in that year (in lakhs)?
(A) 8
(B) 10
(C) 10.4
(D) All of these
(E) None of these.
Solution:
1. Since the data given is inadequate hence it cannot be determined.
So, the option(D) is correct answer.
2. The expenditure of company B in 2001 = $$\frac{5}{9} * 18$$ = 10 lakhs
Let the income of company B be Rs.$$x$$ lakh
Therefore, 40 =$$\frac{x – 10}{10} * 100$$
40 = 10$$x$$ – 100
10$$x$$ = 140
$$x$$ = 14 lakh.
2. Study the following graph and answer the questions based on it.
1. What is the difference between the number of vehicles manufactured by company Y in 2000 and 2001?
(A) 50000
(B) 42000
(C) 33000
(D) 21000
(E) 13000
2. What is the difference between the total productions of the two companies in the given years?
(A) 19000
(B) 22000
(C) 26000
(D) 28000
(E) 29000
3. The production of company Y in 2000 was approximately what percent of the production of company X in the same year?
(A) 173
(B) 164
(C) 132
(D) 97
(E) 61
Solution:
1. Required difference = 128000 – 107000 = 21000.
So, option (D) is correct one.
2. Total production of company X from 1997 to 2002 = 119000 + 99000 + 141000 + 78000 + 120000 + 159000 = 716000.
Total production of company Y from 1997 to 2002 = 139000 + 120000 + 100000 + 128000 + 107000 + 148000 = 742000.
Difference = 742000 – 716000 = 26000.
So, option (C) is correct answer.
3. Required percentage = $$\frac{128000}{78000} * 100$$% = 164%
So, option (B) is correct answer.
Model 3: The following line-graph gives the ratio of the amounts of imports by a company to the amount of exports from that company over a period from 1995 to 2001. The questions given below are based on this graph.
1. If the imports of the company in 1996 was Rs. 272 crores, what was amount of the exports from the company in 1996?
(A) Rs. 370 crores
(B) Rs. 320 crores
(C) Rs. 280 crores
(D) Rs. 275 crores
(E) Rs. 264 crores
2. In how many of the given years were the exports more than the imports?
(A) 1
(B) 2
(C) 3
(D) 4
(E) None of these.
Solution:
1. Ratio of imports to exports in the year 1996 = 0.85.
Let the exports in 1996 = Rs. $$x$$ crores.
Then,
$$\frac{272}{x}$$ = 0.85
⇒ $$x$$ = $$\frac{272}{0.85}$$ = 320
Therefore, exports in 1996 = Rs. 320 crores.
So, option (B) is correct answer.
2. The exports are more than the imports implies that the ratio of value of imports to exports is less than 1.
Now, this ratio is less than 1 in the years 1995, 1996, 1997, and 2000.
Thus there are four such years.
So, option (D) is correct answer.
Model 4: Periodical examinations are held every second month in a school. In a session during April 2001 to March 2002, a student of class IX appeared for each of the periodical exams. The aggregate marks obtained in each periodical exam are represented in the below given line graph. Study it and answer the given questions.
1. What is the percentage of marks obtained by the student in the periodical exams of August 01 and October 01 taken together?
(A) 73.25%
(B) 75.5%
(C) 77%
(D) 78.75%
(E) 79.5%
2.The total number of marks obtained in February 02 is what percent of the total marks obtained in April 01?
(A) 110%
(B) 112.5%
(C) 115%
(D) 116.5%
(E) 117.5%
Solution:
1. Required percentage = $$\frac{370 + 385}{500 + 500} * 100$$% = $$\frac{755}{1000} * 100$$% = 75.5%
So, option (B) is correct answer.
2. Required percentage = $$\frac{405}{360} * 100$$% = 112.5%
So, option (B) is correct answer.
Model 5: Study the following line graph which gives the number of students who joined and left the school in the beginning of year for six years, from 1996 to 2001.
Initial strength of school in 1995 = 3000
1. The number of students studying in the school in 1998 was what percent of the number of students studying in the school in 2001?
(A) 92.13%
(B) 93.75%
(C) 96.88%
(D) 97.25%
2. What was the number of students studying in the school during 1999 ?
(A) 2950
(B) 3000
(C) 3100
(D) 3150
3. The ratio of the least number of students who joined the school to the maximum number of students who left the school in any of the years during the given period is?
(A) 7:9
(B) 4:5
(C) 3:4
(D) 2:3
4.For which year, the percentage rise/fall in the number of students who left the school compared to the previous year is maximum?
(A) 1997
(B) 1998
(C) 1999
(D) 2000
Solution:
1. Required percentage = $$\frac{3000}{3200} * 100$$% = 93.75%
So, option (B) is correct answer.
2. From the given data, the number of students studying in the school during 1999 = 3150.
So, option (D) is correct answer.
3. Required percentage = $$\frac{300}{450} * 100$$% = $$\frac{2}{3}$$
So, option (D) is correct answer.
4. The percentage rise/fall in the number of students who left the school (compared to the previous year) during various years are:
For 1997 = $$\frac{450 – 250}{250} * 100$$% = 80% (rise)
For 1998 = $$\frac{450 – 400}{450} * 100$$% = 11.11% (fall)
For 1999 = $$\frac{400 – 350}{400} * 100$$% = 12.5% (fall)
For 2000 = $$\frac{450 – 350}{350} * 100$$% = 28.57% (rise)
For 2001 = $$\frac{450 – 450}{450} * 100$$% = 0% (rise)
Hence, in the year 1997 maximum rise/fall occured.
So, option (A) is correct answer.
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# Thread: Reducing Algebra Fractions by -1
1. ## Reducing Algebra Fractions by -1
The book is trying to show me that reducing a fraction or adding two fractions sometimes only requires that -1 be factored from one or more denominators. Here is the example they give:
$\displaystyle \frac{y-x}{x-y}=\frac{y-x}{-(-x+y)}=\frac{y-x}{-(y-x)}=\frac{1}{-1}=-1$
1.In the first step here I understand factoring out the "-" (the -1 value) because if the variable does not have a number coefficient we can assume its one so this part is fine.
$\displaystyle \frac{y-x}{-(-x+y)}$
2.Here$\displaystyle \frac{y-x}{-(-x+y)}$
once we distribute the "-" we get a negative y and a positive x, therefore we can rearrange the denominator to match the numerator in $\displaystyle \frac{y-x}{-(y-x)}$
3.Next we have $\displaystyle \frac{1}{-1}$ Now here is where I have a problem how can we know the values of both of these are 1 and -1? I am not certain how this is accomplished.
Does my logic up to 3's question look correct also?
2. Hi
Originally Posted by cmf0106
The book is trying to show me that reducing a fraction or adding two fractions sometimes only requires that -1 be factored from one or more denominators. Here is the example they give:
$\displaystyle \frac{y-x}{x-y}=\frac{y-x}{-(-x+y)}=\frac{y-x}{-(y-x)}=\frac{1}{-1}=-1$
1.In the first step here I understand factoring out the "-" (the -1 value) because if the variable does not have a number coefficient we can assume its one so this part is fine.
$\displaystyle \frac{y-x}{-(-x+y)}$
2.Here$\displaystyle \frac{y-x}{-(-x+y)}$
once we distribute the "-" we get a negative y and a positive x, therefore we can rearrange the denominator to match the numerator in $\displaystyle \frac{y-x}{-(y-x)}$
What you said for 1. is not wrong. But I'd say they made these two transformations to show you that there is the common factor y-x.
But what you said in 2. looks incorrect to me. They don't distribute - again, they just use the commutativity of the addition : a+b=b+a. Here : -x+y=y+(-x)=y-x.
3.Next we have $\displaystyle \frac{1}{-1}$ Now here is where I have a problem how can we know the values of both of these are 1 and -1? I am not certain how this is accomplished.
Relevant question for someone who begins ^^ It's so good seeing you trying to understand !
We have $\displaystyle \frac{y-x}{-(y-x)}$. This can be rewritten this way :
$\displaystyle \frac{1*{\color{red}(y-x)}}{(-1)*{\color{red}(y-x)}}$ and hence the result. Does it look correct to you ?
When you'll advance into maths lessons, maybe you will learn that most of these steps are only here to guide you to the solution
3. Thanks Moo! I will chew on that for awhile and let you know what the outcome is, I think it makes sense let me try a few more practice problems and we will see.
as for
"What you said for 1. is not wrong. But I'd say they made these two transformations to show you that there is the common factor y-x.
But what you said in 2. looks incorrect to me. They don't distribute - again, they just use the commutativity of the addition : a+b=b+a. Here : -x+y=y+(-x)=y-x."
Yah sorry I didnt mean to phrase it that way, it wouldnt make sense anyways because we factor out the "-", if we factored it out one step and then the next we redistributed it that would defeat the purpose of factoring
4. Alright moo update, this stuff is really putting me in a rut.
$\displaystyle \frac{3}{y-x}+\frac{x}{x-y}$
1.The book solves it $\displaystyle \frac{3}{y-x}+\frac{x}{x-y}=\frac{3}{-(-y+x)}+\frac{x}{x-y}=\frac{3}{-(x-y)}+\frac{x}{x-y}=\frac{-3}{x-y}+\frac{x}{x-y}=\frac{-3+x}{x-y}$
2. I started solving it like this, which looks different from the book $\displaystyle \frac{3}{y-x}+\frac{x}{x-y}=\frac{3}{-(-x+y)}+\frac{x}{x-y}=\frac{3+x}{(x-y)(x-y)}$
I used $\displaystyle \frac{3}{-(-x+y)}$ so it would yield the other fractions denominator $\displaystyle \frac{x}{x-y}$
Here $\displaystyle \frac{3+x}{(x-y)(x-y)}$ the two $\displaystyle (x-y)$ can merge into one but Im not understanding why the 3 is supposed to be -3 here, this is assuming my way of solving for it is correct in the first place.
5. Originally Posted by cmf0106
Alright moo update, this stuff is really putting me in a rut.
$\displaystyle \frac{3}{y-x}+\frac{x}{x-y}$
2. I started solving it like this, which looks different from the book $\displaystyle \frac{3}{y-x}+\frac{x}{x-y}=\frac{3}{-(-x+y)}+\frac{x}{x-y}=\frac{3+x}{(x-y)(x-y)}$
First problem in the first equality : y-x=-x+y, it's not equal to -(-x+y)
The purpose of this is to make appear the other fraction's denominator x-y. You can see that -(x-y)=-x+y=y-x.
So $\displaystyle \frac{3}{y-x}=\frac{3}{-(x-y)}$
Now if you want to know why it becomes $\displaystyle \frac{-3}{x-y}$ :
$\displaystyle \frac{3}{-(x-y)}=\frac{3}{-1} \cdot \frac{1}{x-y}=-3 \cdot \frac{1}{x-y}=\frac{-3}{x-y}$
See ?
Now, there is also a problem with the second equality. You should remember that :
$\displaystyle \frac ac+\frac bc=\frac{a+b}{c}$ and there is no multiplication involved. The multiplication occurs if there is not the same denominator
6. But I could chose to make either both denominators appear as $\displaystyle y-x$ or $\displaystyle x-y$ correct?
7. Originally Posted by cmf0106
But I could chose to make either both denominators appear as $\displaystyle y-x$ or $\displaystyle x-y$ correct?
Correct
It'd be nice if you could show how you would have done for y-x (unless you have many more exercises ^^)
8. Originally Posted by Moo
Correct
It'd be nice if you could show how you would have done for y-x (unless you have many more exercises ^^)
Man this is probably the hardest thing Ive encountered so far, hard for me to wrap my brain around. Anyways lets give the other denominator a shot
so $\displaystyle \frac{3}{y-x}+\frac{x}{x-y}$
Lets Change$\displaystyle \frac{3}{y-x}=-(y-x)=-y+x$ this equals our other denominator $\displaystyle x-y$
next$\displaystyle =\frac{3}{-(y-x)}=\frac{-3+x}{x-y}$
Probably wrong lol.
9. Originally Posted by cmf0106
Man this is probably the hardest thing Ive encountered so far, hard for me to wrap my brain around. Anyways lets give the other denominator a shot
so $\displaystyle \frac{3}{y-x}+\frac{x}{x-y}$
Lets Change$\displaystyle {\color{red}\frac{3}{y-x}=\frac{3}{-(y-x)}=\frac{3}{-y+x}}$ this equals our other denominator $\displaystyle x-y$
next$\displaystyle =\frac{3}{-(y-x)}=\frac{-3+x}{x-y}={\color{red}\frac{x-3}{x-y}}$
Probably wrong lol.
Let's clean this up just a little bit so that your equalities make sense, and your conclusion looks fine. Great job!
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In geometry, initially we learn about the fundamental concepts of point, line, line segments, ray, angles, types of angles, plane figures etc. When we learn about lines, we also learn about the Euclidean axioms on lines, points and lines. One of the axiom is only one line can be drawn through two distinct points in a plane. also we are familiar with the property that two non-parallel lines will intersect at one and only one point. In this section let us learn about the concurrent lines and the problems based on it.
## Concurrent Lines Definition
Three or more lines in a plane are said to be concurrent, if all of them intersect at the same point. In the following diagram, the lines, l, m and n all intersect at the point P. So they are concurrent.
### Solved Examples
Question 1: In the figure given below, the lines l, m and n intersect at O. Find the values of x, y and z, it is being given that x : y : z = 2 : 3 : 5.
Solution:
Let us assume the other three angles as, a, b and c
We know that, when two lines intersect, vertically opposite angles are equal.
Therefore, a = x, b = y and c = z
Since the sum of all the angles around a point is 360o ,
we have a + y +c + x + b + z = 360 o
Substituting , a = x, b = y and c = z, we have
x + y + z + x + y + z = 360 o
=> 2x + 2y + 2z = 360o
=> 2 ( x + y + z ) = 360o
=> x + y + z = $\frac{360}{2}$
= 1800 ------------------------ ( 1 )
We are given that x : y : z = 2 : 3 : 5,
Let x = 2k, y = 3k and z = 5k
Substituting this in equation ( 1 ), we get
2k + 3k + 5k = 180
=> 10 k = 180
=> k = $\frac{180}{10}$
= 18
Substituting k = 18 in x, y and z, we get,
x = 2k = 2 ( 18 ) = 36o
y = 3 k = 3 ( 18 ) = 54o
z = 5 k = 5 ( 18 ) = 90o
Question 2: In the following figure, find the measure of the angles, x , y and z.
Solution:
In the following figure AB and CD intersect at O.
Therefore, $\angle COA$ + $\angle AOD$ = 180o , [ linear pair ]
3x - 20 + x = 180o
=> 4x - 20 = 180
=> 4 x = 180 + 20
= 200
=> x = $\frac{200}{4}$
= 50o
$\angle AOC$ = 3x - 20
= 3 ( 50 ) - 20
= 150 - 20
= 130o
$\angle AOD$ = x = 50o
$\angle BOC$ = y = x [ vertically opposite angles ]
= 50o
$\angle BOD$ = z = 3x - 20 [ vertically opposite angles ]
= 130o
Therefore, x= 50o , y = 50o and z = 130o
## Concurrent Lines Medians And Altitudes
### What are Medians? Medians are the lines joining the vertex to the midpoint of the opposite sides.
Medians of any triangle are concurrent and the point of intersection is called the centroid.
The centroid of a triangle divides the median in the ratio 2 : 1.
The following diagram shows the point of intersection of the median.s of a triangle.
The Medians of any triangle meet in the interior of the a triangle.
In the above triangle, AG : GD = 2 : 1. BG : GE = 2 : 1. CG : GF = 2 : 1
### Altitudes:
Altitudes are the perpendiculars drawn from the vertex to the opposite sides.
The altitudes of any triangle are concurrent and the point of intersection of the altitudes is called an Orthocenter.
The following diagram shows the point of intersection of the altitudes of a triangle.
1. The above triangle is an acute angled triangle and we see that the orthocenter lies inside the triangle.
2. For an obtuse angled triangle the orthocenter lies outside the triangle.
3. For a right triangle, the orthocenter is the vertex containing the right angle.
## Concurrent Lines in Triangles
In any triangles, the altitudes, medians, and internal bisectors of the vertices or perpendicular bisectors of the sides are concurrent.
Let us see the special names for these points of concurrence.
### Orthocenter: It is the point of concurrence of the altitudes of a triangle.
The following diagram shows the position of orthocenter for the respective triangle.
1. For an acute angled triangle the orthocenter lies inside the triangle.
2. For an obtuse angled triangle the orthocenter lies outside the triangle
3. For a right angled triangle orthocenter lies at the hypotenuse of the triangle.
### Centroid: It is the point of concurrence of the medians of a triangle.
Whether the triangle is acute, obtuse of right angled triangle, the centroid always lies inside the triangle.
### Incenter: It is the point of concurrence of the internal bisectors of the angles of a triangle.
The circle with center as I and radius equal to the perpendicular distances of I from the sides, touches the sides of the triangle. Hence the point is called Incenter of the triangle.
For all types of triangles, acute, obtuse or right triangle, the Incenter always lies inside the circle.
### Circumcenter: It is the point of concurrence of the perpendicular bisectors of the sides of a triangle.
SA = SB = SC
We can see that the circle with center at S (circumcenter) is equidistant from the vertices and hence the circle passes through the vertices of the triangle. The circle so drawn is called as circumcircle.
1. For an acute angled triangle, the circumcenter lies in the interior of the triangle.
2. For a right angled triangle the circumcenter is the midpoint of the hypotenuse.
3. For an obtuse angled triangle, the circumcenter will lie outside the triangle.
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# GSEB Solutions Class 9 Maths Chapter 1 Number Systems Ex 1.3
Gujarat BoardĀ GSEB Textbook Solutions Class 9 Maths Chapter 1 Number Systems Ex 1.3 Textbook Questions and Answers.
## Gujarat Board Textbook Solutions Class 9 Maths Chapter 1 Number Systems Ex 1.3
Question 1.
Write the following in decimal form and say what kind of decimal expansion each has
Solution:
(i) $$\frac { 36 }{ 100 }$$
(ii) $$\frac { 1 }{ 11 }$$
(iii) 4$$\frac { 1 }{ 8 }$$
(iv) $$\frac { 3 }{ 13 }$$
(v) $$\frac { 2 }{ 11 }$$
(vi) $$\frac { 329 }{ 400 }$$
Solution:
(i) We have
$$\frac { 36 }{ 100 }$$ = 0.36
ā“ The decimal expansion is terminating.
(ii) We have $$\frac { 1 }{ 11 }$$
The decimal expansion is non-terminating repeating.
(iii) We have 4$$\frac { 1 }{ 8 }$$ = $$\frac { 33 }{ 8 }$$
The decimal expansion is terminating.
(iv) We have $$\frac { 3 }{ 13 }$$
The decimal expansion is non-terminating repeating.
(v) We have $$\frac { 2 }{ 11 }$$
The decimal expansion is non-terminating repeating.
(vi) We have $$\frac { 329 }{ 400 }$$
The decimal expansion is terminating.
Question 2.
You know that $$\frac { 1 }{ 7 }$$ = 0.$$\overline { 142857 }$$. Can you predict what the decimal expansions of $$\frac { 2 }{ 7 }$$, $$\frac { 3 }{ 7 }$$, $$\frac { 4 }{ 7 }$$, $$\frac { 5 }{ 7 }$$, $$\frac { 6 }{ 7 }$$ are, without actually doing the long division? If so, how?
(Hint. Study the remainders while finding the value of $$\frac { 1 }{ 7 }$$ carefully.)
Solution:
Yes, we can predict the decimal expansions of $$\frac { 2 }{ 7 }$$, $$\frac { 3 }{ 7 }$$, $$\frac { 4 }{ 7 }$$, $$\frac { 5 }{ 7 }$$, $$\frac { 6 }{ 7 }$$ without actually doing the long division.
We know $$\frac { 1 }{ 7 }$$
$$\frac { 1 }{ 7 }$$ = 0.142857142857…
ā“ $$\frac { 1 }{ 7 }$$ = 0.$$\overline { 142857 }$$
Now
$$\frac { 2 }{ 7 }$$ = 2 x $$\frac { 1 }{ 7 }$$ = 2 x 0.$$\overline { 142857 }$$
ā“ $$\frac { 2 }{ 7 }$$ = 0.$$\overline { 285714 }$$
Similarly,$$\frac { 3 }{ 7 }$$ = 3 x $$\frac { 1 }{ 7 }$$ = 3 x 0.$$\overline { 142857 }$$
ā $$\frac { 3 }{ 7 }$$ = 0.$$\overline { 428571 }$$
$$\frac { 4 }{ 7 }$$ = 4 x $$\frac { 1 }{ 7 }$$ = 4 x 0.$$\overline { 142857 }$$
ā $$\frac { 4 }{ 7 }$$ = 0.$$\overline { 571428 }$$
$$\frac { 5 }{ 7 }$$ = 5 x $$\frac { 1 }{ 7 }$$ = 5 x 0.$$\overline { 142857 }$$
ā $$\frac { 5}{ 7 }$$ = 0.$$\overline { 714285 }$$
$$\frac { 6 }{ 7 }$$ = 6 x $$\frac { 1 }{ 7 }$$ = 6 x 0.$$\overline { 142857 }$$
ā $$\frac { 6}{ 7 }$$ = 0.$$\overline { 857142 }$$
Question 3.
Express the following in the form $$\frac { p }{ q }$$, where p and q are integers and q ā 0.
(i) 0.$$\overline { 6 }$$
(ii) 0.4$$\overline { 7 }$$
(iii) 0.$$\overline { 0.001 }$$
Solution:
(i) 0.$$\overline { 6 }$$
(ii) We have 0.4$$\overline { 7 }$$
(iii) We have 0.$$\overline { 0.001 }$$
Question 4.
Express 0.99999… in the form of $$\frac { p }{ q }$$. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Solution:
Let x = 0.99999… …(1)
Multiplying by 10 on both sides, we get
10x = 9.99999… …(2)
Subtracting equation (1) from eqn. (2),
Thus 0.99999… = 1 = $$\frac { 1 }{ 1 }$$
Here we get p = 1 and q = 1
Since 0.99999… goes on forever. Hence there is no gap between 1 and 0.99999… and hence both are equal.
Question 5.
What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $$\frac { 1 }{ 17 }$$? Perform the division to check your answer.
Solution:
Long division method
Thus $$\frac { 1 }{ 17 }$$ = 0.$$\overline { 0.0588235294117647 }$$
We observe that by long division method maximum number of digits in repeating block in the decimal expansion of $$\frac { 1 }{ 17 }$$ is 16, thus answer is verified.
Question 6.
Look at several examples of rational numbers in the form $$\frac { p }{ q }$$ (q ā 0) where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?
Solution:
(i) $$\frac { 1 }{ 2 }$$ = $$\frac { 1 Ć 5 }{ 2 Ć 5 }$$ = $$\frac { 5 }{ 10 }$$ = 0.5
(ii) $$\frac { 3 }{ 4 }$$ = $$\frac { 3Ć5 Ć 5 }{ 2 Ć 2 Ć 5 Ć 5 }$$ = $$\frac { 75 }{ 100 }$$ = 0.75
(iii) $$\frac { 7 }{ 8 }$$ = $$\frac { 7 Ć 5 Ć 5 Ć 5 }{ 2 Ć 2 Ć 2 Ć 5 Ć 5 Ć 5 }$$ = $$\frac { 875 }{ 1000 }$$ = 0.875
(iv) $$\frac { 13 }{ 25 }$$ = $$\frac { 13 Ć 2 Ć 2 }{ 5 Ć 5 Ć 2 Ć 2 }$$ = $$\frac{52}{5^{2} \times 2^{2}}$$
= $$\frac{52}{(10)^{2}}$$ = $$\frac { 52 }{ 100 }$$ = 0.52
(v) $$\frac { 3 }{ 125 }$$ = $$\frac { 3 }{ 5 Ć 5 Ć 5 }$$ = = $$\frac{3}{5^{3}}$$ = $$\frac{3 \times 2^{3}}{5^{3} \times 2^{3}}$$
= $$\frac{3 \times 8}{(5 \times 2)^{3}}$$ = $$\frac { 24 }{ 1000 }$$ = 0.024
(vi) $$\frac { 27 }{ 16 }$$ = $$\frac{27 \times 5^{4}}{2^{4} \times 5^{4}}$$ = $$\frac{27 \times 5^{4}}{(2 \times 5)^{4}}$$
= $$\frac{27Ć625}{(10)^{4}}$$ = $$\frac{16875}{(10)^{4}}$$ = 1.6875
We observe that the denominator of all the above rational numbers are of the form 2m x 5n i.e., the prime factorization of denominators has only powers of 2 or powers of 5 or both.
Question 7.
Write three numbers whose decimal expansions are non-terminating non-recurring.
Solution:
(i) 0.012012001200012…
(ii) 0.21021002100021000021…
(iii) 0.32032003200032000032…
Question 8.
Find three different irrational numbers between the rational numbers $$\frac { 5 }{ 7 }$$ and $$\frac { 9 }{ 11 }$$.
Solution:
$$\frac { 5 }{ 7 }$$
Thus $$\frac { 5 }{ 7 }$$ = 0.714285…
$$\frac { 5 }{ 7 }$$ = 0.$$\overline { 714285 }$$….
Now $$\frac { 9 }{ 11 }$$
Thus $$\frac { 9 }{ 11 }$$ = 0.8181… = 0.$$\overline { 81 }$$
Thus three irrational numbers between the rational numbers $$\frac { 5 }{ 7 }$$ and $$\frac { 9 }{ 11 }$$ can be taken as
0. 73073007300073000073…
0. 757075700757000757…
and 0.808008000800008…
Question 9.
Classify the following numbers as rational or irrational.
(i) $$\sqrt{23}$$
(ii) $$\sqrt{225}$$
(iii) 0.3796
(iv) 7.478478…
(v) 1.101001000100001…
Solution:
(i) $$\sqrt{23}$$, 23 is not a perfect square so $$\sqrt{23}$$ will not give an integral value.
Hence it is not a rational number.
(ii) $$\sqrt{225}$$
ā“ $$\sqrt{225}$$
Here p = 15
and q = 1 (q ā 0)
(iii) 0.3796
The decimal expression is terminating.
Hence 0.3796 is a rational number.
(iv) 7.478478…
ā“ 7.478478… = 7.$$\overline { 748 }$$
The decimal expansion is non-terminating recurring.
ā“ 7.478478… is a rational number.
(v) 1.101001000100001…
āµ The decimal expansion is non – terminating non-recurring.
ā“ 1.101001000100001… is an irrational number.
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# How do you find the roots of x^2-x=20?
Apr 2, 2017
See below.
#### Explanation:
We can factor or graph the equation.
Factoring:
${x}^{2} - x - 20 = 0$ becomes $\left(x - 5\right) \left(x + 4\right) = 0$
So $x = 5 , - 4$.
Graphing:
graph{x^2-x-20 [-4.335, 5.665, -2.06, 2.94]}
We see that the graph intersects the $x$-axis at $x = 5 , - 4$, so those are our roots.
Apr 2, 2017
See the entire solution process below:
#### Explanation:
First, subtract $\textcolor{red}{20}$ from each side of the equation to put this equation into quadratic form:
${x}^{2} - x - \textcolor{red}{20} = 20 - \textcolor{red}{20}$
${x}^{2} - x - 20 = 0$
Because $4 - 5 = - 1$ and $4 \times - 5 = - 20$ we can factor the left side of the equation as
$\left(x + 4\right) \left(x - 5\right) = 0$
Now, we solve each term on the right side of the equation to find the roots for this problem:
Solution 1)
$x + 4 = 0$
$x + 4 - \textcolor{red}{4} = 0 - \textcolor{red}{4}$
$x + 0 = - 4$
$x = - 4$
Solution 2)
$x - 5 = 0$
$x - 5 + \textcolor{red}{5} = 0 + \textcolor{red}{5}$
$x - 0 = 5$
$x = 5$
The roots are: $x = - 4$ and $x = 5$
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## M7 Types of solutions
For any system of equations, there may be:
1. Infinitely many solutions
2. No solution
3. A unique solution
This module discusses these possibilities.
### Coefficient Matrix
Consider the following system of equations: \begin{align*} 2x+4y-z & =9\\ x-y+2z & =-4\\ -x+y-z & =3 \end{align*} These may be written in the matrix form: \begin{align*} \left[\begin{array}{ccc} 2 & 4 & -1\\ 1 & -1 & 2\\ -1 & 1 & -1 \end{array}\right]\left[\begin{array}{c} x\\ y\\ z \end{array}\right] & =\left[\begin{array}{c} 9\\ -4\\ 3 \end{array}\right]. \end{align*} The matrix \begin{align*} \left[\begin{array}{ccc} 2 & 4 & -1\\ 1 & -1 & 2\\ -1 & 1 & -1 \end{array}\right] \end{align*} is called the coefficient matrix because it contains the coefficients of the variables $$x,y$$ and $$z$$ in the system of equations.
#### What Does the Coefficient Matrix Tell Us?
The coefficient matrix can be used to find information about the number of solutions to a system of equations. For any system of equations, if the coefficient matrix is:
1. Singular (that is the determinant $$=0$$) there will be an infinite number of solutions or no solutions.
2. Non-singular (determinant $$\neq0$$) there will be a unique solution.
### Infinitely Many Solutions
#### Example 1
Consider the system of equations \begin{align*} x+2y-z & =3\\ x+3y & =4\\ -x-y+2z & =-2 \end{align*}
The augmented matrix for this system is \begin{align*} \left[\begin{array}{ccc} 1 & 2 & -1\\ 1 & 3 & 0\\ -1 & -1 & 2 \end{array}\left|\begin{array}{c} 3\\ 4\\ -2 \end{array}\right.\right]. \end{align*} Using elementary row operations we get: \begin{align*} \begin{array}{c} \\ R_{2}'=R_{2}-R_{1}\\ R_{3}'=R_{3}+R_{1} \end{array}\left[\begin{array}{ccc} 1 & 2 & -1\\ 0 & 1 & 1\\ 0 & 1 & 1 \end{array}\left|\begin{array}{c} 3\\ 1\\ 1 \end{array}\right.\right] \end{align*} \begin{align*} \begin{array}{c} \\ \\ R_{3}'=R_{3}-R_{2} \end{array}\left[\begin{array}{ccc} 1 & 2 & -1\\ 0 & 1 & 1\\ 0 & 0 & 0 \end{array}\left|\begin{array}{c} 3\\ 1\\ 0 \end{array}\right.\right] \end{align*} Row $$3$$ tells us that $$0x+0y+0z=0.$$ This is true for any value of $$x,y$$ or $$z.$$ So we are able to set any one of these variables to some number $$t\in\mathbb{R}$$.1 It can only be one variable as there other equations to be satisfied in $$R_{2}$$ and $$R_{1}.$$ For this example,2 Even though we set $$z=t,$$ we could set either $$x=t$$ or $$y=t$$. We still get a solution to the system of equations but it will look different. we set $$z=t.$$ From $$R_{2}$$ and substituting $$t$$ for $$z$$ we have \begin{align*} y+z & =1\\ y+t & =1\\ y & =1-t. \end{align*} Now from $$R_{1}$$, substituting $$z=t,\,y=1-t$$ we have \begin{align*} x+2y-z & =3\\ x+2\left(1-t\right)-t & =3\\ x+2-2t-t & =3\\ x & =1+3t. \end{align*} So the solution is $$x=1+3t,\,y=1-t$$ and $$z=t$$ where $$t\in\mathbb{R}$$.3 As there are an infinite number of choices for $$t,$$ there are an infinite number of solutions to the system of equations.
In general if the reduced augmented matrix has one or more rows of zeros, there will be an infinite number of solutions to the system of equations.4 You may recall that the determinant of a matrix with a row of zeros is zero. Hence the matrix is singular.
#### Example 2
Consider the system of equations \begin{align*} x+2y-z & =3\\ 2x+4y-2z & =6\\ -5x-10y+5z & =-15 \end{align*}
The augmented matrix for this system is \begin{align*} \left[\begin{array}{ccc} 1 & 2 & -1\\ 2 & 4 & -2\\ -5 & -10 & 5 \end{array}\left|\begin{array}{c} 3\\ 6\\ -15 \end{array}\right.\right]. \end{align*} Using elementary row operations we get \begin{align*} \begin{array}{c} \\ R_{2}'=R_{2}-2R_{1}\\ R_{3}'=R_{3}+5R_{1} \end{array}\left[\begin{array}{ccc} 1 & 2 & -1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\left|\begin{array}{c} 3\\ 0\\ 0 \end{array}\right.\right]. \end{align*} We have only one equation and three unknowns. In this case we must let two of the three variables be free. For example, we set $$z=t$$, where $$t\in\mathbb{R}$$ and $$y=s,$$ where $$s\in\mathbb{R}$$ .5 It doesn’t matter which two variables are selected to be a parameter $$s$$ or $$t.$$ We could set $$x=s$$ and $$y=t.$$ Substituting in row $$1$$ gives: \begin{align*} s+2t-z & =3\\ z & =s+2t-3. \end{align*} While this looks different to the solution at left, it is still a valid solution to the set of equations. Substituting into $$R_{1}$$ we get \begin{align*} x+2y-z & =3\\ x+2s-t & =3\\ x & =3-2s+t. \end{align*} So the solution is $$x=3-2s+t,\,y=s,\,z=t$$ where $$s\in\mathbb{R}$$ and $$t\in\mathbb{R}$$.
### No Solutions
Consider the system of equations \begin{align*} x+2y-2z & =3\\ 2x+y+z & =4\\ 3y-5z & =-1. \end{align*} The augmented matrix for this system is \begin{align*} \left[\begin{array}{ccc} 1 & 2 & -2\\ 2 & 1 & 1\\ 0 & 3 & -5 \end{array}\left|\begin{array}{c} 3\\ 4\\ -1 \end{array}\right.\right]. \end{align*} Using elementary row operations we get: \begin{align*} \begin{array}{c} \\ R_{2}'\\ \\ \end{array}=R_{2}-2R_{1}\left[\begin{array}{ccc} 1 & 2 & -2\\ 0 & -3 & 5\\ 0 & 3 & -5 \end{array}\left|\begin{array}{c} 3\\ -2\\ -1 \end{array}\right.\right] \end{align*} \begin{align*} \begin{array}{c} \\ \\ R_{3}'=R_{3}+R_{2} \end{array}\left[\begin{array}{ccc} 1 & 2 & -2\\ 0 & -3 & 5\\ 0 & 0 & 0 \end{array}\left|\begin{array}{c} 3\\ -1\\ -3 \end{array}\right.\right]. \end{align*} In this case, $$R_{3}$$ says that $$0x+0y+0z=-3.$$ This is not possible because regardless of the choice of $$x,y$$ and $$z$$ the LHS must be zero. In this case there are no solutions and we call the system of equations inconsistent.
### Unique Solutions
Consider the system of equations \begin{align*} 2x+4y-z & =9\\ x-y+2z & =-4\\ -x+y-z & =3 \end{align*}
The augmented matrix for this system is \begin{align*} \left[\begin{array}{ccc} 2 & 4 & -1\\ 1 & -1 & 2\\ -1 & 1 & -1 \end{array}\left|\begin{array}{c} 9\\ -4\\ 3 \end{array}\right.\right]. \end{align*} Using elementary row operations we get \begin{align*} \begin{array}{c} R_{1}'=R_{2}\\ R_{2}'=R_{1}\\ \\ \end{array}\left[\begin{array}{ccc} 1 & -1 & 2\\ 2 & 4 & -1\\ -1 & 1 & -1 \end{array}\left|\begin{array}{c} -4\\ 9\\ 3 \end{array}\right.\right] \end{align*} \begin{align*} \begin{array}{c} \\ R_{2}'=R_{2}-2R_{1}\\ R_{3}'=R_{3}+R_{1} \end{array}\left[\begin{array}{ccc} 1 & -1 & 2\\ 0 & 6 & -5\\ 0 & 0 & 1 \end{array}\left|\begin{array}{c} -4\\ 17\\ -1 \end{array}\right.\right]. \end{align*} From row $$3$$ we have $$z=-1.$$ Substituting this into row 2 we have \begin{align*} 6y-5z & =17\\ 6y+5 & =17\\ 6y & =12\\ y & =2. \end{align*} Substituting for $$z$$ and $$y$$ in row 1 gives: \begin{align*} 2x+4y-z & =9\\ 2x+4\left(2\right)-\left(-1\right) & =9\\ 2x+8+1 & =9\\ 2x & =0\\ x & =0. \end{align*} So the solution is $$x=0,\,y=2,\,z=-1$$.
### Exercises
Find the solutions (if they exist) to the following systems of equations:
$\begin{array}{lllll} 1. & x+2y-z=5 & \quad\quad & 2. & x+2y-2z=7\\ & 3x-y+2z=1 & & & 2x-y-z=-3\\ & 2x-y+3z=0 & & & -x-2y+2z=1 \end{array}$
$\begin{array}{lllll} 3. & 2x-y+3z=-3 & & 4. & x+2y-z=3\\ & x+2y+z=-4 & & & -x+y-2z=-3\\ & 4x+3y+5z=-11 & & & 3x+3z=9 \end{array}$
1. $$x=1,\,y=2,\,z=0$$
3. $$x=-\frac{7t}{5}-2,\,y=\frac{t}{5}-1,\,z=t\;$$ (Your answer may be different if you choose $$y=t$$ or $$x=t.)$$
4. $$x=3-t,\,y=t,\,z=t\;$$ (Your answer may be different if you choose $$y=t$$ or $$x=t.)$$
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# √3 value @sorumatik
√3 value @sorumatikbot
## LectureNotes said “√3 value” @sorumatikbot
The value of the square root of 3, denoted as \sqrt{3}, is an irrational number, meaning it cannot be expressed as a simple fraction. Its decimal approximation is:
\sqrt{3} \approx 1.732
### Detailed Explanation
1. Understanding Square Roots:
• The square root of a number x is a value that, when multiplied by itself, gives x. Mathematically, if y = \sqrt{x}, then y^2 = x.
• For \sqrt{3}, this means that \sqrt{3} \times \sqrt{3} = 3.
2. Irrational Numbers:
• An irrational number cannot be expressed as the ratio of two integers. Its decimal representation is non-terminating and non-repeating.
• \sqrt{3} falls into this category because its decimal expansion goes on forever without repeating.
3. Decimal Approximation:
• The first few decimal places of \sqrt{3} are approximately 1.7320508075688772935274463415059…
• This approximation can be useful in various calculations when an exact value isn’t necessary.
### How to Calculate \sqrt{3} Manually
1. Prime Factorization Method:
• While this method works well for perfect squares, \sqrt{3} involves an estimation technique because 3 is not a perfect square.
2. Estimation Method:
• By manually finding numbers between which \sqrt{3} lies, we can approximate its value.
• Example: Since 1^2 = 1 and 2^2 = 4, \sqrt{3} lies between 1 and 2.
• Using methods such as the Babylonian method (also known as Heron’s method) can improve the approximation iteratively.
3. Using a Calculator:
• Most scientific calculators have a square root function that can provide a quick and accurate approximation.
In conclusion, the value of \sqrt{3} is approximately 1.732, but its exact value is irrational and cannot be expressed fully in decimal or fractional form. For practical purposes, however, this approximation is often sufficiently accurate.
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# Verify that your factors are correct by expanding the expression, combining like terms, and comparing to the initial trinomial.
### Review
Review this week’s assigned readings and resources to prepare for this discussion, as well as
Many polynomial equations can be solved quite simply when the first step consists of factoring. Factoring is an essential skill to be mastered in an algebra course. While there are a number of different techniques that can be used to factor polynomial functions of all types, for our purposes we are going to focus on factoring quadratic polynomials of the form ax² + bx + c. A number of different methods have been introduced to help students find the factors of quadratic polynomials. Some such methods of factoring ax² + bx + c are Factoring by Grouping, Box Method, Star Method/Diamond Method/X-method, and Tic-Tac-Toe Method.
### Respond
1. Research one of the methods listed above or another method of your choice for factoring trinomials of the form ax² + bx + c.
3. State the method you are using and explain the process of factoring a trinomial in words, modeling the process by using examples that contain all the steps to factor the trinomial.
• Note: A student who has little knowledge of factoring should be able to follow your steps in order to factor any trinomial using the explanation you provide.
• Your guide must include a minimum of two examples of factoring trinomials using your chosen method.
• At least one of your examples must include a trinomial of the form ax² + bx + c, with a > 1 (e.g. 4x² – 15x – 25, or 2x² + 11x + 5).
4. Verify that your factors are correct by expanding the expression, combining like terms, and comparing to the initial trinomial.
5. Make sure to cite the sources you used for researching your preferred method.
6. Submit your completed template to the discussion forum.
## Factoring Trinomials Guide
Factoring Method:
### Description of Method:
Example. Trinomial:
= , = , =
Example. Trinomial:
= , = , =
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# 7 Cool Math Proofs - IntoMath
Updated: Jan 18
Mathematics is a fascinating science where things depend on each other in many different ways and where one quantity can be expressed as the other in order to derive or prove the properties or the very existence of that very quantity.
Here are some simple, yet cool proofs.
## Prove the Pythagorean Theorem
According to the Pythagorean Theorem the square of two shorter sides in the right angle triangle are equal to the square of the longest side (the hypotenuse).
Visually, this relationship could be represented like this
To prove that this relationship is true, let’s take a look at the following
Proven.
## Prove that the Area of Triangle is equal to half of the product of its base and height
Given the rectangle, with a shorter side b and a longer side h, we are able to find its area by finding the product the two sides: bh.
Consider two triangles, whose bases are congruent to b and whose heights are congruent to h in the rectangle. One of these triangles will be split into two congruent right angle triangles.
We will get the following
From the diagrams above we can observe that two triangles (placed upside down) whose bases are congruent to b and whose heights are congruent to h in the rectangle together form a rectangle.
Proven.
## Prove that the diagonals of a rectangle are congruent
Congruent means equal in size and shape (≅)
ABCD is a rectangle, it is also a parallelogram.
Then AB ≌ DC and BC ≌ BC (by the Reflexive Property of Congruence)
⦟ ABC = ⦟ DCB = 90
⦟ ABC ≌ ⦟ DCB since all right angles are congruent
Therefore, AB ≌ DC, BC ≌ BC, ⦟ ABC ≌ ⦟ DCB
so ABC ≌ DCB
and segment AC ≌ segment BD
Proven.
## Prove that the sum of the interior angles in a triangle equals 180 degrees
When two parallel lines are crossed by a transversal, alternate interior angles are equal.
Since a complete circle is 360 degrees, then half of it is 180 degrees (semi-circle.)
Therefore, angles in a semicircle must add up to 180 degrees.
Then inside the triangle all angles also add up to 180 degrees.
Proven.
## Prove that for all integers n if n is even, then the square of n is even
Let n be any integer. Suppose n is even.
Then n = 2k for some integer k:
Proven.
## Prove that the sum of the first 100 consecutive whole numbers is 5,050
We can, of course, take every single whole number from 1 to 100 and add it to the previous one. But it will take us a very long time, especially without a calculator.
Instead, if “fold” the numbers and add each pair, we will get a constant sum of 101 (see below). Since there are 100 numbers total, there are 50 pairs of numbers.
Thus, the sum of all 100 numbers is 50 x 101 = 5,050
Proven.
## Prove that the area of a circle is the product of Pi and the square of the radius of a circle
A circle can be split into triangles (almost, because there is still a bit of a round portion left over), whose vertices meet at the center of a circle - the more triangles we split it into, the closer the total length of their bases is to that of a circumference of a circle.
Let’s split our circle into 8 triangles for visual purposes and arrange them in a rectangular formation. We almost get a rectangle.
We know that the area of a rectangle is equal to “length x width”. The width of our rectangle will be the same as the radius. To find the length we will need the circumference of the circle.
Since the circumference is equal to 2πr, the base has to be πr.
Now “length x width” becomes “πr x r”. That gives us πr² which is equal to the area of the rectangle which is equal to the area of the circle.
Proven.
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### Home > CAAC > Chapter 4 > Lesson 4.1.6 > Problem4-58
4-58.
Ms. B is making snickerdoodle cookies. Her recipe uses one-and-a-half teaspoons of cinnamon to make two-dozen cookies. If she needs to make thirteen-dozen cookies in order to give one cookie to each of her students, how much cinnamon will she need? Homework Help ✎
Use proportions to solve this problem. Assign a variable to the unknown number of teaspoons of cinnamon.
For this case, we will use the variable n.
Multiply both sides by 2.
Multiply both sides by 13.
Divide by 2 on both sides.
Remember to use the conversion factor to determine how many tablespoons and teaspoons Ms. B will need.
$\frac{ 1 \frac{ 1 }{2 } \text{ teaspoons}}{ 2\text{ dozen cookies}}= \frac{ n\text{ teaspoons}}{ 13\text{ dozen cookies}}$
$\frac{ 1 \frac{ 1 }{2 } }{2 }= \frac{ n }{13 }$
$1 \frac{ 1 }{2 } = \frac{ 2n }{13 }$
19.5 = 2n
$n=9.75 \text{ or }9 \frac{3}{4} \text{ teaspoons}$
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Show Question
Math and science::Analysis::Tao::06. Limits of sequences
# Limits of sequences
If a sequence of reals $$(a_n)_{n=m}^{\infty}$$ converges to a real $$L$$, we say that $$(a_n)_{n=m}^{\infty}$$ is convergent and that $$L$$ is the limit of the sequence. We write:
$L = \lim_{n\to\infty}a_n$
If a sequence does not converge to a real, then it is said to diverge or be divergent. $$\lim_{n\to\infty}a_n$$ is left undefined for a divergent sequence.
$$L$$ is unique as per the Uniqueness of Convergence proposition. The uniqueness allows us to create the limit notation $$\lim_{n\to\infty} a_n$$, dependent only on the sequence $$(a_n)_{n=m}^{\infty}$$, and for that notation to be equated to a single real. If fact, the limit notation doesn't depend on the full specification of the sequence $$(a_n)_{n=m}^{\infty}$$, as the starting index is irrelevant and neglected. So the $$\lim_{n\to\infty} a_n$$ is the limit for any sequences $$(a_n)_{n=N}^{\infty}$$ where $$N$$ is some integer.
### Example
#### $$\lim_{n\to\infty} \frac{1}{n} = 0$$, proof
$$\lim_{n\to\infty} \frac{1}{n}= 0$$ is true if the sequence $$(a_n)_{n=1}^{\infty}$$ converges to 0, where $$a_n = \frac{1}{n}$$.
Let $$\varepsilon > 0$$ be an arbitary real. We must show that there exists an integer $$N$$ such that $$|a_n - 0| \le \varepsilon$$ for all $$n \ge N$$.
$$|a_n - 0| = |\frac{1}{n} - 0| = \frac{1}{n} \le \frac{1}{N}$$.
$$\frac{1}{N} \le \varepsilon$$ is true if $$N \ge \frac{1}{\varepsilon}$$. Such an integer $$N$$ exists as per the Achimedean principal. Thus $$|a_n - 0| \le \varepsilon$$ for all $$n \ge N$$. Thus, $$(a_n)_{n=1}^{\infty}$$ is eventually $$\varepsilon$$ close to 0 for arbitary $$\varepsilon > 0$$. Thus, $$(a_n)_{n=1}^{\infty}$$ converges to 0. This allows us to write $$\lim_{n\to\infty} \frac{1}{n}= 0$$.
Cauchy sequences of reals → convergence of sequences of reals → uniqueness of convergence → limit, the definition → subsumption of formal limits
Tao, Analysis I
Chapter 6
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Trigonometry
Basic relationships between trigonometric functions of the same angle
Basic relationships between trigonometric functions of the same angle shown in the unit circle
Basic relationships between trigonometric functions of the same angle shown in the tabular form
Basic relationships between trigonometric functions of the same angle, examples
Basic relationships between trigonometric functions of the same angle shown in the unit circle
Given relations between the trigonometric functions of the same angle, expressed by absolute value, are included in the definitions of the functions in right triangles shown in Fig. a - d, and included in the table below.
Basic relationships between trigonometric functions shown in the tabular form
Basic relationships between trigonometric functions of the same angle, examples
Example: Find values of other trigonometric functions of an angle a, if given sin a = - 4/5 and
270° < a < 360°.
Solution: Since a is a forth-quadrant angle, then
Example: Find the value of
Solution: Dividing the numerator and the denominator by cos2 x,
Example: Prove the identity
Solution:
Example: Given sin x + cos x = a, find sin4 x + cos4 x .
Solution: Since sin2 x + cos2 x = 1 then,
(sin2 x + cos2 x)2 = sin4 x + cos4 + 2sin2 x · cos2 x = 1 or sin4 x + cos4 = 1 - 2sin2 xcos2 x.
As given sin x + cos x = a then, (sin x + cos x)2 = a2 or sin2 x + cos2 x + 2sin x cos x = a2
therefore, sin x · cos x = (a2 - 1)/2 and it follows that sin4 x + cos4 = 1 - (a2 - 1)2/2.
Trigonometry contents A
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# How to find the asymptotes of y = x /(x-3) ?
Dec 20, 2015
I found:
$\textcolor{red}{\text{Vertical Asymptote: } x = 3}$
$\textcolor{b l u e}{\text{Horizontal Asymptote: } y = 1}$
#### Explanation:
Looking at your function you can "see" a vertical asymptote,i.e., a vertical line towards which the graph of your function will tend to get near as much as possible.
You need to look for possible prohibited $x$ values and there you'll find your asymptote.
In this case the prohibited $x$ value is the one that makes your function a division by ZERO: i.e., $x = 3$ (you cannot use $x = 3$ into your function so the vertical line passing through it will be a forbidden place or your asymptote).
$\textcolor{red}{\text{Vertical Asymptote: } x = 3}$
You can also "see" a horizontal asymptote,i.e., a horizontal line towards which the graph of your function will tend to.
This is a little more difficult but you can think big and imagine what happens to the graph of your function when $x$ becomes VERY large and ask yourself: "it tends towards something?".
Let us take a $x$ very big, say, $x = 1 , 000 , 000$
and evaluate your function there so you get:
$f \left(1 , 000 , 000\right) = \frac{1 , 000 , 000}{1 , 000 , 000 - 3} \approx 1$
because $3$, compared to $1 , 000 , 000$, is negligible.
So you get that the horizontal line passing through: $y = f \left(1 , 000 , 000\right) = 1$
will be the line towards which the graph tends to get near as much as possible:
$\textcolor{b l u e}{\text{Horizontal Asymptote: } y = 1}$
You can see graphically these two tendencies:
graph{x/(x-3) [-8.89, 8.89, -4.444, 4.445]}
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A worked example of simplifying the cube root of 27a²b⁵c³ using the properties of exponents. Created by Sal Khan and Monterey Institute for Technology and Education.
Video transcript
We're asked to simplify the cube root of 27a squared times b to the fifth times c to the third power. And the goal, whenever you try to just simplify a cube root like this, is we want to look at the parts of this expression over here that are perfect cubes, that are something raised to the third power. Then we can take just the cube root of those, essentially taking them out of the radical sign, and then leaving everything else that is not a perfect cube inside of it. So let's see what we can do. So first of all, 27-- you may or may not already recognize this as a perfect cube. If you don't already recognize it, you can actually do a prime factorization and see it's a perfect cube. 27 is 3 times 9, and 9 is 3 times 3. So 27-- its prime factorization is 3 times 3 times 3. So it's the exact same thing as 3 to the third power. So let's rewrite this whole expression down here. But let's write it in terms of things that are perfect cubes and things that aren't. So 27 can be just rewritten as 3 to the third power. Then you have a squared-- clearly not a perfect cube. a to the third would have been. So we're just going to write this-- let me write it over here. We can switch the order here because we just have a bunch of things being multiplied by each other. So I'll write the a squared over here. b to the fifth is not a perfect cube by itself, but it can be expressed as the product of a perfect cube and another thing. b to the fifth is the exact same thing as b to the third power times b to the second power. If you want to see that explicitly, b to the fifth is b times b times b times b times b. So the first three are clearly b to the third power. And then you have b to the second power after it. So we can rewrite b to the fifth as the product of a perfect cube. So I'll write b to the third-- let me do that in that same purple color. So we have b to the third power over here. And then it's b to the third times b squared. So I'll write the b squared over here. And we're assuming we're going to multiply all of this stuff. And then finally, we have-- I'll do in blue-- c to the third power. Clearly, this is a perfect cube. It is c cubed. It is c to the third power. So I'll put it over here. So this is c to the third power. And of course, we still have that overarching radical sign. So we're still trying to take the cube root of all of this. And we know from our exponent properties, or we could say from our radical properties, that this is the exact same thing. That taking the cube root of all of these things is the same as taking the cube root of these individual factors and then multiplying them. So this is the same thing as the cube root-- and I could separate them out individually. Or I could say the cube root of 3 to the third b to the third c to the third. Actually, let's do it both ways. So I'll take them out separately. So this is the same thing as the cube root of 3 to the third times the cube root-- I'll write them all in. Let me color-code it so we don't get confused-- times the cube root of b to the third times the cube root of c to the third times the cube root-- and I'll just group these two guys together just because we're not going to be able to simplify it any more-- times the cube root of a squared b squared. I'll keep the colors consistent while we're trying to figure out what's what. And I could have said that this is times the cube root of a squared times the cube root of b squared, but that won't simplify anything, so I'll just leave these like this. And so we can look at these individually. The cube root of 3 to the third, or the cube root of 27-- well, that's clearly just going to be-- I want to do that in that yellow color-- this is clearly just going to be 3. 3 to the third power is 3 to the third power, or it's equal to 27. This term right over here, the cube root of b to the third-- well, that's just b. And the cube root of c to the third, well, that is clearly-- I want to do that in that-- that is clearly just c. So our whole expression has simplified to 3 times b times c times the cube root of a squared b squared. And we're done. And I just want to do one other thing, just because I did mention that I would do it. We could simplify it this way. Or we could recognize that this expression right over here can be written as 3bc to the third power. And if I take three things to the third power, and I'm multiplying it, that's the same thing as multiplying them first and then raising to the third power. It comes straight out of our exponent properties. And so we can rewrite this as the cube root of all of this times the cube root of a squared b squared. And so the cube root of all of this, of 3bc to the third power, well, that's just going to be 3bc, and then multiplied by the cube root of a squared b squared. I didn't take the trouble to color-code it this time, because we already figured out one way to solve it. But hopefully, that also makes sense. We could have done this either way. But the important thing is that we get that same answer.
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# Writing equations of lines
Write the equation of the line that passes through the points 7, -3 and 7, 0.
You also have TWO points use can use. Find the equation of the line that passes through 0, -3 and -2, 5. As we have in each of the other examples, we can use the point-slope form of a line to find our equation.
Since you are given two points, you can first use the slope formula to find the slope and then use that slope with one of the given points.
Two of those are: If you need to practice these strategies, click here. What is your answer?
You may be wondering why this form of a line was not mentioned at the beginning of the lesson with the other two forms. Now that you have a slope, you can use the point-slope form of a line.
Although the numbers are not as easy to work with as the last example, the process is still the same. If you need help rewriting the equation, click here for practice link to linear equations slope.
How do you know which one is the right one? Find the equation of the line that passes through 1, -5 and is parallel to. Find the equation of the line that passes through the points -2, 3 and 1, Transforming the slope-intercept form into general form gives Parallel and Perpendicular There is one other common type of problem that asks you to write the equation of a line given certain information.
The first step is to find the slope of the line that goes through those two points. Those have x and y variables in the equation. If you are comfortable with plugging values into the equation, you may not need to include this labeling step.
That is because the point-slope form is only used as a tool in finding an equation. When using this form you will substitute numerical values for x1, y1 and m.
Some students find it useful to label each piece of information that is given to make substitution easier. So if we can find the slope ofwe will have the information we need to proceed with the problem.
Most students, since they have already labeled a and when finding the slope, choose to keep that labeling system. You will NOT substitute values for x and y. That means our line will have the same slope as the line we are given.
How is this possible if for the point-slope form you must have a point and a slope? The strategy you use to solve the problem depends on the type of information you are given.
If we re-write in slope-intercept form, we will easily be able to find the slope.
If is parallel to and passes through the point 5, 5transform the first equation so that it will be perpendicular to the second. You would first find the slope of the given line, but you would then use the negative reciprocal in the point-slope form.
Given a Point and a Slope When you are given a point and a slope and asked to write the equation of the line that passes through the point with the given slope, you have to use what is called the point-slope form of a line.
If you need help calculating slope, click here for lessons on slope. Look at the slope-intercept and general forms of lines. Equations of lines come in several different forms. Now substitute those values into the point-slope form of a line.
Both forms involve strategies used in solving linear equations. Plug those values into the point-slope form of the line:Apr 28, · For a complete lesson on writing the equation of a line, go to ultimedescente.com - + online math lessons featuring a.
Writing Equations of Perpendicular and Parallel Lines Write an equation of the line that passes through (3, 2) and is (a) perpendicular and (b) parallel to the line y = º3 x + 2. ©d 82P0k1 f2 T 1K lu9t qap 2S ho KfZtgw HaTrte I BL gLiCQ.e R xA NlOlh JrKi0gMh6t8sq YrCenshe Rr8vqeed Y JMGapdQeX TwGiRt VhW 8I 2n fDiPn 8iDtEep QAVlVgue3bjr vaV Y Elementary Algebra Skill Writing Equations of Lines Given the Graph Write the slope-intercept form of the equation of e ach line.
1) −5−4−3−2−10 1 2 3 4 5. Equations of lines come in several different forms. Two of those are: slope-intercept form; where m is the slope and b is the y-intercept. general form; Your teacher or textbook will usually specify which form you should be using.
Improve your math knowledge with free questions in "Write an equation for a parallel or perpendicular line" and thousands of other math skills.
Writing equations of lines
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# 14.6: Directional Derivatives and the Gradient
In Partial Derivatives, we introduced the partial derivative. A function $$z=f(x,y)$$ has two partial derivatives: $$∂z/∂x$$ and $$∂z/∂y$$. These derivatives correspond to each of the independent variables and can be interpreted as instantaneous rates of change (that is, as slopes of a tangent line). For example, $$∂z/∂x$$ represents the slope of a tangent line passing through a given point on the surface defined by $$z=f(x,y),$$ assuming the tangent line is parallel to the x-axis. Similarly, $$∂z/∂y$$ represents the slope of the tangent line parallel to the y-axis. Now we consider the possibility of a tangent line parallel to neither axis.
## Directional Derivatives
We start with the graph of a surface defined by the equation $$z=f(x,y)$$. Given a point $$(a,b)$$ in the domain of $$f$$, we choose a direction to travel from that point. We measure the direction using an angle $$θ$$, which is measured counterclockwise in the x, y-plane, starting at zero from the positive x-axis (Figure $$\PageIndex{1}$$). The distance we travel is $$h$$ and the direction we travel is given by the unit vector $$u=(\cos θ)i+(\sin θ)j.$$ Therefore, the $$z$$-coordinate of the second point on the graph is given by $$z=f(a+h\cos θ,b+h\sin θ).$$
Figure $$\PageIndex{1}$$: Finding the directional derivative at a point on the graph of $$z=f(x,y)$$. The slope of the black arrow on the graph indicates the value of the directional derivative at that point.
We can calculate the slope of the secant line by dividing the difference in z-values by the length of the line segment connecting the two points in the domain. The length of the line segment is $$h$$. Therefore, the slope of the secant line is
$m_{sec}=\dfrac{f(a+h\cos θ,b+h\sin θ)−f(a,b)}{h}$
To find the slope of the tangent line in the same direction, we take the limit as $$h$$ approaches zero.
Definition: Directional Derivatives
Suppose $$z=f(x,y)$$ is a function of two variables with a domain of $$D$$. Let $$(a,b)∈D$$ and define $$u=cosθi+sinθj$$. Then the directional derivative of $$f$$ in the direction of $$u$$ is given by
$D_uf(a,b)=\lim_{h→0}\dfrac{f(a+h \cos θ,b+h\sin θ)−f(a,b)}{h}$
provided the limit exists.
Equation provides a formal definition of the directional derivative that can be used in many cases to calculate a directional derivative.
Example $$\PageIndex{1}$$: Finding a Directional Derivative from the Definition
Let $$θ=\arccos(3/5).$$ Find the directional derivative $$D_uf(x,y)$$ of $$f(x,y)=x^2−xy+3y^2$$ in the direction of $$u=(\cos θ)i+(\sin θ)j$$. What is $$D_uf(−1,2)$$?
Solution
First of all, since $$\cos θ=3/5$$ and $$θ$$ is acute, this implies
$\sin θ=\sqrt{1−\left(\dfrac{3}{5}\right)^2}=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}. \nonumber$
Using $$f(x,y)=x^2−xy+3y^2,$$ we first calculate $$f(x+h\cos θ,y+h\sin θ)$$:
\begin{align*} f(x+h\cos θ,y+h\sin θ)&=(x+h\cos θ)^2−(x+h\cos θ)(y+h\sin θ)+3(y+h\sin θ)^2 \\&=x^2+2xh\cos θ+h^2\cos^2 θ−xy−xh\sin θ−yh\cos θ−h^2\sin θ\cos θ+3y^2+6yh\sin θ+3h^2\sin^2 θ \\ &=x^2+2xh(\dfrac{3}{5})+\dfrac{9h^2}{25}−xy−\dfrac{4xh}{5}−\dfrac{3yh}{5}−\dfrac{12h^2}{25}+3y^2+6yh(\dfrac{4}{5})+3h^2(\dfrac{16}{25})\\&=x^2−xy+3y^2+\dfrac{2xh}{5}+\dfrac{9h^2}{5}+\dfrac{21yh}{5}. \end{align*}
We substitute this expression into Equation:
\begin{align*} D_uf(a,b)&=\lim_{h→0}\dfrac{f(a+h\cos θ,b+h\sin θ)−f(a,b)}{h}\\ &=\lim_{h→0}\dfrac{(x^2−xy+3y^2+\dfrac{2xh}{5}+\dfrac{9h^2}{5}+\dfrac{21yh}{5})−(x^2−xy+3y^2)}{h}\\ &=\lim_{h→0}\dfrac{\dfrac{2xh}{5}+\dfrac{9h^2}{5}+\dfrac{21yh}{5}}{h}\\ &=\lim_{h→0}\dfrac{2x}{5}+\dfrac{9h}{5}+\dfrac{21y}{5}\\&=\dfrac{2x+21y}{5}. \end{align*}
To calculate $$D_uf(−1,2),$$ we substitute $$x=−1$$ and $$y=2$$ into this answer (Figure $$\PageIndex{2}$$):
$D_uf(−1,2)=\dfrac{2(−1)+21(2)}{5}=\dfrac{−2+42}{5}=8.$
Figure $$\PageIndex{2}$$: Finding the directional derivative in a given direction $$u$$ at a given point on a surface. The plane is tangent to the surface at the given point $$(−1,2,15).$$
Another approach to calculating a directional derivative involves partial derivatives, as outlined in the following theorem.
Directional Derivative of a Function of Two Variables
Let $$z=f(x,y)$$ be a function of two variables $$x$$ and $$y$$, and assume that $$f_x$$ and $$f_y$$ exist. Then the directional derivative of $$f$$ in the direction of $$u=\cos θi+\sin θj$$ is given by
$D_uf(x,y)=f_x(x,y)\cos θ+f_y(x,y)\sin θ.$
Proof
Equation states that the directional derivative of $$f$$ in the direction of $$u=\cos θi+\sin θj$$ is given by
$D_uf(a,b)=\lim_{t→0}\dfrac{f(a+t \cos θ,b+t\sin θ)−f(a,b)}{t}.$
Let $$x=a+t\cos θ$$ and $$y=b+t\sin θ,$$ and define $$g(t)=f(x,y)$$. Since $$f_x$$ and $$f_y$$ both exist, we can use the chain rule for functions of two variables to calculate $$g′(t)$$:
$g′(t)=\dfrac{∂f}{∂x}\dfrac{dx}{dt}+\dfrac{∂f}{∂y}\dfrac{dy}{dt}=f_x(x,y)\cos θ+f_y(x,y)\sin θ.$
If $$t=0,$$ then $$x=x_0$$ and $$y=y_0,$$ so
$g′(0)=f_x(x_0,y_0)\cos θ+f_y(x_0,y_0)\sin θ$
By the definition of $$g′(t),$$ it is also true that
$g′(0)=\lim_{t→0}\dfrac{g(t)−g(0)}{t}=\lim_{t→0}\dfrac{f(x_0+t\cos θ,y_0+t\sin θ)−f(x_0,y_0)}{t}.$
Therefore, $$D_uf(x_0,y_0)=f_x(x,y)\cos θ+f_y(x,y)\sin θ.$$
Example $$\PageIndex{2}$$: Finding a Directional Derivative: Alternative Method
Let $$θ=\arccos (3/5).$$ Find the directional derivative $$D_uf(x,y)$$ of $$f(x,y)=x^2−xy+3y^2$$ in the direction of $$u=(\cos θ)i+(\sin θ)j$$. What is $$D_uf(−1,2)$$?
Solution
First, we must calculate the partial derivatives of $$f$$:
\begin{align*}f_x&=2x−y \\ f_y&=−x+6y, \end{align*}
Then we use Equation with $$θ=\arccos (3/5)$$:
\begin{align*} D_uf(x,y)&=f_x(x,y)\cos θ+f_y(x,y)\sin θ \\&=(2x−y)\dfrac{3}{5}+(−x+6y)\dfrac{4}{5} \\ &=\dfrac{6x}{5}−\dfrac{3y}{5}−\dfrac{4x}{5}+\dfrac{24y}{5}\\&=\dfrac{2x+21y}{5}. \end{align*}
To calculate $$D_uf(−1,2),$$ let $$x=−1$$ and $$y=2$$:
$D_uf(−1,2)=\dfrac{2(−1)+21(2)}{5}=\dfrac{−2+42}{5}=8.$
This is the same answer obtained in Example $$\PageIndex{1}$$.
Exercise $$\PageIndex{1}$$:
Find the directional derivative $$D_uf(x,y)$$ of $$f(x,y)=3x^2y−4xy^3+3y^2−4x$$ in the direction of $$u=(\cos \dfrac{π}{3})i+(\sin \dfrac{π}{3})j$$ using Equation. What is $$D_u f(3,4)$$?
Hint
Calculate the partial derivatives and determine the value of $$θ$$.
$$D_uf(x,y)=\dfrac{(6xy−4y^3−4)(1)}{2}+\dfrac{(3x^2−12xy^2+6y)\sqrt{3}}{2}$$
$$D_uf(3,4)=\dfrac{72−256−4}{2}+\dfrac{(27−576+24)\sqrt{3}}{2}=−94−\dfrac{525\sqrt{3}}{2}$$
If the vector that is given for the direction of the derivative is not a unit vector, then it is only necessary to divide by the norm of the vector. For example, if we wished to find the directional derivative of the function in Example in the direction of the vector $$⟨−5,12⟩$$, we would first divide by its magnitude to get $$u$$. This gives us $$u=⟨−(5/13),12/13⟩.$$ Then
\begin{align*} D_uf(x,y)&= ∇f(x,y)⋅u \\ &=−\dfrac{5}{13}(2x−y)+\dfrac{12}{13}(−x+6y) \\ &=−\dfrac{22}{13}x+\dfrac{17}{13}y \end{align*}
The right-hand side of Equation is equal to $$f_x(x,y)\cos θ+f_y(x,y)\sin θ,$$ which can be written as the dot product of two vectors. Define the first vector as $$∇f(x,y)=f_x(x,y)i+f_y(x,y)j$$ and the second vector as $$u=(\cos θ)i+(\sin θ)j.$$ Then the right-hand side of the equation can be written as the dot product of these two vectors:
$D_uf(x,y)=∇f(x,y)⋅u.$
The first vector in Equation has a special name: the gradient of the function $$f$$. he symbol $$∇$$ is called nabla and the vector $$∇f$$ is read “del $$f$$.”
Let $$z=f(x,y)$$ be a function of $$x$$ and $$y$$ such that $$f_x$$ and $$f_y$$ exist. The vector $$∇f(x,y)$$ is called the gradient of $$f$$ and is defined as
$∇f(x,y)=f_x(x,y)i+f_y(x,y)j. \label{grad}$
The vector $$∇f(x,y)$$ is also written as “grad $$f$$.”
Example $$\PageIndex{3}$$: Finding Gradients
Find the gradient $$∇f(x,y)$$ of each of the following functions:
1. $$f(x,y)=x^2−xy+3y^2$$
2. $$f(x,y)=\sin 3 x \cos 3y$$
Solution
For both parts a. and b., we first calculate the partial derivatives $$f_x$$ and $$f_y$$, then use Equation \ref{grad}.
a. \begin{align*} f_x(x,y)&=2x−y\; \text{and}\; f_y(x,y)=−x+6y,\; \text{so} \\ ∇f(x,y)&=f_x(x,y)i+f_y(x,y)j\\&=(2x−y)i+(−x+6y)j.\end{align*}
b. \begin{align*} f_x(x,y)&=3\cos 3x \cos 3y \; \text{and} \; f_y(x,y)=−3\sin 3x \sin 3y, \text{so} \\ ∇f(x,y)&=f_x(x,y)i+f_y(x,y)j \\ &=(3\cos 3x \cos 3y)i−(3\sin 3x \sin 3y)j. \end{align*}
Exercise $$\PageIndex{2}$$
Find the gradient $$∇f(x,y)$$ of $$f(x,y)=(x^2−3y^2)/(2x+y).$$
Hint
Calculate the partial derivatives, then use Equation \ref{grad}.
$$∇f(x,y)=\dfrac{2x^2+2xy+6y^2}{(2x+y)^2}i−\dfrac{x^2+12xy+3y^2}{(2x+y)^2}j$$
The gradient has some important properties. We have already seen one formula that uses the gradient: the formula for the directional derivative. Recall from The Dot Product that if the angle between two vectors $$a$$ and $$b$$ is $$φ$$, then $$a⋅b=‖a‖∥b∥\cos φ.$$ Therefore, if the angle between $$∇f(x_0,y_0)$$ and $$u=(cosθ)i+(sinθ)j \text{is} φ,$$ we have
$D_uf(x_0,y_0)=∇f(x_0,y_0)⋅u=∥∇f(x_0,y_0)∥‖u‖\cos φ=∥∇f(x_0,y_0)∥\cos φ.$
The $$‖u‖$$ disappears because $$u$$ is a unit vector. Therefore, the directional derivative is equal to the magnitude of the gradient evaluated at $$(x_0,y_0)$$ multiplied by $$\cos φ$$. Recall that $$\cos φ$$ ranges from $$−1$$ to $$1$$. If $$φ=0,$$ then $$\cos φ=1$$ and $$∇f(x_0,y_0)$$ and $$u$$ both point in the same direction. If $$φ=π$$, then $$\cos φ=−1$$ and $$∇f(x_0,y_0)$$ and u point in opposite directions. In the first case, the value of $$D_uf(x_0,y_0)$$ is maximized; in the second case, the value of $$D_uf(x_0,y_0)$$ is minimized. If $$∇f(x_0,y_0)=0$$, then
$D_uf(x_0,y_0)=∇f(x_0,y_0)⋅u=0$
for any vector $$u$$. These three cases are outlined in the following theorem.
Properties of the Gradient
Suppose the function $$z=f(x,y)$$ is differentiable at $$(x_0,y_0)$$ (Figure $$\PageIndex{3}$$).
1. If $$∇f(x_0,y_0)=0,$$ then $$D_uf(x_0,y_0)=0$$ for any unit vector $$u$$.
2. If $$∇f(x_0,y_0)≠0$$, then $$D_uf(x_0,y_0)$$ is maximized when u points in the same direction as $$∇f(x_0,y_0)$$. The maximum value of $$D_uf(x_0,y_0)$$ is $$∥∇f(x_0,y_0)∥.$$
3. If $$∇f(x_0,y_0)≠0,$$ then $$D_uf(x_0,y_0)$$ is minimized when u points in the opposite direction from $$∇f(x_0,y_0)$$. The minimum value of $$D_uf(x_0,y_0)$$ is $$−∥∇f(x_0,y_0)∥.$$
Figure $$\PageIndex{3}$$: The gradient indicates the maximum and minimum values of the directional derivative at a point.
Example $$\PageIndex{4}$$: Finding a Maximum Directional Derivative
Find the direction for which the directional derivative of $$f(x,y)=3x^2−4xy+2y^2$$ at $$(−2,3)$$ is a maximum. What is the maximum value?
Solution:
The maximum value of the directional derivative occurs when $$∇f$$ and the unit vector point in the same direction. Therefore, we start by calculating $$∇f(x,y$$):
$f_x(x,y)=6x−4y \; \text{and}\; f_y(x,y)=−4x+4y, \; \text{so}$
$∇f(x,y)=f_x(x,y)i+f_y(x,y)j=(6x−4y)i+(−4x+4y)j.$
Next, we evaluate the gradient at $$(−2,3)$$:
$∇f(−2,3)=(6(−2)−4(3))i+(−4(−2)+4(3))j=−24i+20j.$
We need to find a unit vector that points in the same direction as $$∇f(−2,3),$$ so the next step is to divide $$∇f(−2,3)$$ by its magnitude, which is $$\sqrt{(−24)^2+(20)^2}=\sqrt{976}=4\sqrt{61}−$$. Therefore,
$\dfrac{∇f(−2,3)}{∥∇f(−2,3)∥}=\dfrac{−24}{4\sqrt{61}}i+\dfrac{20}{4\sqrt{61}}j=−\dfrac{6\sqrt{61}}{61}i+\dfrac{5\sqrt{61}}{61}j.$
This is the unit vector that points in the same direction as $$∇f(−2,3).$$ To find the angle corresponding to this unit vector, we solve the equations
$\cos θ=\dfrac{−6\sqrt{61}}{61}\; \text{and}\; \sin θ=\dfrac{5\sqrt{61}}{61}$
for $$θ$$. Since cosine is negative and sine is positive, the angle must be in the second quadrant. Therefore, $$θ=π−\arcsin((5\sqrt{61})/61)≈2.45rad.$$
The maximum value of the directional derivative at $$(−2,3)$$ is $$∥∇f(−2,3)∥=4\sqrt{61}$$ (see the Figure $$\PageIndex{4}$$).
Figure $$\PageIndex{4}$$: The maximum value of the directional derivative at $$(−2,3)$$ is in the direction of the gradient.
Exercise $$\PageIndex{3}$$:
Find the direction for which the directional derivative of $$g(x,y)=4x−xy+2y^2$$ at $$(−2,3)$$ is a maximum. What is the maximum value?
Hint
Evaluate the gradient of $$g$$ at point $$(−2,3)$$.
The gradient of $$g$$ at $$(−2,3)$$ is $$∇g(−2,3)=i+14j$$. The unit vector that points in the same direction as $$∇g(−2,3)$$ is $$\dfrac{∇g(−2,3)}{∥∇g(−2,3)∥}=\dfrac{1}{\sqrt{197}}i+\dfrac{14}{\sqrt{197}}j=\dfrac{\sqrt{197}}{197}i+\dfrac{14\sqrt{197}}{197}j,$$ which gives an angle of $$θ=\arcsin ((14\sqrt{197})/197)≈1.499rad$$. The maximum value of the directional derivative is $$∥∇g(−2,3)∥=\sqrt{197}$$.
Figure $$\PageIndex{5}$$ shows a portion of the graph of the function $$f(x,y)=3+\sin x \sin y$$. Given a point $$(a,b)$$ in the domain of $$f$$, the maximum value of the gradient at that point is given by $$∥∇f(a,b)∥$$. This would equal the rate of greatest ascent if the surface represented a topographical map. If we went in the opposite direction, it would be the rate of greatest descent.
Figure $$\PageIndex{5}$$: A typical surface in $$R^3$$. Given a point on the surface, the directional derivative can be calculated using the gradient.
When using a topographical map, the steepest slope is always in the direction where the contour lines are closest together (Figure $$\PageIndex{6}$$). This is analogous to the contour map of a function, assuming the level curves are obtained for equally spaced values throughout the range of that function.
Figure $$\PageIndex{6}$$: Contour map for the function $$f(x,y)=x^2−y^2$$ using level values between $$−5$$ and $$5$$.
## Gradients and Level Curves
Recall that if a curve is defined parametrically by the function pair $$(x(t),y(t)),$$ then the vector $$x′(t)i+y′(t)j$$ is tangent to the curve for every value of $$t$$ in the domain. Now let’s assume $$z=f(x,y)$$ is a differentiable function of $$x$$ and $$y$$, and $$(x_0,y_0)$$ is in its domain. Let’s suppose further that $$x_0=x(t_0)$$ and $$y_0=y(t_0)$$ for some value of $$t$$, and consider the level curve $$f(x,y)=k$$. Define $$g(t)=f(x(t),y(t))$$ and calculate $$g′(t)$$ on the level curve. By the chain Rule,
$g′(t)=f_x(x(t),y(t))x′(t)+f_y(x(t),y(t))y′(t).$
But $$g′(t)=0$$ because $$g(t)=k$$ for all $$t$$. Therefore, on the one hand,
$f_x(x(t),y(t))x′(t)+f_y(x(t),y(t))y′(t)=0;$
on the other hand,
$f_x(x(t),y(t))x′(t)+f_y(x(t),y(t))y′(t)=∇f(x,y)⋅⟨x′(t),y′(t)⟩.$
Therefore,
$∇f(x,y)⋅⟨x′(t),y′(t)⟩=0.$
Thus, the dot product of these vectors is equal to zero, which implies they are orthogonal. However, the second vector is tangent to the level curve, which implies the gradient must be normal to the level curve, which gives rise to the following theorem.
Gradient Is Normal to the Level Curve
Suppose the function $$z=f(x,y)$$ has continuous first-order partial derivatives in an open disk centered at a point $$(x_0,y_0)$$. If $$∇f(x_0,y_0)≠0$$, then $$∇f(x_0,y_0)$$ is normal to the level curve of $$f$$ at $$(x_0,y_0).$$
We can use this theorem to find tangent and normal vectors to level curves of a function.
Example $$\PageIndex{5}$$: Finding Tangents to Level Curves
For the function $$f(x,y)=2x^2−3xy+8y^2+2x−4y+4,$$ find a tangent vector to the level curve at point $$(−2,1)$$. Graph the level curve corresponding to $$f(x,y)=18$$ and draw in $$∇f(−2,1)$$ and a tangent vector.
Solution:
First, we must calculate $$∇f(x,y):$$
$f_x(x,y)=4x−3y+2 \;\text{and}\; f_y=−3x+16y−4 \;\text{so}\; ∇f(x,y)=(4x−3y+2)i+(−3x+16y−4)j.$
Next, we evaluate $$∇f(x,y)$$ at $$(−2,1):$$
$∇f(−2,1)=(4(−2)−3(1)+2)i+(−3(−2)+16(1)−4)j=−9i+18j.$
This vector is orthogonal to the curve at point $$(−2,1)$$. We can obtain a tangent vector by reversing the components and multiplying either one by $$−1$$. Thus, for example, $$−18i−9j$$ is a tangent vector (see the following graph).
Figure $$\PageIndex{7}$$: Tangent and normal vectors to $$2x^2−3xy+8y^2+2x−4y+4=18$$ at point $$(−2,1)$$.
Exercise $$\PageIndex{4}$$:
For the function $$f(x,y)=x^2−2xy+5y^2+3x−2y+4$$, find the tangent to the level curve at point $$(1,1)$$. Draw the graph of the level curve corresponding to $$f(x,y)=8$$ and draw $$∇f(1,1)$$ and a tangent vector.
Hint
Calculate the gradient at point $$(1,1)$$.
$$∇f(x,y)=(2x−2y+3)i+(−2x+10y−2)j$$
$$∇f(1,1)=3i+6j$$
Tangent vector: $$6i−3j$$ or $$−6i+3j$$
## Three-Dimensional Gradients and Directional Derivatives
The definition of a gradient can be extended to functions of more than two variables.
Definition: Gradients in 3D
Let $$w=f(x,y,z)$$ be a function of three variables such that $$f_x,f_y,$$ and $$f_z$$ exist. The vector $$∇f(x,y,z)$$ is called the gradient of $$f$$ and is defined as
$∇f(x,y,z)=f_x(x,y,z)i+f_y(x,y,z)j+f_z(x,y,z)k.$
$$∇f(x,y,z)$$ can also be written as grad $$f(x,y,z).$$
Calculating the gradient of a function in three variables is very similar to calculating the gradient of a function in two variables. First, we calculate the partial derivatives $$f_x,f_y,$$ and $$f_z$$, and then we use Equation.
Example $$\PageIndex{6}$$: Finding Gradients in Three Dimensions
Find the gradient $$∇f(x,y,z)$$ of each of the following functions:
1. $$f(x,y)=5x^2−2xy+y^2−4yz+z^2+3xz$$
2. $$f(x,y,z)=e^{−2z}\sin 2x \cos 2y$$
Solution:
For both parts a. and b., we first calculate the partial derivatives $$f_x,f_y,$$ and $$f_z$$, then use Equation.
a.
\begin{align*} f_z(x,y,z)&=10x−2y+3z,f_y(x,y,z)=−2x+2y−4z \; \text{and} \; f_z(x,y,z)=3x−4y+2z, \text{so} \\ ∇f(x,y,z)&=f_x(x,y,z)i+f_y(x,y,z)j+f_z(x,y,z)k \\ &=(10x−2y+3z)i+(−2x+2y−4z)j+(−4x+3y+2z)k. \end{align*}
b.
\begin{align*} f_x(x,y,z)&=−2e^{−2z}\cos 2x \cos 2y, f_y(x,y,z)=−2e^{−2z} \sin 2x \sin 2y \; \text{and} \\ f_z(x,y,z)&=−2e^{−2z}\sin 2x \cos 2y, \; \text{so} \\ ∇f(x,y,z)&=f_x(x,y,z)i+f_y(x,y,z)j+f_z(x,y,z)k \\&=(2e^{−2z}\cos 2x \cos 2y)i+(−2e^{−2z})j+(−2e^{−2z}) \\ & =2e^{−2z}(\cos 2x \cos 2y i−\sin 2x \sin 2yj−\sin 2x \cos2yk). \end{align*}
Exercise $$\PageIndex{5}$$:
Find the gradient $$∇f(x,y,z)$$ of $$f(x,y,z)=\dfrac{x^2−3y^2+z^2}{2x+y−4z.}$$
$∇f(x,y,z)=\dfrac{2x^2+2xy+6y^2−8xz−2z^2}{(2x+y−4z)^2}i−\dfrac{x^2+12xy+3y^2−24yz+z^2}{(2x+y−4z)^2}j+\dfrac{4x^2−12y^2−4z^2+4xz+2yz}{(2x+y−4z)^2}k$
The directional derivative can also be generalized to functions of three variables. To determine a direction in three dimensions, a vector with three components is needed. This vector is a unit vector, and the components of the unit vector are called directional cosines. Given a three-dimensional unit vector $$u$$ in standard form (i.e., the initial point is at the origin), this vector forms three different angles with the positive $$x−,y−,$$ and $$z-$$axes. Let’s call these angles $$α,β,$$ and $$γ$$. Then the directional cosines are given by $$\cos α,\cos β,$$ and $$\cos γ$$. These are the components of the unit vector $$u$$; since $$u$$ is a unit vector, it is true that $$\cos^2 α+\cos^2 β+\cos^2 γ=1.$$
Definition: Directional Derivative of a Function of Three variables
Suppose $$w=f(x,y,z)$$ is a function of three variables with a domain of $$D$$. Let $$(x_0,y_0,z_0)∈D$$ and let $$u=\cos αi+\cos βj+\cos γk$$ be a unit vector. Then, the directional derivative of $$f$$ in the direction of $$u$$ is given by
$D_uf(x_0,y_0,z_0)=\lim_{t→0}\dfrac{f(x_0+t \cos α,y_0+t\cos β,z_0+t\cos γ)−f(x_0,y_0,z_0)}{t}$
provided the limit exists.
We can calculate the directional derivative of a function of three variables by using the gradient, leading to a formula that is analogous to Equation.
Directional Derivative of a Function of Three Variables
Let $$f(x,y,z)$$ be a differentiable function of three variables and let $$u=\cos αi+\cos βj+\cos γk$$ be a unit vector. Then, the directional derivative of $$f$$ in the direction of $$u$$ is given by
$D_uf(x,y,z)=∇f(x,y,z)⋅u=f_x(x,y,z)\cos α+f_y(x,y,z)\cos β+f_z(x,y,z)\cos γ.$
The three angles $$α,β,$$ and $$γ$$ determine the unit vector $$u$$. In practice, we can use an arbitrary (nonunit) vector, then divide by its magnitude to obtain a unit vector in the desired direction.
Example $$\PageIndex{7}$$: Finding a Directional Derivative in Three Dimensions
Calculate $$D_uf(1,−2,3)$$ in the direction of $$v=−i+2j+2k$$ for the function
$f(x,y,z)=5x^2−2xy+y^2−4yz+z^2+3xz.$
Solution:
First, we find the magnitude of $$v$$:
$‖v‖=\sqrt{(−1)^2+(2)^2}=3.$
Therefore, $$\dfrac{v}{‖v‖}=\dfrac{−i+2j+2k}{3}=−\dfrac{1}{3}i+\dfrac{2}{3}j+\dfrac{2}{3}k$$ is a unit vector in the direction of $$v$$, so $$\cos α=−\dfrac{1}{3},\cos β=\dfrac{2}{3},$$ and $$\cos γ=\dfrac{2}{3}$$. Next, we calculate the partial derivatives of f:
\begin{align*} f_x(x,y,z)&=10x−2y+3z \\ f_y(x,y,z)&=−2x+2y−4z \\ f_z(x,y,z)&=−4y+2z+3x, \end{align*}
then substitute them into Equation:
\begin{align*} D_uf(x,y,z)&=f_x(x,y,z)\cos α+f_y(x,y,z)\cos β+f_z(x,y,z)\cos γ \\ &=(10x−2y+3z)(−\dfrac{1}{3})+(−2x+2y−4z)(\dfrac{2}{3})+(−4y+2z+3x)(\dfrac{2}{3}) \\ &=−\dfrac{10x}{3}+\dfrac{2y}{3}−\dfrac{3z}{3}−\dfrac{4x}{3}+\dfrac{4y}{3}−\dfrac{8z}{3}−\dfrac{8y}{3}+\dfrac{4z}{3}+\dfrac{6x}{3} \\ &=−\dfrac{8x}{3}−\dfrac{2y}{3}−\dfrac{7z}{3}. \end{align*}
Last, to find $$D_uf(1,−2,3),$$ we substitute $$x=1,y=−2,$$ and $$z=3:$$
\begin{align*} D_uf(1,−2,3)&=−\dfrac{8(1)}{3}−\dfrac{2(−2)}{3}−\dfrac{7(3)}{3} \\ &=−\dfrac{8}{3}+\dfrac{4}{3}−\dfrac{21}{3} \\&=−\dfrac{25}{3}. \end{align*}
Exercise $$\PageIndex{6}$$:
Calculate $$D_uf(x,y,z)$$ and $$D_uf(0,−2,5)$$ in the direction of $$v=−3i+12j−4k$$ for the function $$f(x,y,z)=3x^2+xy−2y^2+4yz−z^2+2xz.$$
Hint
First, divide $$v$$ by its magnitude, calculate the partial derivatives of $$f$$, then use Equation.
$$D_uf(x,y,z)=−\dfrac{3}{13}(6x+y+2z)+\dfrac{12}{13}(x−4y+4z)−\dfrac{4}{13}(2x+4y−2z)$$
$$D_uf(0,−2,5)=\dfrac{384}{13}$$
## Key Concepts
• A directional derivative represents a rate of change of a function in any given direction.
• The gradient can be used in a formula to calculate the directional derivative.
• The gradient indicates the direction of greatest change of a function of more than one variable.
## Key Equations
• directional derivative (two dimensions)
$$D_uf(a,b)=\lim_{h→0}\dfrac{f(a+h\cos θ,b+h\sin θ)−f(a,b)}{h}$$
or
$$D_uf(x,y)=f_x(x,y)\cos θ+f_y(x,y)\sin θ$$
• gradient (two dimensions)
$$∇f(x,y)=f_x(x,y)i+f_y(x,y)j$$
• gradient (three dimensions)
$$∇f(x,y,z)=f_x(x,y,z)i+f_y(x,y,z)j+f_z(x,y,z)k$$
• directional derivative (three dimensions)
$$D_uf(x,y,z)=∇f(x,y,z)⋅u=f_x(x,y,z)\cos α+f_y(x,y,z)\cos β+f_x(x,y,z)\cos γ$$
## Glossary
directional derivative
the derivative of a function in the direction of a given unit vector
the gradient of the function $$f(x,y)$$ is defined to be $$∇f(x,y)=(∂f/∂x)i+(∂f/∂y)j,$$ which can be generalized to a function of any number of independent variables
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# Tutor profile: Kayla R.
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## Questions
### Subject:Pre-Algebra
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Question:
2a + a = 18. Solve for a
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Kayla R.
2a + a = 18. What is a? Combine like terms so that 2a + a = 3a 3a = 18. Then we will need to divide 3 from each side. 3/3 = 1, so that leaves the variable a by itself. Then we will need to divide 18 and 3, which equals to 6. a = 6.
### Subject:Basic Math
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Question:
I have 20 dollars to spend at the super market. I want to buy a gallon of milk for 5 dollars, a loaf of bread for 3 dollars, and a can of soup for 1 dollar. How much will my total be? Then, calculate how much change I will receive from the cashier after my purchase.
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If I have 20 dollars to spend, I will need to carefully plan out my purchases at the super market in order no to overspend. First, I will need to add up the total amount of the three items that I need to buy: gallon of milk = $5, loaf of bread =$3, and a can of soup = $1. 5 + 3 + 1 =$9. $9 dollars will be my total amount for the three items that I need to purchase. Next I will calculate how much change I will receive in return for the purchased items. I know I have$20, so I need to subtract the $9 total amount from the$20 I have to make this purchase. $20-$9 = $11. I have$11 remaining from my change.
### Subject:Algebra
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Evaluate 5x + 5y, when x = 3 and y = 4
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5x + 5y 5(3) + 5(4) 15 + 20 Answer: 35
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# Exploring 0.6875 as a Fraction: Examples, Solutions, and FAQs
SECTIONS
Because they represent a portion or division of a whole, fractions are a crucial component of mathematics. They are utilized in many facets of daily life, including percentage calculations and cookery instructions. In this article, we will examine the fractional representation of a particular decimal, 0.6875. We’ll give multiple illustrations, thorough solutions, and responses to frequently asked problems about this mathematical idea.
## Understanding 0.6875 as a Decimal
Let’s first examine what 0.6875 signifies as a decimal before converting it to a fraction. The decimal point is followed by the figures 6, 8, 7, 5, and the phrase “six hundred eighty-seven and five thousandths” to represent the number 0.6875. You must express this decimal in a way that demonstrates its fractional equivalent in order to convert it to a fraction.
## Expressing 0.6875 as a Fraction
You can do the following operations to convert a decimal to a fraction:
Find the place value of the digit that comes after the decimal point. It is in the thousandths place in this instance.
With the same value as the decimal number, represent the decimal as a fraction. Depending on the place value of the last digit, the denominator will be a power of 10.
If required, simplify the fraction.
Let’s apply these procedures to find 0.6875 as a Fraction.
## Example 1: Converting 0.6875 as a Fraction
The final digit is in the thousandths place, so please indicate the place value.
Put the decimal in fraction form:
$$0.6875 = \frac{6875}{10000}$$
If you can, make the fraction simpler. Both the numerator and the denominator in this situation are divisible by 125:
$$\frac{6875} {125} =55$$
$$\frac{10000}{125} = 80$$
The simple fraction is as follows:
$$\frac{55}{80}$$
Now that we have successfully reduced 0.6875 as a Fraction, 55/80, we can go on.
## Example 2: Simplifying the Fraction
It’s crucial to remember that fractions can frequently be made even simpler. By determining the greatest common divisor (GCD) of the numerator and denominator, which is 5 in the previous example, 55/80 can be made simpler:
$$\frac{\frac{55}{5}} {\frac{80}{5}} = \frac{11}{16}$$
$$\frac{55}{80}$$ is therefore reduced to $$\frac{11}{16}$$. This is the fraction that represents $$0.6875$$ in its most basic form.
## Example 3: Converting to a Mixed Number
In some cases, you may prefer to express the fraction as a mixed number. To do this, you divide the numerator by the denominator to find the whole number part and the remainder becomes the numerator of the fractional part. Here’s how you can convert 11/16 to a mixed number:
• Divide the numerator (11) by the denominator (16):
• $$\frac{11} {16} = 0$$ with a remainder of 11
• Write the whole number part (0) and the remainder (11) as the numerator of the fractional part over the denominator:
• $$0\frac{11}{16}$$
So, $$\frac{11}{16}$$ as a mixed number is $$0 \frac{11}{16}$$.
So 0.6875 as a Fraction is equal to $$\frac{11}{16}$$..
Let’s address some common questions related to fractions and the conversion of decimals to fractions.
##### Q1: Why is it important to convert decimals to fractions?
A1: The ability to represent decimal values as rational numbers makes the conversion of decimals to fractions crucial. In mathematical computations, fractions are frequently simpler to use and offer a clearer grasp of the relationship between various quantities.
##### Q2: Are there decimals that cannot be exactly expressed as fractions?
A2: Certain decimals cannot be precisely translated into fractions. For instance, it is impossible to precisely represent (pi) as a fraction since its decimal expansion has an infinite and non-repeating pattern. Irrational numbers are what these decimals are known as.
##### Q3: How can I convert other decimals into fractions?
A3: Use the procedures outlined before in this article to turn other decimals into fractions. Determine the place value, represent the decimal as a fraction, and, if necessary, simplify it. For every decimal, follow these instructions.
Q4: Can fractions be simplified further than the examples shown here?
A4: Yes, fractions may frequently be made even simpler by determining their greatest common divisor (GCD). In mathematics, breaking down fractions into their simplest forms is an effective technique.
##### Q5: What is the significance of mixed numbers?
A5: Mixed numbers come in handy when you need to represent a value as both a whole number and a fraction in the actual world. For instance, you might need to measure ingredients in recipes using mixed amounts.
##### Q6: Are there shortcuts or tricks for converting decimals to fractions?
A6: The method described in this blog is applicable to all decimals, even though there are some shortcuts for simple decimals. Finding the place value and writing the decimal as a fraction over the right power of 10 are the critical steps.
## Conclusion
Percentage comprehension is a fundamental mathematical skill that is essential to everyday life. Remembering that percentages are fractions of 100 will help you solve problems involving percentages, such as “What percentage is 15 of 60?” You can quickly compute percentages and use this information in a variety of situations by using the percentage formula.
Stay tuned with our latest math posts
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# Tangent Ratio Problems
In these lessons, we will learn how to find angles and sides using the tangent ratio and how to solve word problems using the tangent ratio.
Hints on solving trigonometry problems:
• If no diagram is given, draw one yourself.
• Mark the right angles in the diagram.
• Show the sizes of the other angles and the lengths of any lines that are known
• Mark the angles or sides you have to calculate.
• Consider whether you need to create right triangles by drawing extra lines. For example, divide an isosceles triangle into two congruent right triangles.
• Decide whether you will need Pythagorean theorem, sine, cosine or tangent.
• Check that your answer is reasonable. The hypotenuse is the longest side in a right triangle.
How to use the tangent ratio to find missing sides or angles?
Example:
Calculate the length of the side x, given that tan θ = 0.4
Solution:
Solving Problems with the Tangent Ratio
Examples:
1. Find the opposite side given the adjacent side of a right triangle.
2. Find the adjacent side given the opposite side of a right triangle.
How to apply the tangent ratio?
A trigonometric ratio is a ratio of the lengths of two sides in a right triangle.
Let triangle ABC be a right triangle with acute angle A. The tangent of angle A is defined as
Tan A = (leg opposite angle A)/(leg adjacent to angle A)
Find missing sides and angle of right triangles.
How to find the opposite side or adjacent side using the tangent ratio?
This video shows you how to use the Tangent Ratio to find the unknown side of a right angle triangle.
What is the tangent ratio and how to use it to find angles and sides in right triangles?
How to use the Tangent Ratio to solve Word Problems?
How to calculate the height of a flag pole using tangent?
The measure of corresponding angles of similar triangle are always equal, so are the ratios of the corresponding sides.
How to calculate the height of a tree using tangent?
How to use the tangent ratio to find the angle of elevation?
Example:
A toy ladder is set against a 68mm tall stack of coins. If the base of the ladder is 22mm away from the base of the coins, what angle of elevation does the ladder form?
How to solve Trigonometric Word Problems?
1. From a point on the ground 96 m from a tree, the angle to the top of the tree is 38 degrees. What is the height of the tree?
2. The angle form the ground to the top of the Statue of Liberty is 7 degrees at a distance of 1220 ft from the building. Find the height of the statue.
3. A ladder is leaning up against a house. The bottom of the ladder is 3 ft away from the building and the ladder makes an angle of 75 degrees with the ground.
a) How high up the building does the ladder reach?
b) How long is the ladder?
4. A surveyor measured BC to be 125 ft. Find the distance AB across the lake.
5. After takeoff, an airplane maintained a flight angle of 8 degrees with the ground. Find the elevation after it covered after it covered a ground distance of 1200 m.
6. For the airplane in problem 5, find the distance it traveled in the air along the flight path while covering the ground distance of 1200 m.
7. A submarine maintains a diving angle of 22. How far has it travelled when it is directly under a point 350 m along the surface from the point where it submerged?
8. A surveyor wants to find the distance between peaks A and B. He finds point C, 288 ft from peak A, so that ACB is a right angle. The measure BAC is 89. Find the distance AB.
9. Sears Tower is 1454 ft tall. Suppose point A is 1000 ft from the base of the tower. What is the tangent of the angle at A formed by the ground and the line of vision to the top of the tower?
10. Diving at a constant angle A, a submarine descends 102 m while travelling 300 m. Find the degree measure of A.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# Finding Outliers in a Data Set Online Quiz
Following quiz provides Multiple Choice Questions (MCQs) related to Finding Outliers in a Data Set. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.
Q 1 - Find the outlier(s) in the given data set below.
68, 6, 74, 70, 82
### Explanation
Step 1:
The data that is different from other numbers in set is 6.
Step 2:
So the outlier for this data set is 6.
Q 2 - Find the outlier(s) in the given data set below.
28, 26, 29, 30, 81, 32, 37
### Explanation
Step 1:
The data that is different from other numbers in the given set is 81.
Step 2:
So the outlier for this data set is 81.
Q 3 - Find the outlier(s) in the given data set below.
2, 5, 31, 29, 40, 44, 39
### Explanation
Step 1:
The data that is different from other numbers in the given set are 2 and 5.
Step 2:
So the outliers for this data set are 2 and 5.
Q 4 - Find the outlier(s) in the given data set below.
15, 11, 9, 15, 63
### Explanation
Step 1:
The data that is different from other numbers in the given set is 63.
Step 2:
So the outlier for this data set is 63.
Q 5 - Find the outlier(s) in the given data set below.
89, 85, 88, 8, 56, 75, 62, 65
### Explanation
Step 1:
The data that is different from other numbers in the given set is 8.
Step 2:
So the outlier for this data set is 8.
Q 6 - Find the outlier(s) in the given data set below.
42, 47, 48, 40, 37, 84, 49
### Explanation
Step 1:
The data that is different from other numbers in the given set is 84.
Step 2:
So the outlier for this data set is 84.
Q 7 - Find the outlier(s) in the given data set below.
16, 14, 3, 12, 15, 17, 22, 15, 52
### Explanation
Step 1:
The data that is different from other numbers in the given set is 52.
Step 2:
So the outlier for this data set is 52.
Q 8 - Find the outlier(s) in the given data set below.
30, 29, 28, 26, 1, 2, 42, 44
### Explanation
Step 1:
The data that is different from other numbers in the given set are 1, and 2.
Step 2:
So the outliers for this data set are 1, 2.
Q 9 - Find the outlier(s) in the given data set below.
27, 20, 1, 3, 30, 31, 33
### Explanation
Step 1:
The data that is different from other numbers in the given set are 1 and 3.
Step 2:
So the outliers for this data set are 1 and 3.
Q 10 - Find the outlier(s) in the given data set below.
21, 15, 18, 11, 13, 71
### Explanation
Step 1:
The data that is different from other numbers in the given set is 71.
Step 2:
So the outlier for this data set is 71.
finding_outliers_in_data_set.htm
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# Non - verbal reasoning (Series)
Non-Verbal reasoning appears in Bank exams, Infosys, MAT exams constantly. There are 5 Problem Figures (PF) will be given with 5 Answer Figures (AF). We need to determine the next figure in the series. There are certain rules which make solving these problems easy. So study the rules and solved examples.
How to answer these questions:
Step 1:
For all the series problems the following rules apply. If problem figures A and E are equal our answer is problem figure B. Similarly, the other rules as follows.
1. PF(A) = PF(E) then answer is PF(B)
2. PF(D) = PF(E) then answer is PF(C)
3. PF(A) = PF(C) = PF(E) then answer is PF(B)
4. PF(A) = PF(D), PF(B) = PF(E) then answer is PF(C)
5. PF(D) = inverse of PF(A) and PF(E) = inverse of PF(2) then answer is inverse of PF(C)
Step 2:
In general, the items in the box takes different positions in the subsequent figures. They may rotate certain degrees either clock wise or anti-clockwise. Look at the following diagram. In some problems new items add to the existing figures and some existing figures vanish.
Solved Examples
1.
In this problem If PF(A) = PF(E) then answer is PF(B). In the answer options AF(4) is same as PF(B) so option 4
2.
Here PF(C) and PF(E) are equal. So Answer figure should be PF(B). So correct option is c.
3.
The arrow is changing its positions clock wise 90o, 45o, 135o, 45o, ....next should be 180o. So option 3.
4.
Simple one. A new arrow and a new line are adding alternatively. In PF(E) a new line has added. So in the next figure a new arrow must be added. And total lines should be 6. Option 5
5.
Small hand is moving anticlock wise 90o, 45o, 90o, 45o,... and Big hand is moving clock wise 135o constantly. So in the next figure, small hand must move 90o anti clockwise, and big hand must move 135o. So option 4
6.
Here the symbol is changing positions anti clockwise by 45o and every time a new symbol is adding. The "C"s in the middle are rotating clock wise by 90o. So the next figure must be option 4
7.
This is a simple analogy. There is a relationship between 1 and 2, 3 and 4. the small figures in the first diagram are getting bigger and vice-versa. So Option 3
8.
All the three symbols in the dice are rotating clockwise. So option 3
Alternative method:
We know that if PF(A) = PF(D), PF(B) = PF(E) then answer is PF(C). So option 3
9.
A new symbols is appearing in the middle of the previous figure and the previous figure is getting bigger. So option 4 is the right option. 3 and 5 options are ruled out as the figures in the middle are appeared already.
10.
A dot and line are adding constantly to the figures in left and right sides alternatively. So option 3
11.
There appears to be no pattern on immediate look, but his problem can be solved by simple observation. Have a look at the diagram below..
The positions of two symbols are not changing in 2 consecutive figures. So option 5
12.
the arrow and small line inside the small square are rotating constantly anti clockwise and clockwise respectively by 90o, 45o, 90o, 45o,... and 45o, 90o, 45o, 90o . So next figure would be option 3.
13.
The line is rotating anti clock wise by 90o, 180o, 270o, 360o so next figure should be 90o from figure E and a new symbol must appear. So option 1 is the correct.
14.
Symbols X is rotating clockwise by 45o, 90o, 45o, 90o. So our options will be either 1 or 3 as in the next figure symbol X must move 45o. A new symbols is being added to X each time one at front and next time at back. So option 3 is right one.
15.
Simple. Observe PF(A) and PF(E) are equal. So next figure will be PF(B). So option 5
16.
the symbols are changing constantly in clockwise direction and a new symbol is being added.
The red rounded circle is a place whenever a symbol appear in that position must not appear in the next. And remaining positions are moving clockwise by 90o. A new symbol must come at the place shown by green arrow. So our option will be 1. Option 2 is ruled out as + symbol appeared earlier.
17.
Circle is moving diagonally and triangle is moving clockwise by 90o. So option 1 is correct one.
18.
Here you can easily observe that the lines are rotating 90o clockwise. also in PF(B) and PF(D), half line has added at the right most side and in figures PF(C) and PF(E) a new line has added. So in our answer half line has to be added and lines should rotate 90o. So answer option 2.
19.
Simple one. Figures A and B changed their symbols opposite them. C and D also did So. So option 1
20.
Symbols in A, B are same except Symbols at bottom. A new symbol is coming there. Similarly in C, D. So option 3. Option 2 is ruled out as C appeared earlier.
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Rectilinear Motion and Tangent Lines Katie Faith Laura Woodlee
# Rectilinear Motion and Tangent Lines Katie Faith Laura Woodlee
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### Text of Rectilinear Motion and Tangent Lines Katie Faith Laura Woodlee
Rectilinear Motion and
Tangent LinesKatie Faith
Laura Woodlee
Let’s Review some
Calculus!ARF
What is Rectilinear Motion?
• Definition: It is any motion along a straight line.
• Application to Calculus:– In the AP Calculus AB course, this method is
used to solve problems dealing with particle motion. Three equations are found and utilized to find the position, velocity, and acceleration of the particle at different points of time.
Rectilinear Motion (Continued)
• The position of the particle is found using the position function, which is the original function. This is usually written as x(t) or p(t) .It describes the particle’s location at a specific point in relation to time (t). If given velocity, this function is found by taking the integral of the velocity function and then plugging in a known point to find the constant (C).
Rectilinear Motion (Continued)
• The velocity of the particle is found by taking the derivative of the original function (position function). It can also be found by taking the integral of the acceleration function and then plugging in a known point to find the value of the constant (C). The function is written as v(t) . The velocity describes the movement of the particle, either left, right, or stopped. This direction is determined using a sign line.
STOP!!! SIGN LINE REVIEW!!!
• In this application of calculus, sign lines are used to show when the velocity of the particle is moving to the right (positively), moving to the left (negatively), or changing direction (the point when it switches from positive to negative).
• To make a sign line, first the zeroes of the function must be found. For example: v(t)= t²-3t + 2 v(t)= (t-1)(t-2)
• Then the sign line is set up: + - +The zeroes are written on the (t-1)Side (t-2)
1 2 a time line with the numbers is written on the bottom• This sign line shows that the function is increasing from negative infinityto 1, resting at 1, decreasing from 1 to 2, resting at 2, and increasing from 2 to infinity.
Dots represent the zeroes (this is where it is at
rest)Things to remember:
Negative (dotted) lines are always to the left of the zero (point) unless there is a negative in front of t
Positive (solid) lines are always to the right unless there is a negative in front of t
V(t)
Rectilinear Motion (Continued)
• The acceleration of the particle is found by taking the second derivative of the position function, or the first derivative of the velocity function. It is written as a(t). The acceleration of the particle describes the absolute value of distance that the particle travels during a certain interval of time. The zeroes of this function describe when the particle is at rest.
Example Problem 1
1) 1983 AB 2: A particle moves along the x-axis so that at time t its position is given by:
x(t)= t³ - 6t²+ 9t + 11a) What is the velocity of the particle at t=0?b) During what time intervals is the particle
moving to the left?c) What is the total distance traveled by the
particle from t=0 to t=5?
Step-by-Step Solution 1
a)To solve, you must first find the formula for velocity, which is done by taking the derivative of the position formula. Using chain rule, this would be:
v(t)= 3t² - 12t + 9To find the velocity when t=0, plug in 0 for t:
v(0)= 3(0)²- 12 (0) + 9v(0)=9
b)To find the time intervals that the particle is moving to the left, you must make a sign line of the velocity function. First find the zeroes of the velocity function:
v(t)= 3t² - 12t +9 v(t)= 3(t-3)(t-1)Then make the sign line: + - + 3 (t-3) (t-1) 1 3The sign line shows the particle is moving to the left from 1 to 3.
V(t)
Step-by-Step Solution 1
c)To find the total distance traveled, you will be looking at the position that the particle is at different times. The sign line, in the last part of the problem, showed that the particle was at rest at t=1, and changed direction at t=3. These are two important points that the position must be found at, along with the endpoints, which are t=0 and t=5.
First we must find the position of the particle at these critical points using the position formula: p(t)= t³ - 6t² + 9t + 11Then find the total distance traveled at each interval and add to find total distance:
p(0)=11p(1)=15 +p(3)=-43 +p(5)=143
698 units total
4 58 186
Example Problem 2
2) 1982 AB 1: A particle moves along the x-axis in such a way that its acceleration at time t for t > 0 is given by:
a(t)= 3/t²When t=1, the position of the particle is 6 and
the velocity is 2.a) Write an equation for the velocity, v(t), of the
particle for all t>0.b) Write an equation for the position, x(t), of the
particle for all t>0.c) Find the position of the particle when t=e.
Step-by-Step Solution 2
a) To find the equation for velocity, you must first take the integral of the acceleration equation and then plug in the known point v(1) = 2 to find the value for C:
a(t)= 3 t² v(t)= -6 + C t 2= -6 + C C= 8 1
Final Velocity Equation:
v(t)= -6 + 8 t
Step-by-Step Solution 2b) To find the equation for position, you must take the
integral of the velocity equation and then plug in the known point x(1)=6 to find the value for C:
v(t) = -6 + 8 tx(t) = -6ln t + 8t + C 6 = -6ln1 + 8(1) + C6 = -6(0) + 8 + C6 = 8 + C C = -2
c) To find the position when t =e, plug in e for t in the position formula:
x(e) = -6lne + 8e –2 = -6 + 8e –2 = 8e – 8 or 8(e – 1)
Final Position Equation:
x(t) = -6lnt + 8t - 2
Try Me Problem
3) 1987 AB 1: A particle moves along the x-axis so that its acceleration at any time t is given by a(t)= 6t – 18. At time t=0 the velocity of the particle is v(0)=24, and at time t=1, its position is x(1)= 20.
a) Write an expression for the velocity v(t) of the particle at any time t.
b) For what values of t is the particle at rest?c) Write an expression for the position x(t) of
that particle at any time t.d) Find the total distance traveled by the particle
from t=1 to t=3.
Solution
a) v(t) = 3t² - 18t + 24b) t = 2, t = 4c) x(t) = t³ - 9t² + 24t + 4d) x(1) = 20 x(2) = 24 x(3) = 22
426 total units
Work (If Needed)
What are Tangent Lines?
• Definition: A line that corresponds to a function and only intersects it at one specific point.
• General Equation: y-y1=m(x-x1)
Tangent Lines (Continued)
• To solve tangent line problems, certain components must be given or solved for:– The original function– The derivative of that function– X-coordinate (x1)– Y-coordinate(y1)– The solution of the derivative by plugging in
the original x- and y- coordinatesOnce these components are solved for, plug
them into the general equation.
Example Problem 1
• 1978 AB 1: given the function f defined by f(x)= x³ - x² - 4x + 4. Write an equation of the line tangent to the graph of f at x= -1
Step-by-Step Solution 2
• To find the equation of the line tangent, you must first find the y- coordinate and the slope:
f(x)= x3-x2-4x+4f(-1) = (-1)³ - (-1)² - 4(-1) + 4f(-1) = -1 – 1 + 4 +4f(-1) = 6f’(x) = 3x² - 2x – 4f’(-1) = 3(-1)² - 2(-1) –4 f’(-1) = 3 + 2 –4 = 1
When the elements areplugged into the general
equation, the tangent line is:
y – 6 = 1 ( x + 1) or
y = x + 7
Example Problem 2
• 1985 AB 1: Let f be the function given by :
f(x)= 2x – 5 x² - 4
Write an equation for the line tangent to the graph of f at the point (0, f(0)).
Step-by-Step Solution 2• To find the equation of the line tangent, you must first find the
y- coordinate and the slope:
f(x)= 2x – 5 x² - 4f(0) = 2(0) –5 (0)² - 4f(0) = 5 4f’(x) = (x² -4)(2) – (2t – 5)(2t) (t² -4)²f’(0) = (0² - 4)(2) – (2(0) – 5)(2(0)) (0² - 4)²f’(0) = (-4)(2) 16
= -8
When the elements are plugged into the general
equation, the tangent line is:
y-(5/4) = -8x or
y = -8x + (5/4)
Try Me Problem
• 1986 AB 1: Let f be defined by: f(x) = 7 – 15x + 9x² - x³
Write an equation of the line tangent to the graph of f at x=2.
Solution
y- 5 = 9(x – 2) ory =9x –13 ory – 9x = -13
Work (If Needed)
What are Normal Lines?
• Normal lines are similar to tangent lines except their slope is perpendicular to the tangent lines’ slope.
• The general equation is the same:y-y1=m(x-x1)
except the slope is the opposite reciprocal
Normal Lines (Continued)
• To solve normal line problems, certain components must be given or solved for:– The original function– The derivative of that function
– X-coordinate (x1)
– Y-coordinate(y1)
– The solution of the derivative by plugging in the original x- and y- coordinates, then finding the opposite reciprocal of the solution
STOP!!! WHAT DOES PERPENDICULAR
MEAN?!?!?• A perpendicular line has the opposite
reciprocal slope of the line it is perpendicular to. This means it crosses the line at a 90º angle.
• Opposite reciprocal means that the slope is “flipped” and the sign is changed. For example:
1) m= -2 m=½ 2) m= 7 m= -
12 12 7
Example Problem
• Find the equation of the normal line to the curve:
f(x)=x2-2x+1 at the point x=3.
Step-by-Step Solution• To find the equation of the line normal to the graph, you must first
find the y- coordinate and the slope:
f(x)=x2-2x+1f(3)=(3)2-2(3)+1f(3)=4 f’(x)=2x-2f’(3)=2(3)-2 f’(3)=4 This is the slope of the tangent line and needs to be made
perpendicular for the line to be normal to the graph. m= -¼
When the elements are plugged into the general equation, the normal line is:
y-4=-¼(x-3)y-4=-¼x+(3/4)
y=-¼x+19/4
Try Me Problem
• Find the equation of the normal line to the graph of:
f(x)=(x2+3)½
at the point (−1, 2).
Solution
y-2=2(x+1) or2x-y=-4
Work (If Needed)
BibliographyBibliography• "Rectilinear motion." Connexions - Sharing Knowledge and Building "Rectilinear motion." Connexions - Sharing Knowledge and Building
Communities. N.p., n.d. Web. 1 Mar. 2011. Communities. N.p., n.d. Web. 1 Mar. 2011. <http://cnx.org/content/m13612/latest/>.. <http://cnx.org/content/m13612/latest/>..
• Picture. 1 Mar. 2011. Picture. 1 Mar. 2011. <http://www.analyzemath.com/calculus/Problems/tangent_2.gif>.<http://www.analyzemath.com/calculus/Problems/tangent_2.gif>.
• Picture. 2 Mar. 2011. Picture. 2 Mar. 2011. <http://www.capecodshops.com/id-524/ImgUpload/P_507686_1284420.JP<http://www.capecodshops.com/id-524/ImgUpload/P_507686_1284420.JPG>.G>.
• "Tangents and Normals." Series Math Study. N.p., n.d. Web. 6 Mar. 2011. "Tangents and Normals." Series Math Study. N.p., n.d. Web. 6 Mar. 2011. <http://www.seriesmathstudy.com/tangentexample.htm>.<http://www.seriesmathstudy.com/tangentexample.htm>.
•KAYE AUTREYKAYE AUTREY© Laura Woodlee and Katie Faith, 3/7/11© Laura Woodlee and Katie Faith, 3/7/11
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# Unit 1 Review Part 1 3 combined Handout KEY.notebook. September 26, 2013
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1 Math 10c Unit 1 Factors, Powers and Radicals Key Concepts 1.1 Determine the prime factors of a whole number Explain why the numbers 0 and 1 have no prime factors. 0 and 1 have no prime factors because there are no prime numbers small enough to fit in them. Prime numbers have only two factors- 1 and itself. ZERO is not a prime number because it has infinite factors (e.g. 0x1 = 0, 0x2 = 0, 0x3= 0, etc.). ONE is not prime because it has only 1 factor: 1 (e.g. 1x1 = 1) 1.3 Determine, using a variety of strategies, the greatest common factor or least common multiple of a set of whole numbers, and explain the process. Find the GCF (answer is a smaller # than your original #s) Find the LCM (answer is a bigger # than your original #s)
2 1.1 Determine, concretely, whether a given whole number is a perfect square, a perfect cube or neither. 1.2 Determine, using a variety of strategies, the square root of a perfect square, and explain the process. (Hint: USE PRIME FACTORIZATION) 1.3 Determine, using a variety of strategies, the cube root of a perfect cube, and explain the process. (Hint: USE PRIME FACTORIZATION) Solve problems that involve prime factors, greatest common factors, least common multiples, square roots or cube roots A farmer has a rectangular plot of land measuring 400 m by 640 m. He wants to subdivide this land into congruent square pieces. What is the side length of the largest possible square? What are the dimensions of the smallest square that could be tiled using 18 cm by 24 cm tile? Assume the tiles cannot be cut. The largest possible side length the square could have is 80 m =2x3 2 =2 3 x3 LCM = 2 3 x3 2 = 72 The smallest square that could be tiled would have a side length of 72 cm. 2
3 2.1 Sort a set of numbers into rational and irrational numbers. Identify as Relational or Irrational a) b) c) d) a) Rational b) Rational c) Rational d) Irrational (non-repeating and non-terminating) 2.2 Determine an approximate value of a given irrational number. (Estimate without calculator) a) b) c) = 4.1 =4.45 = Approximate the locations of irrational numbers on a number line, using a variety of strategies, and explain the reasoning. 2.4 Order a set of irrational numbers on a number line. a) b) c) d) e) f) 2.5 Express a radical as a mixed radical in simplest form (limited to numerical radicands). = a) b) c) 4 d) 3
4 2.6 Express a mixed radical as an entire radical (limited to numerical radicands) a) b) c) d) 2.7 Explain, using examples, the meaning of the index of a radical. In this example, 3 is the index. This means it is a cubed root. In other words, if this root was multiplied to itself three times, you would get the radicand. 2.8 Represent, using a graphic organizer, the relationship among the subsets of the real numbers (natural, whole, integer, rational, irrational). 4
5 3.1 Explain, using patterns, why a n = 1/a n, a 0 / 3.2 Explain, using patterns, why a 1/n = n, n> = 10 2 = 10 1 = 10 0 = 10 1 = 10 2 = /10 1/100 a 3 = a 3 a 2 = a 2 a 1 = a a 0 = 1 a -1 = 1/a 1 a -2 = 1/a 2 a -3 = 1/a Apply the exponent laws to expressions with rational and variable bases and integral and rational exponents, and explain the reasoning: (a m )(a n ) = a m+n x 8 2 x 8 4 x 8 7 = 12x 9 = y 4 = 8 2 = 64 a m a n = a m n, a = 8y 9 = x 4 z 6 (a m ) n = a mn 1. = 4 15 (ab) m = a m b m = c 4 d 4 = 3 4 c -4 d 8 = 81d 8 c 4 (a/b) n = a n /b n, b 0 1. = y 6 z 9 5
6 3.4 Express powers with rational exponents as radicals and vice versa. = or = 3.5 Solve a problem that involves exponent laws or radicals. A square has an area of 1134 m 2. Determine the perimeter of the square. Write the answer as a radical in simplest form. P = s+s+s+s or P = 4s A = s * s or A = s = s 2 s = simplify 1134 s = = 9 14 P = 4(9 14) = The perimeter is m. 3.6 Identify and correct errors in a simplification of an expression that involves powers did not flip the equation to the reciprocal and incorrectly put a negative base forgot index 3 6
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Ex 11.4
Chapter 11 Class 11 Conic Sections
Serial order wise
Get live Maths 1-on-1 Classs - Class 6 to 12
### Transcript
Ex 11.4, 15 Find the equation of the hyperbola satisfying the given conditions: Foci (0,±√10), passing through (2, 3) Since Foci is on the y−axis So required equation of hyperbola is 𝑦2/𝑎2 – 𝑥2/𝑏2 = 1 Now, Co-ordinates of foci = (0, ± c) & given foci = (0, ±√10) So, (0, ± c) = (0, ±√10) c = √𝟏𝟎 Also, c2 = a2 + b2 Putting value of c (√10)2 = a2 + b2 10 = a2 + b2 a2 + b2 = 10 b2 = 10 − a2 Since point (2,3) passes through the hyperbola, it satisfies the equation of hyperbola Putting values x = 2, y = 3 in equation 𝑦^2/𝑎^2 − 𝑥^2/𝑏^2 = 1 a 3^2/𝑎^2 − 2^2/𝑏^2 = 1 3^2/𝑎^2 − 2^2/𝑏^2 = 1 Also putting b2 = 10 − a2 3^2/𝑎^2 − 2^2/(10 − 𝑎^2 ) = 1 9/𝑎^2 − 4/(10 − 𝑎^2 ) = 1 ((10 − 𝑎^2 ) 9 − 4𝑎^2)/(𝑎^2 (10 − 𝑎^2)) = 1 (90 − 9𝑎^2 − 4𝑎^2)/(𝑎^2 (10 − 𝑎^2)) = 1 90 − 13a2 = a2(10 − a2) 90 − 13a2 = 10a2 − a4 a4 − 23a2 + 90 = 0 Let a2 = t So, our equation becomes t2 − 23t + 90 = 0 t2 − 18t − 5t + 90 = 0 t(t − 18) − 5(t − 18) = 0 (t − 5) (t − 18) = 0 So, t = 5 or t = 18 Now, a2 = t Thus, a2 = 5, 18 Putting t = 5 a2 = 5 Putting t = 18 a2 = 18 For a2 = 5 b2 = 10 – a2 b2 = 10 – 5 b2 = 5 For a2 = 18 b2 = 10 – a2 b2 = 10 – 18 b2 = –8 Since square cannot be negative a2 = 18 is not possible Thus, a2 = 5, b2 = 5 Required equation of hyperbola 𝑦^2/𝑎^2 − 𝑥^2/𝑏^2 =1 Putting values 𝒚^𝟐/𝟓 − 𝒙^𝟐/𝟓 =𝟏
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# What is the Measure of Angle M?
What is the Measure of Angle M?
Have you ever wondered what the measure of angle M is? Angle M is a common geometric concept that is used in a variety of applications, from carpentry to surveying. In this article, we will explore the different ways to measure angle M, as well as some of the properties of angles in general.
We will start by defining what an angle is and then discuss the different ways to measure angles. We will then explore some of the properties of angles, including the sum of the interior angles of a triangle and the exterior angles of a polygon. Finally, we will provide some examples of how angle M is used in real-world applications.
By the end of this article, you will have a solid understanding of angle M and how it is used in mathematics and other fields.
Angle Measure Description
Angle M 120 An acute angle
Angle N 90 A right angle
Angle O 60 An obtuse angle
Determining the Measure of Angle M
## What is an angle?
An angle is formed by two rays that share a common endpoint. The common endpoint is called the vertex of the angle, and the rays are called the sides of the angle. The measure of an angle is the amount of rotation between the two rays.
## How to measure an angle?
There are several ways to measure an angle. One common method is to use a protractor. A protractor is a tool that is shaped like a half-circle. It is divided into 180 degrees, with each degree marked by a small line. To measure an angle with a protractor, place the vertex of the angle at the center of the protractor and align one of the sides of the angle with the 0-degree line. The other side of the angle will intersect the protractor at a point that indicates the measure of the angle.
Another method of measuring an angle is to use a ruler and a compass. To measure an angle with a ruler and a compass, draw a line that represents one side of the angle. Then, use the compass to draw an arc that intersects the line at two points. The distance between the two points is equal to the measure of the angle.
## Types of angles
There are four main types of angles:
• Right angles measure 90 degrees.
• Obtuse angles measure more than 90 degrees but less than 180 degrees.
• Acute angles measure less than 90 degrees.
• Straight angles measure 180 degrees.
## Formulas for finding the measure of an angle
There are several formulas that can be used to find the measure of an angle. One common formula is the sine rule, which states that:
sin(A) / a = sin(B) / b = sin(C) / c
where A, B, and C are the angles of a triangle and a, b, and c are the lengths of the sides opposite those angles.
Another common formula is the cosine rule, which states that:
c^2 = a^2 + b^2 – 2ab cos(C)
where a, b, and c are the lengths of the sides of a triangle and C is the angle opposite side c.
Examples of Angles
## Right angles
A right angle is an angle that measures 90 degrees. Right angles are often indicated by a small box in the corner of the angle.
## Obtuse angles
An obtuse angle is an angle that measures more than 90 degrees but less than 180 degrees. Obtuse angles are often indicated by a small arc in the corner of the angle.
## Acute angles
An acute angle is an angle that measures less than 90 degrees. Acute angles are often indicated by a small dot in the corner of the angle.
## Straight angles
A straight angle is an angle that measures 180 degrees. Straight angles are often indicated by a small line in the corner of the angle.
## Applications of Angles
Angles are used in a variety of applications in geometry, trigonometry, physics, and everyday life.
### In geometry
Angles are used to measure the size of shapes. For example, the measure of an angle is used to find the area of a triangle. Angles are also used to define the properties of shapes. For example, a right angle is an angle that measures 90 degrees.
### In trigonometry
Angles are used to solve trigonometric problems. Trigonometry is the study of the relationships between angles and sides of triangles. Angles are used to find the sine, cosine, and tangent of an angle.
### In physics
Angles are used to describe the motion of objects. For example, the angle of elevation is the angle between the horizontal and the line of sight to an object. The angle of inclination is the angle between the horizontal and the direction of motion of an object.
### In everyday life
Angles are used in a variety of everyday activities. For example, angles are used to measure the slope of a hill, the direction of a road, and the angle of a door. Angles are also used in construction, carpentry, and other trades.
Angles are an important concept in mathematics and science. They are used to measure the size of shapes, to solve trigonometric problems, and to describe the motion of objects. Angles are also used in a variety of everyday activities.
What is the measure of angle m?
The measure of angle m is 50 degrees. This can be determined using the following steps:
1. Draw a line segment AB.
2. Construct an angle at point A with measure m.
3. Label the other point on the ray as C.
4. Use a protractor to measure the angle ACB. The measure of angle ACB will be equal to the measure of angle m.
How do I find the measure of angle m if I only know the measure of angle A?
If you only know the measure of angle A, you can find the measure of angle m using the following equation:
m = 180 – A
For example, if angle A is 30 degrees, then the measure of angle m will be 180 – 30 = 150 degrees.
What is the relationship between angle m and angle n?
Angle m and angle n are supplementary angles. This means that they add up to 180 degrees.
What is the measure of angle m if it is inscribed in a circle with radius r?
If angle m is inscribed in a circle with radius r, then the measure of angle m will be (r * r) / 2.
What is the measure of angle m if it is an exterior angle of a triangle?
If angle m is an exterior angle of a triangle, then the measure of angle m will be equal to the sum of the measures of the two opposite interior angles.
**What is the measure of angle m if it is a central angle of a circle?
If angle m is a central angle of a circle, then the measure of angle m will be equal to the measure of the arc that it intercepts.
the measure of angle M is 135 degrees. This can be determined using the law of sines, which states that the ratio of the sine of an angle to the length of its opposite side is equal for all angles in a triangle. In this case, the opposite side of angle M is 10 cm and the sine of angle M is 0.866. Therefore, the measure of angle M is 135 degrees.
#### Author Profile
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Hatch, established in 2011 by Marcus Greenwood, has evolved significantly over the years. Marcus, a seasoned developer, brought a rich background in developing both B2B and consumer software for a diverse range of organizations, including hedge funds and web agencies.
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Limit by epsilon-delta proof: Example 1
We have discussed extensively the meaning of the $\epsilon-\delta$ definition. In this post, we are going to learn some strategies to prove limits of functions by definition. The meat of the proof is finding a suitable $\delta$ for all possible $\epsilon$ values.
Recall that the definition states that the limit of $f(x) = L$ as $x$ approaches $a$, if for all $\epsilon > 0$, however small, there exists a $\delta > 0$ such that if $0 < | x - a| < \delta$, then $|f(x) - L| < \epsilon$.
Example 1: Let $f(x) = 3x + 5$. Prove that $\lim_{x \to 2} f(x) = 11$
If we are going to study definition limit above, and apply it to the given function, we have $\lim_{x \to 2} 3x + 5 = 11$, if for all $\epsilon > 0$, however small, there exists a $\delta > 0$ such that if $0 < |x - 2| < \delta$, then $|3x + 5 - 11| < \epsilon$. We want to find the value of $\delta$, in terms of $\epsilon$; therefore, we can manipulate one of the inequalities to the other’s form. In particular, we will manipulate $|3x + 5 - 11| < \epsilon$ to an expression such that the expression inside the absolute value sign will become $x - 2$.
Simplifying, we have $|3x - 6| = 3|x-2|< \epsilon$. This expression is equivalent to $3|x-2|< \epsilon$. Dividing both sides by $3$, we have $|x-2|< \frac{\epsilon}{3}$. Now, both inequalities have the same form.
That means that whatever value of $\epsilon$ is, we can find a $\delta = \frac{\epsilon}{3}$ satisfying the conditions above. To explain further, let us have a specific example.
Let us choose a small $\epsilon = 0.6$. From the definition, for $\epsilon = 0.6$, there exists a $\delta=\frac{\epsilon}{3} =\frac{0.6}{3}= 0.2$ such that if $0 < | x - 2| < 0.2$, then $|3x + 5 - 11| < 0.6$. The geometric interpretation of this statement is shown below.
Click the picture to view GeoGebra applet by Sylvain Bérubé
Now, let us intepret the definition. While reading these statements, look at the third diagram above:
1. Our $L=11$, we have chosen $\epsilon = 0.6$, and our computation gives us $\delta = \frac{\epsilon}{3} = \frac{0.6}{3} =0.2$.
2. The statement, let $\epsilon = 0.6$ means that we are creating an interval which is $(L - \epsilon, L + \epsilon) =(11-0.6,11+0.6)= (10.4,11.6)$. This can be seen in the y-axis.
3. The statement $0 < |x - 2| < 0.2$ means that we are creating an interval $(a - \delta, a + \delta)=(2 -0.2,2+0.2)= (1.8, 2.2)$. This can be seen in the x-axis
4. In layman’s term, for $\epsilon = 0.6$, there exists a $\delta=0.2$ such that if we take the value a particular $x$ between $1.8$ and $2.2$, we are sure that the corresponding $f(x)$ is between $10.4$ and $11.6$. Recall, however, that $0.6$ is a particular value. The definition states that we can make it as small as we want and still find a suitable $\delta$, however small, our $\epsilon$ is.
5. In fact a game can be developed where player A gives a particular $\epsilon$, and player B searches for a suitable $\delta$ satisfying the definition. No matter small a value player A assigns to $\epsilon$, we are sure that player B can always find a suitable $\delta$ satisfying the definition.
The general strategy in proving limits by $\epsilon-\delta$ definition is to manipulate the inequality $|f(x) - L| < \epsilon$ such that the expression $|f(x)-L|$ is simplified to $|x - a|$.
To explain the$\epsilon-\delta$ definition of limits further, I will give you three or four more examples in the near future.
15 thoughts on “Limit by epsilon-delta proof: Example 1”
1. You might want to fix up #1 in the interpretation of the definition. Your Limit and epsilon values are reversed.
2. The epsilon-delta limit definition represent a big learning step for most students, and sure is a real challenge. It is quite important to have a visual representation of the formal and not intuitive definition. An animation might help too, to see how the value of delta is influenced by the value of epsilon.
By the way, you have a small mistake in statement #1 above.
Sylvain Bérubé
Sherbrooke, Québec
3. @Nick and Silvain:
Thank you for pointing out the typo error. Error fixed. 🙂
4. Other than that, I love watching your page. I subscribe through RSS feed on Google Reader. This is one of my favorites you’ve posted.
5. Hey thanks for including my GeoGebra links on your page. I’ve been looking for new ideas to put together on GeoGebra as a hobby. Let me know if you have an idea you’d like to see and I’ll try to do it the best I can. Thanks so much.
6. Pingback: March 2011 Top Posts « Mathematics and Multimedia
7. Pingback: Math and Multimedia 2011 Quarter 1 Top Posts « Mathematics and Multimedia
• Thank you Funsho. 🙂
8. Can you please state an example showing that the limit does not exist at a point of discontinuity. I mean you can assume a point of discontinuity like a jump(break) or a point where value of function tends to infinity. That would really help.
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WHAT IS TRIGONOMETRY
WHAT IS TRIGONOMETRY
• Trigonometry is a branch of Mathematics that studies triangles and the relationship between their sides and the angles.
• The Word “ TRIGONOMETRY ” is derived from the Greek words , “ TRI ” (Meaning – Three), “GON ”(Meaning – Sides), “METRON” (Meaning – Measure).
TRIGONOMETRY = TRI + GON + METRON
INTRODUCTION TO TRIGONOMETRY
• Let us consider a right triangle . Here angle A is an acute angle and the position of the side BC with respect to angle A. We will call the side Opposite to angle A. AC is the Hypotenuse of the right triangle and the side AB is the part of angle A. So, we will call it as the side Adjacent to angle A
• The Hypotenuse is the longest side, and is always opposite the right angle.
• The Opposite side is the one opposite the angle we are interested in.
• The Adjacent side is the one next to the angle we are interested in.
TRIGONOMETRIC RATIOS
• This section presents the 3 basic Trigonometric Ratios Sine, Cosine and Tangent. The concept of similar Triangles and the Pythagorean Theorem can be used to develop the Trigonometry of right Triangles.
• The Trigonometric ratios of an acute angle in a right triangle ABC are defined as follows.
• Sin A = Side Opposite to angle A/Hypotenuse = BC/AC
• Cos A = Side Adjacent to angle A/ Hypotenuse = AB/AC
• Tan A = Side Opposite to angle A/Side Adjacent to angle A = BC/AB
• Since hypotenuse is the longest side in a right triangle, therefore, the value of sin A or cos A is always less than or equal to 1.
Relationship between Trigonometrical Ratios
• Cosec A = 1/Sin A = Hypotenuse/Side Opposite to angle A = AC/BC
• Sec A = 1/Cos A = Hypotenuse/Side Adjacent to angle A = AC/AB
• Cot A = 1/Tan A = Side Adjacent to angle A/Side Opposite to angle A = AB/BC
• Tan A = Sin A/Cos A = BC/AB
Trigonometric Identities
An equation involving trigonometric ratios of an angle is called a trigonometric identity if it is true for all values of the angle.
For any acute angle A, we have
1. Sin2A + cos2 A = 1
2. 1 + tan2A = sec2A
3. 1 + cot2A = cosec2A
NOTE: Trigonometry is the study of triangles. Using trigonometry we can solve many problems precisely where we may have only been able to approximate before. Trigonometry is used in everything from Engineering and Architecture to G.P.S
Question. In triangle ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Find the value of sin A and cos A.
Solution:
By using the Pythagoras theorem, we have
AC2 = AB + BC2
AC2 = (24)2 + (7)2
AC2 = 576 + 49
AC = √625 = 25 cm.
. sin A = BC/AC = 7/25 ,
cos A = AB/AC = 24/25
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# Velocity
Have you ever gone bowling? Statistics say you probably have, as more than 67 million people bowl each year here in America. If you are one of the 67 million, you have demonstrated as well as observed the concept of velocity. The action of throwing a bowling ball down a lane until it strikes the pins is a prime example of velocity because the ball is displaced, by the length of the lane, over a specific amount of time. This allows for the velocity of the ball to be determined and this value is often displayed on the screen along with your score. Therefore, let this article introduce the concept of velocity through definitions and examples and demonstrate how velocity and speed are the same, yet different.
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Figure 1; Bowling demonstrates the concept of velocity.
## Definition of Velocity
Velocity is a vector quantity used to describe an object's direction of motion and speed. It is often characterized by two types, average velocity, and instantaneous velocity. Average velocity is a vector quantity that relies on the final and initial position of an object.
Average velocity is an object's change in position with respect to time.
Instantaneous velocity is the velocity of an object at a specific moment in time.
Instantaneous velocity is the derivative of an object's change in position with respect to time.
## Formula for Velocity
The mathematical formula corresponding to the definition of average velocity is
$$v_{avg} = \frac{ \Delta x }{ \Delta t},$$
where $$\Delta x$$ is the displacement measured in meters $$( \mathrm{m} )$$ and $$\Delta t$$ is time measured in seconds $$( \mathrm{s} )$$. Note that if we take the derivative of this, the equation becomes $$v = \frac{ \mathrm{d}x }{ \mathrm{d}t }$$, where $$dx$$ is are infinitely small change in displacement and $$dt$$ is are infinitely small change in time. If we let time go to zero, this equation now gives us the mathematical formula corresponding to the definition of instantaneous velocity.
One can also calculate the average velocity over time using the initial and final values of velocity.
$$v_{\text{avg}}=\frac{v_o + v}{2}$$
where $$v_o$$ is initial velocity and $$v$$ is final velocity.
This equation is derivable from the kinematic equation for average distance as follows:
\begin{aligned}\Delta{x}=& \frac{v_o+v}{2}(t) \\ \frac{\Delta{x}}{t}= & \frac{v_o+v}{2} \\ v_{\text{avg}}= & \frac{v_o+v}{2}. \\ \end{aligned}
Note from the above that $$\frac{\Delta{x}}{t}$$ is the definition of average velocity.
## SI Unit of Velocity
Using the formula for velocity, its SI unit is calculated as follows:
$$v_{\text{avg}}= \frac{ \Delta x }{ \Delta t } = \frac{ \mathrm{m} }{ \mathrm{s} }$$
Therefore, the SI unit for velocity is $$\frac{ \mathrm{m} } { \mathrm{s} }$$.
## Calculating Average Velocity from an Acceleration-Time Graph
Another way to calculate average velocity over time is by means of an acceleration-time graph. When looking at an acceleration-time graph, you can determine the velocity of the object as the area under the acceleration curve is the change in velocity.
$$\text{Area}=\Delta{v}.$$
For example, the acceleration-time graph below represents the function, $$a(t)=0.5t+5$$ between $$0\,\mathrm{s}$$ to $$5\,\mathrm{s}$$. Using this, we can show that the change in velocity corresponds to the area under the curve.
The function indicates that as time increases by one second, the acceleration increases by $$0.5\,\mathrm{\frac{m}{s^2}}$$.
Figure 2: Determining average velocity from an acceleration-time graph.
Using this graph, we can find what the velocity will be after a specific amount of time by understanding that the change in velocity is the integral of acceleration
$$\Delta v=\int_{t_1}^{t_2}a(t)$$
where the integral of acceleration is the area under the curve and represents the change in velocity. Therefore,
\begin{aligned}\Delta v&=\int_{t_1}^{t_2}a(t) \\ \Delta v&=\int_{t_1=0}^{t_2=5}(0.5t +5)dt\\ \Delta v&=\frac{0.5t^2}{2}+5t \\ \Delta v&=\left(\frac{0.5(5)^2}{2}+5(5)\right)-\left(\frac{0.5(0)^2}{2}+5(0)\right)\\ \Delta v&=31.25\,\mathrm{\frac{m}{s}}.\\\end{aligned}
We can double-check this result by calculating the area of two different shapes (a triangle and a rectangle) as the first figure shows.
Start by calculating the area of the blue rectangle:
\begin{aligned}\text{Area}&=(\text{height})(\text{width})=hw \\\text{Area}&=(5)(5)\\ \text{Area}&=25.\\\end{aligned}
Now calculate the area of the green triangle:
\begin{aligned}\text{Area}&=\frac{1}{2}\left(\text{base}\right)\left(\text{height}\right)=\frac{1}{2}bh \\\text{Area}&=\frac{1}{2}\left(5\right)\left(2.5\right)\\ \text{Area}&=6.25.\\\end{aligned}
Now, adding these two together, we retrieve the result for the area under the curve:
\begin{aligned}\text{Area}_{\text{(curve)}}&=\text{Area}_{(\text{rec})}+ \text{Area}_{(\text{tri})} \\{Area}_{(\text{curve})}&= 25 + 6.25\\ \text{Area}_{(\text{curve})}&=31.25.\\\end{aligned}
The values match clearly, showing that in the acceleration-time graph, the area under the curve represents the change in velocity.
## Instantaneous Velocity from a Graph
We can calculate average velocity and instantaneous velocity by means of a position-time graph and a velocity-time graph. Let's familiarize ourselves with this technique, starting with the velocity-time graph below.
Figure 3: A velocity-time graph depicting constant velocity.
From this velocity-time graph, we can see that the velocity is constant with respect to time. Consequently, this tells us that the average velocity and the instantaneous velocity are equal because velocity is constant. However, this is not always the case.
Figure 4: A velocity-time graph depicting a scenario when velocity is not constant with respect to time.
When looking at this velocity-time graph, we can see that the velocity is not constant as it is different at different points. This tells us that average velocity and instantaneous velocity are not equal. However, to better understand instantaneous velocity, let's use the position-time graph below.
Figure 5: A position-time graph depicting instantaneous velocity as slope.
Suppose the blue line on the graph above represents a displacement function. Now using the two points seen on the graph, we could find the average velocity by using the equation, $$v_{avg}=\frac{\Delta{x}}{\Delta{t}}$$ which is simply the slope between those points. However, what will happen if we make one point a fixed point and vary the other, so it gradually approaches the fixed point? In simple terms, what will happen as we make the change in time smaller and smaller? Well, the answer is instantaneous velocity. If we vary one point, we will see that as the time approaches zero, the time interval becomes smaller and smaller. Therefore, the slope between these two points becomes closer and closer to the line tangent at the fixed point. Hence, the line tangent to the point is in fact instantaneous velocity.
## Difference Between Velocity and Speed
In everyday language, people often consider the words velocity and speed as synonyms. However, although both words refer to an object's change in position relative to time, we consider them as two distinctly different terms in physics. To distinguish one from the other, one must understand these 4 key points for each term.
Speed corresponds to how fast an object is moving, accounts for the entire distance an object covers within a given time period, is a scalar quantity, and cannot be zero.
Velocity corresponds to speed with direction, only accounts for an object's starting position and final position within a given time period, is a vector quantity, and can be zero. Their corresponding formulas are as follows:
\begin{aligned} \mathrm{Speed} &= \mathrm{\frac{Total\,Distance}{Time}} \\ \mathrm{Velocity} &= \mathrm{\frac{Displacement}{Time} = \frac{Final\,Position - Starting\,Position}{Time}}.\end{aligned}
Note that the direction of an object's velocity is determined by the object's direction of motion.
A simple way to think about speed and velocity is walking. Let's say you walk to the corner of your street at $$2\,\mathrm{\frac{m}{s}}$$. This only indicates speed because there is no direction. However, if you go north $$2\,\mathrm{\frac{m}{s}}$$ to the corner, then this represents velocity, since it includes direction.
### Instantaneous Velocity and Instantaneous Speed
When defining speed and velocity, it is also important to understand the concepts of instantaneous velocity and instantaneous speed. Instantaneous velocity and instantaneous speed both are defined as the speed of an object at a specific moment in time. However, the definition of instantaneous velocity also includes the object's direction. To better understand this, let us consider an example of a track runner. A track runner running a 1000 m race will have changes in their speed at specific moments in time throughout the entire race. These changes might be most noticeable toward the end of the race, the last 100 m, when runners begin to increase their speed to cross the finish line first. At this particular point, we could calculate the instantaneous speed and instantaneous velocity of the runner and these values would probably be higher than the runner's calculated speed and velocity over the entire 1000m race.
## Velocity Example Problems
When solving velocity problems, one must apply the equation for velocity. Therefore, since we have defined velocity and discussed its relation to speed, let us work through some examples to gain familiarity with using the equations. Note that before solving a problem, we must always remember these simple steps:
1. Read the problem and identify all variables given within the problem.
2. Determine what the problem is asking and what formulas are needed.
3. Apply the necessary formulas and solve the problem.
4. Draw a picture if necessary to help illustrate what is happening and provide a visual aid for yourself.
### Examples
Let's use our newfound knowledge of velocity to complete some examples involving average velocity and instantaneous velocity.
For travel to work, an individual drives $$4200\,\mathrm{m}$$ along a straight road every day. If this trip takes $$720\,\mathrm{s}$$ to complete, what is the average velocity of the car over this journey?
Figure 6: The act of driving can be used to calculate average velocity.
Based on the problem, we are given the following:
• displacement,
• time.
As a result, we can identify and use the equation,
$$v_{\text{avg}}=\frac{\Delta{x}}{\Delta{t}}$$ to solve this problem. Therefore, our calculations are:
\begin{aligned}v_{\text{avg}} &=\frac{\Delta{x}}{\Delta{t}} \\\\ v_{\text{avg}}&=\frac{4200\,\mathrm{m}}{720\,\mathrm{s}} \\\\ v_{\text{avg}}&=5.83\,\mathrm{\frac{m}{s}}. \\\end{aligned}
The average velocity of car is $$5.83\,\mathrm{\frac{m}{s}}.$$
Now, lets complete a slightly more difficult example that will involve some calculus.
An object undergoing linear motion is said to have a displacement function of $$x(t)=at^2 + b,$$ where $$a$$ is given to be $$3\,\mathrm{\frac{m}{s^2}}$$ and b is given to be $$4\,\mathrm{m}.$$ Calculate the magnitude of the instantaneous velocity when $$t= 5\,\mathrm{s}.$$
Based on the problem, we are given the following:
• displacement function,
• values of $$a$$ and $$b.$$
As a result, we can identify and use the equation,$$v=\frac{dx}{dt}$$, to solve this problem. We must take the derivative of the displacement function to find an equation for velocity in terms of time, giving us: \begin{align}v=\frac{dx}{dt}=6t\\\end{align} and now we can insert our value for time to calculate the instantaneous velocity.
\begin{align}v=\frac{dx}{dt}=6t=6(5\,\mathrm{s})=30\,\mathrm{\frac{m}{s}}.\\\end{align}
## Velocity - Key takeaways
• Average velocity is an object's change in position with respect to time.
• The mathematical formula for average velocity is $$v=\frac{\Delta{x}}{\Delta{t}}.$$
• Instantaneous velocity is the derivative of an object's change in position with respect to time.
• The mathematical formula for instantaneous velocity is $$v=\frac{dx}{dt}.$$
• The SI unit for velocity is $$\mathrm{\frac{m}{s}}.$$
• In the acceleration-time graph, the area under the curve represents the change in velocity.
• The line tangent to a point in a position-time graph is the instantaneous velocity at that point.
• Speed indicates how fast an object is moving, while velocity is a speed with direction.
• Instantaneous speed is the speed of an object at a specific moment in time while instantaneous velocity is instantaneous speed with direction.
## References
1. Figure 1 - White Bowling Pins and Red Bowling Ball from (https://www.pexels.com/photo/sport-alley-ball-game-4192/) licensed by (Public Domain)
2. Figure 6 - Cars ahead on road from (https://www.pexels.com/photo/cars-ahead-on-road-593172/) licensed by (Public Domain)
#### Flashcards in Velocity 12
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##### Frequently Asked Questions about Velocity
What is velocity?
Velocity is the change in an object's position over time.
What is an example of velocity?
An example is calculating the average velocity of an object whose displacement is given to be 1000m and the change in time is given to be 100s. Average velocity equals 10 meters per second.
What is the difference between speed and velocity?
Both refer to an object's change in position relative to time, however, speed is a scalar quantity only including magnitude and velocity is a vector quantity, including magnitude and direction.
What is the unit for velocity?
The SI unit for velocity is meters per second, m/s.
What is the formula for calculating velocity?
The formula is velocity equals displacement over time.
## Test your knowledge with multiple choice flashcards
Instantaneous velocity and instantaneous speed refer to an object at a specific moment in time.
It is possible to have a constant speed and a non-constant velocity.
It is possible to have a constant velocity but a non-constant speed.
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# You asked: How many possibilities are there to roll a 7 with two dice?
Contents
For each of the possible outcomes add the numbers on the two dice and count how many times this sum is 7. If you do so you will find that the sum is 7 for 6 of the possible outcomes. Thus the sum is a 7 in 6 of the 36 outcomes and hence the probability of rolling a 7 is 6/36 = 1/6.
## How many ways can you roll a 7 with 2 dice?
Explanation: When two dices are rolled, there are six possibilities of rolling a sum of 7 .
## How many ways can you hit 7 on dice?
Probabilities for the two dice
Total Number of combinations Probability
6 5 13.89%
7 6 16.67%
8 5 13.89%
9 4 11.11%
## How many possibilities are there with 2 dice?
When two dice are rolled, there are now 36 different and unique ways the dice can come up. This figure is arrived at by multiplying the number of ways the first die can come up (six) by the number of ways the second die can come up (six). 6 x 6 = 36.
## What is the probability that the total of the two dice will add up to 7 or 11?
What is the probability of rolling a sum of 7 or 11 with two dice? So, P(sum of 7 or 11) = 2/9.
## What is the probability of rolling a 7 with one dice?
Two (6-sided) dice roll probability table
Roll a… Probability
6 15/36 (41.667%)
7 21/36 (58.333%)
8 26/36 (72.222%)
9 30/36 (83.333%)
## What is the probability of number 7 coming on the dice?
For each of the possible outcomes add the numbers on the two diceand count how many times this sum is 7. If you do so you will find that the sum is 7 for 6 of the possible outcomes. Thus the sum is a 7 in 6 of the 36 outcomes and hence the probability of rolling a 7is 6/36 = 1/6.
## How many ways can you roll 7 in craps?
If you want to know the probability of rolling a 7, you just divide the number of ways you can get a 7 (there are six ways) by the total number of possibilities (36). Six divided by 36 is the same as 1/6, which is also the same at 16.67%.
## How often does a 7 roll in craps?
I know that a 7 will appear every 6 rolls, but with come-out 7-11s and craps, plus the possibility of shooters making multiple points, I think the average number of rolls may be higher than expected. Is there any mathematical reference material on this? The average number of rolls per shooter is 8.525510.
## How many dices are possible?
Note that there are 36 possibilities for (a,b). This total number of possibilities can be obtained from the multiplication principle: there are 6 possibilities for a, and for each outcome for a, there are 6 possibilities for b. So, the total number of joint outcomes (a,b) is 6 times 6 which is 36.
IT IS INTERESTING: How do you pick 3 numbers on the Irish lottery?
## How many combinations of 6 dice are there?
6 dice have 46656 (6x6x6x6x6x6) combinations.
## How many possibilities are there with 4 dice?
My initial reaction is to say that the answer is 64, since 4 dice can have 6 outcomes. In my train of thought, the first dice can have 6 outcomes, same as the second, third and fourth, thus 6∗6∗6∗6 would seem fitting.
## What is the probability of rolling a 7 or 11?
The probability of winning on the first roll is the probability of rolling 7 or 11, which is 1/6 plus 1/18, which equals to 2/9. Suppose we roll 4 on the first roll (the probability of rolling 4 is 1/12). On each successive roll the probability of rolling 7 is 1/6 and the probability of rolling 4 is 1/12.
## What is the probability of getting a sum of either 7 or 12 when rolling two dice?
The probability is 25% .
## When 3 dice are rolled what is the probability of getting a sum of 7?
Probability of a sum of 7: 15/216 = 7.0%
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# Math Expressions Grade 3 Unit 6 Lesson 6 Answer Key Side Lengths with Area and Perimeter
## Math Expressions Common Core Grade 3 Unit 6 Lesson 6 Answer Key Side Lengths with Area and Perimeter
Math Expressions Grade 3 Unit 6 Lesson 6 Homework
Write an equation for the area of each rectangle.
Side Lengths with Area and Perimeter Grade 3 Unit 6 Math Expressions Question 1.
(6×3) + (6×7) = 6×10
Explanation:
Are of the rectangle is the length × breadth. Hence the equation for the given rectangle is 6×10 = 60 or (6×3) + (6×7) = 60
Question 2.
(5×3) +(5×6) = 5×9
Explanation:
Are of the rectangle is the length × breadth. Hence the equation for the given rectangle is 5×9 = 45 or (5×3) + (5×6) = 45
Find the unknown side length in each diagram.
Side Lengths with Area and Perimeter Lesson 6 Grade 3 Unit 6 Question 3.
7 in
Math Expressions Grade 3 Unit 6 Lesson 6 Answer Key Question 4.
11 in
Answer Key Grade 3 Unit 6 Lesson 6 Math Expressions Question 5.
7ft
Unit 6 Lesson 6 Math Expressions Grade 3 Question 6.
9 ft
Solve.
Lesson 6 Answer Key Grade 3 Side Lengths with Area and Perimeter Unit 6 Question 7.
Sarah is lining a square tray with 1 inch square tiles. The side length of the tray is 9 inches. How many tiles does Sarah need?
81 tiles
Side Lengths with Area and Perimeter Grade 3 Unit 6 Math Expressions Question 8.
Mark is gluing a ribbon around the sides of a picture frame. The frame is 11 inches long and 7 inches wide. How much ribbon does Mark need?
36 inches.
Math Expressions Grade 3 Unit 6 Lesson 6 Remembering
Question 1.
649
Explanation:
The addition of both numbers is 649.
Question 2.
81
Explanation:
The subtraction of both numbers is 81.
Question 3.
315
Explanation:
The subtraction of both numbers is 315.
Question 4.
820
Explanation:
The addition of both numbers is 820.
Question 5.
Kenzie bakes 9 oatmeal cookies. Kenzie bakes 7 fewer cookies than Lisa. How many oatmeal cookies does Lisa bake?
Explanation:
Kenzie bakes 9 oatmeal cookies and also bakes 7 fewer cookies than Lisa therefore the oatmeal cookies does Lisa bake are (9+7) =16 oatmeal cookies.
Question 6.
Hayden reads 8 books over the summer. He reads 5 more books than Max. How many books does Max read?
8 books – 5 books =3 books.
Explanation:
Hayden reads 8 books and he also reads 5 more books than max. Therefore Max read 8-5=3 books.
Find the perimeter and area of each figure. Remember to include the correct units in your answers.
Question 7.
Perimeter = ___
Area = ____
Perimeter = 14 cm
Area = 7 sq cm
Explanation:
Question 8.
Perimeter = ___
Area = ____
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```Name ___________________________
9-1 Notes
IB Math SL
Lesson 9-1 Right triangle trigonometric ratios
Learning Goal: How can we use Trig Ratios in a right triangle to determine the length of a side or the angle in a right
triangle?
Today we will review triangle concepts you learned past math courses. We will be pair teaching and looping our
work!
#1 Loop Partner Work
Loop 1: Solving for sides of a right triangle
In your pairs the TEACHER will read the review notes and worked out examples. Then you will collaborate with
each other to solve the practice questions and check answers. Don’t forget to show all work.
A ratio is another way of saying a proportion or a fraction.
In Right triangles we often are interested in looking at the different sides of the triangle.
The Sides of a Right Triangle
The hypotenuse is the longest side of a right triangle, the side opposite
of the right angle.
The opposite side or leg is opposite the given non-right angle.
The adjacent side or leg is next to the given non-right angle.
If we know one side of a right triangle and one angle we can often find one of the other 2 sides through the
use of trigonometric ratios!
Worked out example:
Try one together:
Work on this set of problems individually. DO NOT WORK WITH YOUR PARTNER until both of you are done
with this section. Once both partners are completed, check answers with me then move on to the next loops.
Do not progress without your partner.
1. As shown in the diagram below, a building casts a 72 foot shadow on the ground when the angle of elevation of the
Sun is 40°.
How tall is the building, to the nearest foot?
2. Find all unknown angles and sides, to 3 significant figures of:
3. IB QUESTION: The following diagram shows a sloping roof. The surface ABCD is a rectangle. The angle ADE is 55°. The
vertical height, AF, of the roof is 3 m and the length DC is 7 m.
B
C
A
7m
3m
55°
E
F
D
Loop 2: Solving for angles in a right triangle
In your pairs the TEACHER will read the review notes and worked out examples. Then you will collaborate with
each other to solve the practice questions and check answers. Don’t forget to show all work.
Finding Angles Using SOH-CAH-TOA using inverse trig functions:
To find angles () using Sine, Cosine, and Tangent you must perform the inverse of the trig function
The inverse trig functions are represented as:
− − −
When using the inverse trig functions, you must be in degree mode in your calculator. ( 2ndtrig function)
Finding the angle of elevation or depression
Angle of elevation or depression is the angle between a horizontal line and the line joining a point of observation to
some object above or below the horizontal line.
Angle of elevation is always INSIDE the triangle, starting from the bottom!
Angle of depression is always OUTSIDE the triangle, starting from the top!
Worked out example:
Try together:
2. The diagram below shows the path a bird flies from the top of a 9.5 foot tall sunflower to a point on the ground
5 feet from the base of the sunflower.
To the nearest tenth of a degree what is the measure of angle x?
Work on this set of problems individually. DO NOT WORK WITH YOUR PARTNER until both of you are done
with this section. Once both partners are completed, check answers with me then move on to the next loops.
Do not progress without your partner.
1. A ladder is resting against the side of a building. The bottom of the ladder is 12 feet from the building, and the
ladder reaches 7 feet up the side of the building. Find the measure of the angle of elevation, to the nearest
tenth of a degree.
2. Solve for
3. An observer stands 100 m from the base of a building. The angle of elevation of the top of the building is 65°. How
tall is the building, to the nearest meter?
Putting it all together! Work on these together!
1. IB Question: The diagram shows a cuboid 22.5 cm by 40 cm by 30 cm.
H
G
E
F
40 cm
D
C
30 cm
A
(a)
B
22.5 cm
Calculate the length of [AC]. (HINT: Pythagorean theorem!)
(b)
Calculate the size of GÂC . (Hint: Visualize the triangle that these 3 vertices create!)
2. IB QUESTION: The diagram shows a water tower standing on horizontal ground. The height of the tower is
26.5 m.
xm
A
a) From a point A on the ground the angle of elevation to the top of the tower is 28°. Label Angle A as 28°in the
diagram.
b) Calculate, correct to the nearest metre, the distance x m.
```
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TEXT
# Solving a System of Nonlinear Equations Using Substitution
A system of nonlinear equations is a system of two or more equations in two or more variables containing at least one equation that is not linear. Recall that a linear equation can take the form $Ax+By+C=0$. Any equation that cannot be written in this form in nonlinear. The substitution method we used for linear systems is the same method we will use for nonlinear systems. We solve one equation for one variable and then substitute the result into the second equation to solve for another variable, and so on. There is, however, a variation in the possible outcomes.
## Intersection of a Parabola and a Line
There are three possible types of solutions for a system of nonlinear equations involving a parabola and a line.
### A General Note: Possible Types of Solutions for Points of Intersection of a Parabola and a Line
Figure 2 illustrates possible solution sets for a system of equations involving a parabola and a line.
• No solution. The line will never intersect the parabola.
• One solution. The line is tangent to the parabola and intersects the parabola at exactly one point.
• Two solutions. The line crosses on the inside of the parabola and intersects the parabola at two points.
Figure 2
### How To: Given a system of equations containing a line and a parabola, find the solution.
1. Solve the linear equation for one of the variables.
2. Substitute the expression obtained in step one into the parabola equation.
3. Solve for the remaining variable.
4. Check your solutions in both equations.
### Example 1: Solving a System of Nonlinear Equations Representing a Parabola and a Line
Solve the system of equations.
$\begin{array}{l}x-y=-1\hfill \\ y={x}^{2}+1\hfill \end{array}$
### Solution
Solve the first equation for $x$ and then substitute the resulting expression into the second equation.
$\begin{array}{llll}x-y=-1\hfill & \hfill & \hfill & \hfill \\ \text{ }x=y - 1\hfill & \hfill & \hfill & \text{Solve for }x.\hfill \\ \hfill & \hfill & \hfill & \hfill \\ \text{ }y={x}^{2}+1\hfill & \hfill & \hfill & \hfill \\ \text{ }y={\left(y - 1\right)}^{2}+1\hfill & \hfill & \hfill & \text{Substitute expression for }x.\hfill \end{array}$
Expand the equation and set it equal to zero.
$\begin{array}{l}y={\left(y - 1\right)}^{2}\hfill \\ \text{ }=\left({y}^{2}-2y+1\right)+1\hfill \\ \text{ }={y}^{2}-2y+2\hfill \\ 0={y}^{2}-3y+2\hfill \\ \text{ }=\left(y - 2\right)\left(y - 1\right)\hfill \end{array}$
Solving for $y$ gives $y=2$ and $y=1$. Next, substitute each value for $y$ into the first equation to solve for $x$. Always substitute the value into the linear equation to check for extraneous solutions.
$\begin{array}{l}\text{ }x-y=-1\hfill \\ x-\left(2\right)=-1\hfill \\ \text{ }x=1\hfill \\ \hfill \\ x-\left(1\right)=-1\hfill \\ \text{ }x=0\hfill \end{array}$
The solutions are $\left(1,2\right)$ and $\left(0,1\right),\text{}$ which can be verified by substituting these $\left(x,y\right)$ values into both of the original equations.
Figure 3
### Could we have substituted values for $y$ into the second equation to solve for $x$ in Example 1?
Yes, but because $x$ is squared in the second equation this could give us extraneous solutions for $x$. For $y=1$
$\begin{array}{l}y={x}^{2}+1\hfill \\ y={x}^{2}+1\hfill \\ {x}^{2}=0\hfill \\ x=\pm \sqrt{0}=0\hfill \end{array}$
This gives us the same value as in the solution. For $y=2$
$\begin{array}{l}y={x}^{2}+1\hfill \\ 2={x}^{2}+1\hfill \\ {x}^{2}=1\hfill \\ x=\pm \sqrt{1}=\pm 1\hfill \end{array}$
Notice that $-1$ is an extraneous solution.
### Try It 1
Solve the given system of equations by substitution.
$\begin{array}{l}3x-y=-2\hfill \\ 2{x}^{2}-y=0\hfill \end{array}$
Solution
## Intersection of a Circle and a Line
Just as with a parabola and a line, there are three possible outcomes when solving a system of equations representing a circle and a line.
### A General Note: Possible Types of Solutions for the Points of Intersection of a Circle and a Line
Figure 4 illustrates possible solution sets for a system of equations involving a circle and a line.
• No solution. The line does not intersect the circle.
• One solution. The line is tangent to the circle and intersects the circle at exactly one point.
• Two solutions. The line crosses the circle and intersects it at two points.
Figure 4
### How To: Given a system of equations containing a line and a circle, find the solution.
1. Solve the linear equation for one of the variables.
2. Substitute the expression obtained in step one into the equation for the circle.
3. Solve for the remaining variable.
4. Check your solutions in both equations.
### Example 2: Finding the Intersection of a Circle and a Line by Substitution
Find the intersection of the given circle and the given line by substitution.
$\begin{array}{l}{x}^{2}+{y}^{2}=5\hfill \\ y=3x - 5\hfill \end{array}$
### Solution
One of the equations has already been solved for $y$. We will substitute $y=3x - 5$ into the equation for the circle.
$\begin{array}{c}{x}^{2}+{\left(3x - 5\right)}^{2}=5\\ {x}^{2}+9{x}^{2}-30x+25=5\\ 10{x}^{2}-30x+20=0\end{array}$
Now, we factor and solve for $x$.
$\begin{array}{l}10\left({x}^{2}-3x+2\right)=0\hfill \\ 10\left(x - 2\right)\left(x - 1\right)=0\hfill \\ x=2\hfill \\ x=1\hfill \end{array}$
Substitute the two x-values into the original linear equation to solve for $y$.
$\begin{array}{l}y=3\left(2\right)-5\hfill \\ =1\hfill \\ y=3\left(1\right)-5\hfill \\ =-2\hfill \end{array}$
The line intersects the circle at $\left(2,1\right)$ and $\left(1,-2\right)$, which can be verified by substituting these $\left(x,y\right)$ values into both of the original equations.
Figure 5
### Try It 2
Solve the system of nonlinear equations.
$\begin{array}{l}{x}^{2}+{y}^{2}=10\hfill \\ x - 3y=-10\hfill \end{array}$
Solution
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# Proofs of Trigonometric Identities
## Convert to sine/cosine, use basic identities, and simplify sides of the equation.
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Practice Proofs of Trigonometric Identities
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Verifying a Trigonometric Identity
Verify that sin2xtan2x=1sin2x\begin{align*}\frac{\sin^2x}{\tan^2x}=1 - \sin^2x\end{align*} .
### Guidance
This concept continues where the previous one left off. Now that you are comfortable simplifying expressions, we will extend the idea to verifying entire identities. Here are a few helpful hints to verify an identity:
• Change everything into terms of sine and cosine.
• Use the identities when you can.
• Start with simplifying the left-hand side of the equation, then, once you get stuck, simplify the right-hand side. As long as the two sides end up with the same final expression, the identity is true.
#### Example A
Verify the identity cot2xcscx=cscxsinx\begin{align*}\frac{\cot^2x}{\csc x}=\csc x - \sin x\end{align*} .
Solution: Rather than have an equal sign between the two sides of the equation, we will draw a vertical line so that it is easier to see what we do to each side of the equation. Start with changing everything into sine and cosine.
cot2xcscxcos2xsin2x1sinxcos2xsinxcscxsinx1sinxsinx\begin{align*}\begin{array}{c|c c} \frac{\cot^2x}{\csc x} & \csc x - \sin x \\ \frac{\frac{\cos^2x}{\sin^2x}}{\frac{1}{\sin x}} & \frac{1}{\sin x}- \sin x \\ \frac{\cos^2x}{\sin x}\end{array}\end{align*}
Now, it looks like we are at an impasse with the left-hand side. Let’s combine the right-hand side by giving them same denominator.
1sinxsin2xsinx1sin2xsinxcos2xsinx\begin{align*}\begin{array}{|c} \frac{1}{\sin x}- \frac{\sin^2x}{\sin x} \\ \frac{1- \sin^2x}{\sin x} \\ \frac{\cos^2x}{\sin x}\end{array}\end{align*}
The two sides reduce to the same expression, so we can conclude this is a valid identity. In the last step, we used the Pythagorean Identity, sin2θ+cos2θ=1\begin{align*}\sin^2 \theta+\cos^2 \theta=1\end{align*} , and isolated the cos2x=1sin2x\begin{align*}\cos^2x=1- \sin^2x\end{align*} .
There are usually more than one way to verify a trig identity. When proving this identity in the first step, rather than changing the cotangent to cos2xsin2x\begin{align*}\frac{\cos^2x}{\sin^2x}\end{align*} , we could have also substituted the identity cot2x=csc2x1\begin{align*}\cot^2x=\csc^2x-1\end{align*} .
#### Example B
Verify the identity sinx1cosx=1+cosxsinx\begin{align*}\frac{\sin x}{1- \cos x}=\frac{1+ \cos x}{\sin x}\end{align*} .
Solution: Multiply the left-hand side of the equation by 1+cosx1+cosx\begin{align*}\frac{1+ \cos x}{1+ \cos x}\end{align*} .
sinx1cosx1+cosx1+cosxsinx1cosxsin(1+cosx)1cos2xsin(1+cosx)sin2x1+cosxsinx=1+cosxsinx====
The two sides are the same, so we are done.
#### Example C
Verify the identity sec(x)=secx\begin{align*}\sec(-x)=\sec x\end{align*} .
Solution: Change secant to cosine.
sec(x)=1cos(x)
From the Negative Angle Identities, we know that cos(x)=cosx\begin{align*}\cos (-x)=\cos x\end{align*} .
=1cosx=secx
Concept Problem Revisit Start by simplifying the left-hand side of the equation.
sin2xtan2x=sin2xsin2xcos2x=cos2x
.
Now simplify the right-hand side of the equation. By manipulating the Trigonometric Identity,
sin2x+cos2x=1\begin{align*}\sin^2x + \cos^2x = 1\end{align*} , we get cos2x=1sin2x\begin{align*}\cos^2x = 1 - \sin^2x\end{align*} .
cos2x=cos2x\begin{align*}\cos^2x =\cos^2x\end{align*} and the equation is verified.
### Guided Practice
Verify the following identities.
1. cosxsecx=1\begin{align*}\cos x \sec x=1\end{align*}
2. 2sec2x=1tan2x\begin{align*}2- \sec^2x=1- \tan^2x\end{align*}
3. cos(x)1+sin(x)=secx+tanx\begin{align*}\frac{\cos \left(-x\right)}{1+ \sin \left(-x\right)}=\sec x+ \tan x\end{align*}
1. Change secant to cosine.
cosxsecx=cos1cosx=1
2. Use the identity 1+tan2θ=sec2θ\begin{align*}1+ \tan^2 \theta=\sec^2 \theta\end{align*} .
2sec2x=2(1+tan2x)=21tan2x=1tan2x
3. Here, start with the Negative Angle Identities and multiply the top and bottom by 1+sinx1+sinx\begin{align*}\frac{1+ \sin x}{1+ \sin x}\end{align*} to make the denominator a monomial.
cos(x)1+sin(x)=cosx1sinx1+sinx1+sinx=cosx(1+sinx)1sin2x=cosx(1+sinx)cos2x=1+sinxcosx=1cosx+sinxcosx=secx+tanx
### Explore More
Verify the following identities.
1. cot(x)=cotx\begin{align*}\cot (-x)=- \cot x\end{align*}
2. csc(x)=cscx\begin{align*}\csc (-x)=- \csc x\end{align*}
3. tanxcscxcosx=1\begin{align*}\tan x \csc x \cos x=1\end{align*}
4. sinx+cosxcotx=cscx\begin{align*}\sin x+ \cos x \cot x=\csc x\end{align*}
5. csc(π2x)=secx\begin{align*}\csc \left(\frac{\pi}{2}-x\right)=\sec x\end{align*}
6. tan(π2x)=tanx\begin{align*}\tan \left(\frac{\pi}{2}-x\right)=\tan x\end{align*}
7. cscxsinxcotxtanx=1\begin{align*}\frac{\csc x}{\sin x}- \frac{\cot x}{\tan x}=1\end{align*}
8. tan2xtan2x+1=sin2x\begin{align*}\frac{\tan^2x}{\tan^2x+1}=\sin^2x\end{align*}
9. (sinxcosx)2+(sinx+cosx)2=2\begin{align*}(\sin x- \cos x)^2+(\sin x+ \cos x)^2=2\end{align*}
10. sinxsinxcos2x=sin3x\begin{align*}\sin x- \sin x \cos^2x= \sin^3x\end{align*}
11. tan2x+1+tanxsecx=1+sinxcos2x\begin{align*}\tan^2x+1+\tan x \sec x=\frac{1+ \sin x}{\cos^2x}\end{align*}
12. cos2x=cscxcosxtanx+cotx\begin{align*}\cos^2x=\frac{\csc x \cos x}{\tan x+ \cot x}\end{align*}
13. 11sinx11+sinx=2tanxsecx\begin{align*}\frac{1}{1- \sin x} - \frac{1}{1+ \sin x}=2 \tan x \sec x\end{align*}
14. csc4xcot4x=csc2x+cot2x\begin{align*}\csc^4x- \cot^4x=\csc^2x+\cot^2x\end{align*}
15. (sinxtanx)(cosxcotx)=(sinx1)(cosx1)\begin{align*}(\sin x - \tan x)(\cos x- \cot x)=(\sin x-1)(\cos x-1)\end{align*}
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# In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows: Number of letters 1-4 4-7 7-10 10-13 13-16 16-19 Number of surnames 6 30 40 16 4 4 Determine the number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size of modal in the surnames.
Class Interval Frequency Cumulative Frequency 1-4 6 6 4-7 30 36 7-10 40 76 10-13 16 92 13-16 4 96 16-19 4 100
We have n = 100
$$\Rightarrow \frac{n}{2} \ = \ 50$$
The cumulative frequency just greater than $$\frac{n}{2}$$ is 76 and the corresponding class is 7-10 .
Thus 7-10 is the median class such that
$$\frac{n}{2} \ = \ 50 , l = 7, cf = 36 , f = 40 and h = 3$$
Median $$= \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h$$
$$= \ 7 \ + \ ( \frac{50 \ - \ 36}{40}) \ × \ 3$$
$$= \ 7 \ + \ \frac{14}{40} \ × \ 3$$
$$= \ 7 \ + \ 1.05 \ = \ 8.05$$
Now, to calculate the mean,
Class Interval $$f_i$$ $$x_i$$ $$f_i x_i$$ 1-4 6 2.5 15 4-7 30 5.5 165 7-10 40 8.5 340 10-13 16 11.5 184 13-16 4 14.5 51 16-19 4 17.5 70 $$\sum f_i \ = \ 100$$ $$\sum f_i x_i \ = \ 825$$
$$\therefore$$ Mean = $$\overline{x} \ = \ \frac{ \sum f_i x_i}{ \sum f_i}$$
$$= \ \frac{832}{100}$$
Therefore, Mean = $$8.32$$
Calculation of mode :
The class 7-10 has the maximum frequency therefore, this is the modal class.
Here l = 7 , h = 3, f1 = 40 , f0 = 30 and f2 = 16
Mode $$= \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h$$
$$= \ 7 \ + \ \frac{40 \ - \ 30}{80 \ - \ 30 \ - \ 16} \ × \ 3$$
$$= \ 7 \ + \ \frac{10}{34} \ × \ 3$$
$$= \ 7 \ + \ 0.88 \ = \ 7.88$$
$$\therefore$$ Median = 8.05 , Mean = 8.32 and Mode = 7.88
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9370 minus 87 percent
This is where you will learn how to calculate nine thousand three hundred seventy minus eighty-seven percent (9370 minus 87 percent). We will first explain and illustrate with pictures so you get a complete understanding of what 9370 minus 87 percent means, and then we will give you the formula at the very end.
We start by showing you the image below of a dark blue box that contains 9370 of something.
9370
(100%)
87 percent means 87 per hundred, so for each hundred in 9370, you want to subtract 87. Thus, you divide 9370 by 100 and then multiply the quotient by 87 to find out how much to subtract. Here is the math to calculate how much we should subtract:
(9370 ÷ 100) × 87
= 8151.9
We made a pink square that we put on top of the image shown above to illustrate how much 87 percent is of the total 9370:
The dark blue not covered up by the pink is 9370 minus 87 percent. Thus, we simply subtract the 8151.9 from 9370 to get the answer:
9370 - 8151.9
= 1218.1
The explanation and illustrations above are the educational way of calculating 9370 minus 87 percent. You can also, of course, use formulas to calculate 9370 minus 87%.
Below we show you two formulas that you can use to calculate 9370 minus 87 percent and similar problems in the future.
Formula 1
Number - ((Number × Percent/100))
9370 - ((9370 × 87/100))
9370 - 8151.9
= 1218.1
Formula 2
Number × (1 - (Percent/100))
9370 × (1 - (87/100))
9370 × 0.13
= 1218.1
Number Minus Percent
Go here if you need to calculate any other number minus any other percent.
9380 minus 87 percent
Here is the next percent tutorial on our list that may be of interest.
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# 02. Matrices: examples
## 02. Examples
### Vectors in the plane
In the last video, we saw that a 2-by-2 matrix of numbers a, b; c, d defines a geometric transformation of the plane R 2: x, y mapsto a x plus b y, c x plus d y.
Remark:
Recall that R denotes the real number line. R 2 denotes the 2-dimensional plane of all column vectors of height 2 (i.e. x, y); R 3 denotes the 3-dimensional space of all column vectors of height 3 (i.e. x, y, z; and more generally, R n denotes the n-dimensional space of all column vectors of height n (i.e. x_1, x_2, dot dot dot, x_n).
Just as the coordinates x, y encode points in the plane, we should think of the matrix a, b; c, d as encoding a transformation of the plane. In this lecture, we will take a range of examples and see what the corresponding transformation looks like.
### Example 1
Let M be the 2-by-2 matrix 1, 0; 0, 0. If I apply M to the vector v equals x, y then I get M v equals 1, 0; 0, 0 times x, y, which equals x, 0. This takes v to the point on the x-axis with the same x-coordinate, so M represents a vertical projection map to the x-axis.
### Example 2
Consider the action of the 2-by-2 matrix 1, 0; 0, 1. This sends x, y to x, y; this transformation leaves everything as it was: it is called the identity transformation. We call this matrix the identity matrix, and we often write this matrix as I; it plays the role of the number 1 in the algebra of matrices.
### Useful lemma
Lemma:
Let M be the 2-by-2 matrix a, b; c, d, let e_1 be the vector 1, 0 and e_2 be the vector 0, 1. Then
• M e_1 is the first column of M, i.e. a, c.
• M e_2 is the second column of M, i.e. b, d.
We'll call e_1, e_2 basis vectors, which basically means that any other vector can be written as a combination of e_1 and e_2 in a unique way. More on this in MATH220.
We'll just check it for M e_1: M e_1 equals a, b; c, d times 1, 0, which equals a plus 0, c plus 0, which equals a, c The calculation for M e_2 is similar.
### Example 3
Take M to be the 2-by-2 matrix 0, 1; 1, 0.
• Where does e_1 go? It goes to the first column of M, which is e_2.
• Where does e_2 go? It goes to the second column of M, which is e_1.
So e_1 and e_2 get switched. This corresponds to a reflection in the line y = x:
Let's check that the line y = x is indeed fixed by the action of M. The vectors x, x (and only these ones) lie on this line, so let's compute: M times x, x equals x, x which indeed tells us that the points on the line y = x are fixed.
### Example 4
Take M to be the 2-by-2 matrix 0, minus 1; 1, 0.
• Where does e_1 go? It goes to the first column of M, which is e_2.
• Where does e_2 go? It goes to the second column of M, which is minus e_1.
We see that this looks like a 90 degree (pi over 2 radian) rotation. This makes sense, because the matrix cos theta, minus sine theta, sine theta, cos theta for a rotation by an angle theta specialises to M when theta equals pi over 2, because cos of pi over 2 equals 0 and sine of pi over 2 equals 1.
### Example 5
Take M to be the 2-by-2 matrix 1, 1; 0, 1. We have
• e_1 maps to e_1,
• e_2 maps to 1, 1.
So e_1 is fixed, but e_2 is slanted over in the x-direction. In fact, the whole y-axis gets slanted in the x-direction, for example if we compute M times 0, 2 we get 2, 2
### Example 6
As one final example, we'll take M to be the 2-by-2 matrix minus 3, 16; minus 1, 5. What on earth does this correspond to? I claim that it corresponds to a shear in a different direction. How can we find the fixed direction?
If v equals x, y points in the direction fixed by M then v = M v (that's what it means to be fixed). Therefore x, y equals minus 3, 16; minus 1, 5 times x, y
In other words, the first entries of v and M v must coincide, and so must the second entries. This gives us a pair of linear simultaneous equations: x equals minus 3 times x plus 16 y; y equals minus x plus 5 y. These are both equivalent to y equals x over 4, so the line y equals x over 4 is fixed.
Remark:
Not all matrices have fixed directions, but if they do then this method will find it.
### Outlook
In the next video, we will take a look at bigger matrices and higher-dimensional spaces.
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# Make the internet a better place to learn
## How do I solve 25x^2 - 20x = 11 by completing the square?
AJ Speller
9.915068493150685 years ago
Completing the square is method of solving a quadratic equation that involves finding a value to add to both the left and right side of the equation. This value has the extra benefit of making one side of the equation a perfect trinomial which makes the function easier to identify and/or graph.
Let's begin by factoring $25$ from the left side of the equation.
$25 \left({x}^{2} - \frac{20}{25} x\right) = 11$
Take the coefficient of the $x$ term and divide it by $2$ and square it
${\left(\frac{- \frac{20}{25}}{2}\right)}^{2} = {\left(- \frac{20}{25} \cdot \frac{1}{2}\right)}^{2} = {\left(- \frac{10}{25}\right)}^{2} = \frac{100}{625}$
$\frac{100}{625}$ This the number you add to the left side
$25 \left(\frac{100}{625}\right)$ Is added to the right side because we initially factored out 25 from the left side.
We added these values but the equation remains balanced because they are added to both sides of the equation.
$25 \left({x}^{2} - \frac{20}{25} x + \frac{100}{625}\right) = 11 + 25 \left(\frac{100}{625}\right)$
$25 \left({x}^{2} - \frac{20}{25} x + \frac{100}{625}\right) = 11 + \frac{100}{25}$
$25 \left({x}^{2} - \frac{20}{25} x + \frac{100}{625}\right) = 11 + 4$
$25 \left({x}^{2} - \frac{20}{25} x + \frac{100}{625}\right) = 15$
$\left({x}^{2} - \frac{20}{25} x + \frac{100}{625}\right) = \frac{15}{25}$
${\left(x - \frac{10}{25}\right)}^{2} = \frac{15}{25}$
$\sqrt{{\left(x - \frac{10}{25}\right)}^{2}} = \sqrt{\frac{15}{25}}$
$\left(x - \frac{10}{25}\right) = \sqrt{\frac{15}{25}}$
$x = \sqrt{\frac{15}{25}} + \frac{10}{25}$
$x = \frac{\sqrt{15}}{5} + \frac{10}{25}$
$x = \frac{\sqrt{15}}{5} + \frac{2}{5}$
$x = \frac{\sqrt{15} + 2}{5}$
## How do you determine if -10,20,-40,80 is an arithmetic or geometric sequence?
Don't Memorise
9.126027397260273 years ago
The sequence is a geometric sequence.
#### Explanation:
• In an Arithmetic sequence there is a common difference $d$ between any two consecutive terms
• In a Geometric sequence there is a common ratio $r$ for any two consecutive terms
The sequence given is :
$- 10 , 20 , - 40 , 80$
1) Checking if the sequence is an arithmetic sequence:
color(blue)(d _1= a_2 - a_1) = 20 - (-10) = color(blue)(30
d_2 = a_3 - a_2 = -40 - (20) =color(blue)( -60
As observed ${d}_{1} \ne {d}_{2}$ , so it is not an arithmetic sequence.
2) Checking if the sequence is a geometric sequence:
color(blue)(r _1= a_2/ a_1) = 20 / -10 = color(blue)(-2
color(blue)(r _2 = a_3/ a_2) = (-40 )/ 20 = color(blue)(-2
color(blue)(r _3 = a_4/ a_3) = (80 )/-40 = color(blue)(-2
Since ${r}_{1} = {r}_{2} = r 3$ it forms a geometric sequence.
So the sequence is a geometric sequence.
## What is a piecewise continuous function?
George C.
8.991780821917809 years ago
A piecewise continuous function is a function that is continuous except at a finite number of points in its domain.
#### Explanation:
Note that the points of discontinuity of a piecewise continuous function do not have to be removable discontinuities. That is we do not require that the function can be made continuous by redefining it at those points. It is sufficient that if we exclude those points from the domain, then the function is continuous on the restricted domain.
For example, consider the function:
$s \left(x\right) = \left\{\begin{matrix}- 1 & \text{if x < 0" \\ 0 & "if x = 0" \\ 1 & "if x > 0}\end{matrix}\right.$
graph{(y - x/abs(x))(x^2+y^2-0.001) = 0 [-5, 5, -2.5, 2.5]}
This is continuous for all $x \in \mathbb{R}$ except $x = 0$
The discontinuity at $x = 0$ is not removable. We cannot redefine $s \left(x\right)$ at that point and get a continuous function.
At $x = 0$ the graph of the function 'jumps'. More formally, in the language of limits we find:
${\lim}_{x \to 0 +} s \left(x\right) = 1$
${\lim}_{x \to 0 -} s \left(x\right) = - 1$
So the left limit and right limit disagree with one another and with the value of the function at $x = 0$.
If we exclude the finite set of discontinuities from the domain, then the function restricted to this new domain will be continuous.
In our example, the definition of $s \left(x\right)$ as a function from $\left(- \infty , 0\right) \cup \left(0 , \infty\right) \to \mathbb{R}$ is continuous.
If we graph $s \left(x\right)$ restricted to this domain, it still looks like it is discontinuous at $0$, but $0$ is not part of the domain, so the 'jump' there is irrelevant. At any point, arbitrarily close to $0$, we can choose a little open interval around it in which the function is (constant and therefore) continuous.
Slightly confusingly, the function $\tan \left(x\right)$ is considered continuous - rather than piecewise continuous, because the asymptotes at $x = \frac{\pi}{2} + n \pi$ are excluded from the domain.
graph{tan(x) [-10.06, 9.94, -4.46, 5.54]}
Meanwhile, the sawtooth function $f \left(x\right) = x - \left\lfloor x \right\rfloor$ is not considered piecewise continuous as a function from $\mathbb{R}$ to $\mathbb{R}$, but is piecewise continuous on any finite open interval.
graph{3/5(abs(sin(x * pi/2))-abs(cos(x * pi/2))-abs(sin(x * pi/2)^3)/6+abs(cos(x * pi/2)^3)/6) * tan(x * pi/2)/abs(tan(x * pi/2))+1/2 [-2.56, 2.44, -0.71, 1.79]}
## What is the formula for time from a changing velocity?
Costas C.
8.6 years ago
$t = \frac{u - {u}_{0}}{a}$
$s = {u}_{0} \cdot t + \frac{1}{2} a {t}^{2}$ (Need to solve quadratic)
#### Explanation:
Via changing velocity I pressume you mean an object that accelerates or decelerates.
If acceleration is constant
If you have initial and final speed:
a=(Δu)/(Δt)
$a = \frac{u - {u}_{0}}{t - {t}_{0}}$
Usually ${t}_{0} = 0$, so:
$t = \frac{u - {u}_{0}}{a}$
If the above method does not work because you are missing some values, you can use the equation below. The distance traveled $s$ can be given from:
$s = {u}_{0} \cdot t + \frac{1}{2} a {t}^{2}$
where ${u}_{0}$ is the initial speed
$t$ is the the time
$a$ is the acceleration (note this value is negative if the case is a deceleration)
Therefore, if you know the distance, initial speed and acceleration you can find the time by solving the quadratic equation that is formed. However, if acceleration if not given, you will need the final speed of the object $u$ and can use the formula:
$u = {u}_{0} + a t$
$u - {u}_{0} = a t$
$a = \frac{u - {u}_{0}}{t}$
and substitute to the distance equation, making it:
$s = {u}_{0} \cdot t + \frac{1}{2} \cdot \frac{u - {u}_{0}}{t} \cdot {t}^{2}$
$s = {u}_{0} \cdot t + \frac{1}{2} \cdot \left(u - {u}_{0}\right) \cdot t$
Factor $t$:
$s = t \cdot \left({u}_{0} + \frac{1}{2} \cdot \left(u - {u}_{0}\right)\right)$
$t = \frac{s}{{u}_{0} + \frac{1}{2} \cdot \left(u - {u}_{0}\right)}$
So you got 2 equations. Pick one of them, which will help you solve with the data you are given:
$s = {u}_{0} \cdot t + \frac{1}{2} a {t}^{2}$
$t = \frac{s}{{u}_{0} + \frac{1}{2} \cdot \left(u - {u}_{0}\right)}$
Below are two other cases where acceleration is not constant. FEEL FREE TO IGNORE THEM if acceleration in your case is constant, since you placed it in the Precalculus category and the below contain calculus.
If acceleration is a function of time $a = f \left(t\right)$
The definition of acceleration:
$a \left(t\right) = \frac{\mathrm{du}}{\mathrm{dt}}$
$a \left(t\right) \mathrm{dt} = \mathrm{du}$
${\int}_{0}^{t} a \left(t\right) \mathrm{dt} = {\int}_{{u}_{0}}^{u} \mathrm{du}$
${\int}_{0}^{t} a \left(t\right) \mathrm{dt} = u - {u}_{0}$
$u = {u}_{0} + {\int}_{0}^{t} a \left(t\right) \mathrm{dt}$
If you still don't have enough to solve, that means you have to go to distance. Just use the definition of speed and move on, as if I analyze it further it will only confuse you:
$u \left(t\right) = \frac{\mathrm{ds}}{\mathrm{dt}}$
The second part of this equation means integrading acceleration with respect to time. Doing that gives an equation with only $t$ as the unknown value.
If acceleration is a function of speed $a = f \left(u\right)$
The definition of acceleration:
$a \left(u\right) = \frac{\mathrm{du}}{\mathrm{dt}}$
$\mathrm{dt} = \frac{\mathrm{du}}{a \left(u\right)}$
${\int}_{0}^{t} \mathrm{dt} = {\int}_{{u}_{0}}^{u} \frac{\mathrm{du}}{a \left(u\right)}$
$t - 0 = {\int}_{{u}_{0}}^{u} \frac{\mathrm{du}}{a \left(u\right)}$
$t = {\int}_{{u}_{0}}^{u} \frac{\mathrm{du}}{a \left(u\right)}$
## How do you write the partial fraction decomposition of the rational expression (5x - 1) / ((x - 2)(x + 1))?
Zack M.
8.6 years ago
$\frac{5 x - 1}{\left(x - 2\right) \left(x + 1\right)} = \frac{3}{x - 2} + \frac{2}{x + 1}$
#### Explanation:
The partial fraction decomposition suggests that the function can be broken down into the sum of two other functions, or;
$\frac{5 x - 1}{\left(x - 2\right) \left(x + 1\right)} = \frac{A}{x - 2} + \frac{B}{x + 1}$
Where we need to solve for $A$ and $B$. We can cross multiply to combine the terms on the right hand side over a common denominator. We get;
$\frac{5 x - 1}{\left(x - 2\right) \left(x + 1\right)} = \frac{A \left(x + 1\right) + B \left(x - 2\right)}{\left(x - 2\right) \left(x + 1\right)}$
We can now cancel the denominator on each side, leaving;
$5 x - 1 = A \left(x + 1\right) + B \left(x - 2\right)$
Now we can solve for $A$ and $B$. We can make one of the terms cancel out by choosing the right value for $x$. Lets try x=~1.
5(~1)-1 = A(~1+1) + B(~1-2)
The $A$ term goes away since it is multiplied by zero, leaving;
~6 = ~3B
Solving for $B$;
$B = 2$
We can substitute $B$ and solve for $A$, but it would be easier to do the same trick that we used to solve for $B$. Let $x = 2$.
$5 \left(2\right) - 1 = A \left(2 + 1\right) + B \left(2 - 2\right)$
This time, the $B$ term goes away;
$9 = 3 A$
$A = 3$
Now that we have our values for $A$ and $B$ we can plug into our first function and get;
$\frac{5 x - 1}{\left(x - 2\right) \left(x + 1\right)} = \frac{3}{x - 2} + \frac{2}{x + 1}$
## What are the possible number of positive, negative, and complex zeros of f(x) = –3x^4 – 5x^3 – x^2 – 8x + 4?
George C.
8.594520547945205 years ago
Look at changes of signs to find this has $1$ positive zero, $1$ or $3$ negative zeros and $0$ or $2$ non-Real Complex zeros.
Then do some sums...
#### Explanation:
$f \left(x\right) = - 3 {x}^{4} - 5 {x}^{3} - {x}^{2} - 8 x + 4$
Since there is one change of sign, $f \left(x\right)$ has one positive zero.
$f \left(- x\right) = - 3 {x}^{4} + 5 {x}^{3} - {x}^{2} + 8 x + 4$
Since there are three changes of sign $f \left(x\right)$ has between $1$ and $3$ negative zeros.
Since $f \left(x\right)$ has Real coefficients, any non-Real Complex zeros will occur in conjugate pairs, so $f \left(x\right)$ has exactly $1$ or $3$ negative zeros counting multiplicity, and $0$ or $2$ non-Real Complex zeros.
$f ' \left(x\right) = - 12 {x}^{3} - 15 {x}^{2} - 2 x - 8$
Newton's method can be used to find approximate solutions.
Pick an initial approximation ${a}_{0}$.
Iterate using the formula:
${a}_{i + 1} = {a}_{i} - f \frac{{a}_{i}}{f ' \left({a}_{i}\right)}$
Putting this into a spreadsheet and starting with ${a}_{0} = 1$ and ${a}_{0} = - 2$, we find the following approximations within a few steps:
$x \approx 0.41998457522194$
$x \approx - 2.19460208831628$
We can then divide $f \left(x\right)$ by $\left(x - 0.42\right)$ and $\left(x + 2.195\right)$ to get an approximate quadratic $- 3 {x}^{2} + 0.325 x - 4.343$ as follows:
Notice the remainder $0.013$ of the second division. This indicates that the approximation is not too bad, but it is definitely an approximation.
Check the discriminant of the approximate quotient polynomial:
$- 3 {x}^{2} + 0.325 x - 4.343$
$\Delta = {b}^{2} - 4 a c = {0.325}^{2} - \left(4 \cdot - 3 \cdot - 4.343\right) = 0.105625 - 52.116 = - 52.010375$
Since this is negative, this quadratic has no Real zeros and we can be confident that our original quartic has exactly $2$ non-Real Complex zeros, $1$ positive zero and $1$ negative one.
## What is the standard form of the equation of a circle with centre and radius of the circle x^2 + y^2 - 4x + 8y - 80?
Alan P.
8.591780821917808 years ago
${\left(x - 2\right)}^{2} + {\left(y - \left(- 4\right)\right)}^{2} = {10}^{2}$
#### Explanation:
The general standard form for the equation of a circle is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$
for a circle with center $\left(a , b\right)$ and radius $r$
Given
$\textcolor{w h i t e}{\text{XXX")x^2+y^2-4x+8y-80 (=0)color(white)("XX}}$(note: I added the $= 0$ for the question to make sense).
We can transform this into the standard form by the following steps:
Move the $\textcolor{\mathmr{and} a n \ge}{\text{constant}}$ to the right side and group the $\textcolor{b l u e}{x}$ and $\textcolor{red}{y}$ terms separately on the left.
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{x}^{2} - 4 x} + \textcolor{red}{{y}^{2} + 8 y} = \textcolor{\mathmr{and} a n \ge}{80}$
Complete the square for each of the $\textcolor{b l u e}{x}$ and $\textcolor{red}{y}$ sub-expressions.
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{x}^{2} - 4 x + 4} + \textcolor{red}{{y}^{2} + 8 y + 16} = \textcolor{\mathmr{and} a n \ge}{80} \textcolor{b l u e}{+ 4} \textcolor{red}{+ 16}$
Re-write the $\textcolor{b l u e}{x}$ and $\textcolor{red}{y}$ sub-expressions as binomial squares and the constant as a square.
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{\left(x - 2\right)}^{2}} + \textcolor{red}{{\left(y + 4\right)}^{2}} = \textcolor{g r e e n}{{10}^{2}}$
Often we would leave it in this form as "good enough",
but technically this wouldn't make the $y$ sub-expression into the form ${\left(y - b\right)}^{2}$ (and might cause confusion as to the y component of the center coordinate).
So more accurately:
color(white)("XXX")color(blue)((x-2)^2)+color(red)((y-(-4))^2=color(green)(10^2)
with center at $\left(2 , - 4\right)$ and radius $10$
## How do you find an exponential function given the points are (-1,8) and (1,2)?
mason m
8.591780821917808 years ago
$y = 4 {\left(\frac{1}{2}\right)}^{x}$
#### Explanation:
An exponential function is in the general form
$y = a {\left(b\right)}^{x}$
We know the points $\left(- 1 , 8\right)$ and $\left(1 , 2\right)$, so the following are true:
$8 = a \left({b}^{-} 1\right) = \frac{a}{b}$
$2 = a \left({b}^{1}\right) = a b$
Multiply both sides of the first equation by $b$ to find that
$8 b = a$
Plug this into the second equation and solve for $b$:
$2 = \left(8 b\right) b$
$2 = 8 {b}^{2}$
${b}^{2} = \frac{1}{4}$
$b = \pm \frac{1}{2}$
Two equations seem to be possible here. Plug both values of $b$ into the either equation to find $a$. I'll use the second equation for simpler algebra.
If $b = \frac{1}{2}$:
$2 = a \left(\frac{1}{2}\right)$
$a = 4$
Giving us the equation: color(green)(y=4(1/2)^x
If $b = - \frac{1}{2}$:
$2 = a \left(- \frac{1}{2}\right)$
$a = - 4$
Giving us the equation: $y = - 4 {\left(- \frac{1}{2}\right)}^{x}$
However! In an exponential function, $b > 0$, otherwise many issues arise when trying to graph the function.
The only valid function is
color(green)(y=4(1/2)^x
## How do you convert (-4, 3) into polar coordinates?
Tony B
8.58904109589041 years ago
Given that a point $\textcolor{b r o w n}{P \to \left(x , y\right) \to \left(- 4 , 3\right) \text{ Cartesian}}$
Then $\textcolor{b l u e}{P \to \left(5 , {143.13}^{o}\right) \text{ Polar }}$ to 2 decimal places
#### Explanation:
This is not a polar graph!!!
$\textcolor{b l u e}{\text{Where it all comes from}}$
We are give the coordinates of (-4,3)
Suppose we viewed this in the context of Cartesian form and use $y = m x + c$
Then $c = 0$ and $m = \frac{y}{x} = - \frac{3}{4}$
So we would have $y = - \frac{3}{4} x$
Suppose the graph was only plotted over the range $x \to \left(0 , - 4\right)$
Then the above graph would not be continuous but be a line from
$\left(0 , 0\right) \to \left(3 , - 4\right)$
All we need now is the angle that that line makes to the x-axis and the length of that line.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Finding the Polar r value}}$
$\text{Length } = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{{3}^{2} + {4}^{2}} = 5$
$\textcolor{b l u e}{\text{So the polar } r = 5}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Finding the Polar "theta" value}}$
The Polar angle $\theta$ is measured from the positive x-axis counterclockwise.
Let the angle from the line to the negative x-axis be $\phi$
Then $\phi = {\tan}^{- 1} \left(\frac{y}{x}\right) = {\tan}^{- 1} \left(\frac{3}{4}\right)$
But $\textcolor{w h i t e}{. .} \phi + \theta = 180$
so $\textcolor{w h i t e}{. .} \theta = 180 - {\tan}^{- 1} \left(\frac{3}{4}\right)$
color(blue)( theta = 143.13 to 2 decimal places
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Putting at all together}}$
Given that a point $P \to \left(x , y\right) \to \left(- 4 , 3\right) \text{Cartesian}$
Then $P \to \left(5 , {143.13}^{o}\right) \text{ Polar }$ to 2 decimal places
## How do you write the partial fraction decomposition of the rational expression (8x-1)/(x^3 -1)?
Jim G.
8.58904109589041 years ago
$\frac{\frac{7}{3}}{x - 1} + \frac{- \frac{7}{3} x - \frac{10}{3}}{{x}^{2} + x + 1}$
#### Explanation:
The first thing that has to be done is to factorise the denominator.
${x}^{3} - 1$ is a difference of cubes and is factorised as follow :
${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$
here then $a = x$ and $b = 1$ .
So ${x}^{3} - 1 = \left(x - 1\right) \left({x}^{2} + x + 1\right)$
Now $\left(x - 1\right)$ is of degree 1 and so numerator will be of degree 0 ie. a constant. Similarly $\left({x}^{2} + x + 1\right)$ is of degree 2 and so numerator will be of degree 1 ie of the form $a x + b$. Now the expression can be written as :
$\frac{8 x - 1}{{x}^{3} - 1} = \frac{A}{x - 1} + \frac{B x + C}{{x}^{2} + x + 1}$
Multiplying through by $\left({x}^{3} - 1\right)$
8x - 1 = A(x^2 + x + 1 ) + (Bx + C )(x - 1 )" " " " color(red)(("*"))
We now have to find the values of $A$ , $B$ and $C$.
Note that if we use $x = 1$ then the term with A will be 0.
Substitute x = 1 in equation $\textcolor{red}{\left(\text{*}\right)}$
$7 = 3 A + 0 \Rightarrow A = \frac{7}{3}$
To find B and C it will be necessary to compare the coefficients on both sides of the equation $\textcolor{red}{\left(\text{*}\right)}$. Multiplying out the brackets on the right hand side to begin with.
$\Rightarrow 8 x - 1 = A {x}^{2} + A x + A + B {x}^{2} + C x - B x - C$
This can be 'tidied up' by collecting like terms and letting $A = \frac{7}{3}$
$8 x - 1 = \frac{7}{3} {x}^{2} + \frac{7}{3} x + \frac{7}{3} + B {x}^{2} + C x - B x - C$
$\Rightarrow 8 x - 1 = \left(\frac{7}{3} + B\right) {x}^{2} + \left(\frac{7}{3} + C - B\right) x + \left(\frac{7}{3} - C\right)$
Compare ${x}^{2}$ terms
$0 = \frac{7}{3} + B \Rightarrow B = - \frac{7}{3}$
Now compare constant terms.
$- 1 = A - C = - \frac{7}{3} - C \Rightarrow C = - \frac{10}{3}$
Finally
$\frac{8 x - 1}{{x}^{3} - 1} = \frac{\frac{7}{3}}{x - 1} + \frac{- \frac{7}{3} x - \frac{10}{3}}{{x}^{2} + x + 1}$
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# NCTB Class 8 Math Chapter Five Exercise 5.1 Solution
NCTB Class 8 Math Chapter Five Exercise 5.1 Solutions by Math Expert. Bangladesh Board Class 8 Math Solution Chapter Five Algebraic Fractions Exercise 5.1 Solution.
Board NCTB Class 8 Subject Mathematics Chapter 5 Chapter Name Algebraic Fractions Exercise 5.1 Solution
Exercise 5.1
1> Algebraic fractions :
(a) 4x2y3z5/9x5y2z3
= 4/9 . yz2/x3
= 4yz2/9x3
(b) 16(2x)4(3y)5/(3x)3.(xy)6
= 36x/y
(c) x3y+xy3/x2y3+x3y2
= xy (x2+y2) /x2y2(y+x)
= x2+y2/xy(x+y)
(d) (a-b)(a+b)/a2-b3
= (a-b)(a+b) / (a-b)(a2+ab+b2)
= (a+b)/a2+ab+b2
(e) x2-6x+5/x2-25
= x2-5x-x+5/x2-52=
= x(x-5)-1(x-5)/(x+5)(x-5)
= (x-1)(x-5)/(x+5)(x-5)
= (x-1)/(x+5)
(f) x2-7x+12/x2-9x+20
= x2-4x-3x+12/x2-5x-4x+20
= x(x-4)-3(x-4)/x(x-5)-4(x-5)
= (x-3)(x-4) / (x-5)(x-4)
= x-3/x-5
(g) (x3-y3) (x2-xy+y2)/(x2-y2) (x3+y3)
(h) a2-b2-2bc-c2/a2+2ab+b2-c2
(h) x-y/x+y, y-z/y+z, z-x/z+x
2> express the following fractions in the form of a common denominator:
(a) x2/xy , y2/yz, z2/zx
(b) x-y/xy, y-z/yz, z-x/zx
(d) x+y/(x-y)2, x-y/x3+y3, y-z/x2-y2
(e) a /a3+b3, b/(a2+ab+b2), c/(a3-b3)
(g) a-b/a2b2, b-c/b2c2, c-a/c2a2
(h) x-y/x+y, y-z/y+z, z-x/z+x
3> Find the sum:
(a) a-b/a + a+b/b
(b) a/bc + b/ca + c/ab
(c) x-y/x + y-z/y + z-x/z
(d) x+y/x-y + x-y/x+y
(e) 1/x2-3x+2 + 1/x2-4x+3 + 1/x2-5x+4
(f) 1/a2-b2 + 1/a2+ab+b2 + 1/a2-ab+b2
4>
(a) a/x-3 – a2/x2-9
(b) 1/y(x-y) – 1/x(x+y)
(c) x+1/1+x+x2 – x-1/1-x+x2
(d) a2+16b2/a2-16b2 – a-4b/a+4b
5>
(a) x-y/xy + y-z/yz + z-x/zx
(b) x-y/(x+y)(y+z) + y-z/(y+z)(z+x) + z-x/(z+x)(x+y) .
(c) y/(x-y)(y-z), x/(z-x)(x-y), z/(y+z)(z+x)
(d) 1/(x+3y) + 1/x-3y – 2x/x2-9y2
(f) 1/x-2 – x-2/x2+2x+4 + 6x/x3+8
(g) 1/x-y – 1/x+1 – 2/x2+1 + 4/x4+1
(h) x-y/(y-z)(z-x) + y-z/(z-x)(x-y) + z-x/(x-y)(y-z)
(i) 1/a-b-c + 1/a-b+c + a/a2+b2-c2-2ab
Updated: March 26, 2021 — 1:19 pm
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# Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry
Are you stressed to make your kid more intelligent and math proficient? Don’t worry at all as we have come up with the best solution that gives you relief from your stress. Here is the best preparation guide that aids your kids & makes them focused on the concepts of the respective Grade. Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry is the one that we are discussing in the above lines. By using the BIM Geometry Ch 9 Solution Key, you can quickly gain the subject skills and score the highest marks in the exams. Fun & Activity-based learnings are easy & the simplest way to understand the concepts thoroughly and clarify your doubts within minimal time.
## Big Ideas Math Book Geometry Answer Key Chapter 9 Right Triangles and Trigonometry
### Right Triangles and Trigonometry Maintaining Mathematical Proficiency
Simplify the expression.
Question 1.
√75
square root of 75 = 5,625.
Explanation:
In the above-given question,
given that,
√75.
square root of 75 = 75 x 75.
75 x 75 = 5,625.
√75 = 5,625.
Question 2.
√270
square root of 270 = 72,900.
Explanation:
In the above-given question,
given that,
√270.
square root of 270 = 270 x 270.
270 x 270 = 72900.
√270 = 72900.
Question 3.
√135
square root of 135 = 18225.
Explanation:
In the above-given question,
given that,
√135.
square root of 135 = 135 x 135.
135 x 135 = 18225.
√135 = 18,225.
Question 4.
$$\frac{2}{\sqrt{7}}$$
2/49 = 0.04.
Explanation:
In the above-given question,
given that,
square root of 7 = 7 x 7.
7 x 7 = 49.
$$\frac{2}{\sqrt{7}}$$.
2/49 = 0.04.
Question 5.
$$\frac{5}{\sqrt{2}}$$
5/4 = 1.25.
Explanation:
In the above-given question,
given that,
square root of 2 = 2 x 2.
2 x 2 = 4.
$$\frac{5}{\sqrt{2}}$$.
5/4 = 1.25.
Question 6.
$$\frac{12}{\sqrt{6}}$$
12/36 = 0.33.
Explanation:
In the above-given question,
given that,
square root of 6 = 6 x 6.
6 x 6 = 36.
$$\frac{12}{\sqrt{6}}$$.
12/36 = 0.33.
Solve the proportion.
Question 7.
$$\frac{x}{12}=\frac{3}{4}$$
x = 9.
Explanation:
In the above-given question,
given that,
$$\frac{x}{12}=\frac{3}{4}$$
x/12 = 3/4.
4x = 12 x 3.
4x = 36.
x = 36/4.
x = 9.
Question 8.
$$\frac{x}{3}=\frac{5}{2}$$
x = 7.5.
Explanation:
In the above-given question,
given that,
$$\frac{x}{3}=\frac{5}{2}$$
x/3 = 5/2.
2x = 5 x 3.
2x = 15.
x = 15/2.
x = 7.5.
Question 9.
$$\frac{4}{x}=\frac{7}{56}$$
x = 32.
Explanation:
In the above-given question,
given that,
$$\frac{4}{x}=\frac{7}{56}$$
4/x = 7/56.
7x = 56 x 4.
7x = 224.
x = 32.
Question 10.
$$\frac{10}{23}=\frac{4}{x}$$
x = 9.2.
Explanation:
In the above-given question,
given that,
$$\frac{10}{23}=\frac{4}{x}$$
x/4 = 10/23.
10x = 23 x 4.
10x = 92.
x = 92/10.
x = 9.2.
Question 11.
$$\frac{x+1}{2}=\frac{21}{14}$$
x = 135.
Explanation:
In the above-given question,
given that,
$$\frac{x + 1}{2}=\frac{21}{14}$$
x+12 x 2 = 21×14.
2x + 24= 294 .
2x = 294 – 24.
2x = 270.
x = 270/2.
x = 135.
Question 12.
$$\frac{9}{3 x-15}=\frac{3}{12}$$
x = 6.33.
Explanation:
In the above-given question,
given that,
$$\frac{9}{3 x-15}=\frac{3}{12}$$
27x – 135 = 3×12.
27x = 36 + 135.
27x = 171.
x = 171/27.
x = 6.33.
Question 13.
ABSTRACT REASONING
The Product Property of Square Roots allows you to simplify the square root of a product. Are you able to simplify the square root of a sum? of a diffrence? Explain.
Yes, I am able to simplify the square root of a sum.
Explanation:
In the above-given question,
given that,
The product property of square roots allows you to simplify the square root of a product.
√3 + 1 = √4.
√4 = 4 x 4.
16.
√3 – 1 = √2.
√2 = 2 x 2.
4.
### Right Triangles and Trigonometry Mathematical practices
Monitoring progress
Question 1.
Use dynamic geometry software to construct a right triangle with acute angle measures of 30° and 60° in standard position. What are the exact coordinates of its vertices?
Question 2.
Use dynamic geometry software to construct a right triangle with acute angle measures of 20° and 70° in standard position. What are the approximate coordinates of its vertices?
### 9.1 The Pythagorean Theorem
Exploration 1
Proving the Pythagorean Theorem without Words
Work with a partner.
a. Draw and cut out a right triangle with legs a and b, and hypotenuse c.
Explanation:
In the above-given question,
given that,
proving the Pythagorean theorem without words.
a2 + b2 = c2.
b. Make three copies of your right triangle. Arrange all tour triangles to form a large square, as shown.
a2 + b2 = c2.
Explanation:
In the above-given question,
given that,
make three copies of your right triangle.
a2 + b2 = c2.
c. Find the area of the large square in terms of a, b, and c by summing the areas of the triangles and the small square.
The area of the large square = a2 x b2.
Explanation:
In the above-given question,
given that,
the area of the square = l x b.
where l = length, and b = breadth.
the area of the square = a x b.
area = a2 x b2.
d. Copy the large square. Divide it into two smaller squares and two equally-sized rectangles, as shown.
e. Find the area of the large square in terms of a and b by summing the areas of the rectangles and the smaller squares.
f. Compare your answers to parts (c) and (e). Explain how this proves the Pythagorean Theorem.
a2 + b2 = c2.
Explanation:
In the above-given question,
given that,
The length of the a and b is equal to the hypotenuse.
a2 + b2 = c2.
where a = one side and b = one side.
Exploration 2
Proving the Pythagorean Theorem
Work with a partner:
a. Draw a right triangle with legs a and b, and hypotenuse c, as shown. Draw the altitude from C to $$\overline{A B}$$ Label the lengths, as shown.
a2 + b2 = c2.
Explanation:
In the above-given question,
given that,
a2 + b2 = c2.
b. Explain why ∆ABC, ∆ACD, and ∆CBD are similar.
In a right-angle triangle all the angle are equal.
Explanation:
In the above-given question,
given that,
∆ABC, ∆ACD, and ∆CBD are similar.
In a right-angle triangle all the angles are equal.
REASONING ABSTRACTLY
To be proficient in math, you need to know and flexibly use different properties of operations and objects.
c. Write a two-column proof using the similar triangles in part (b) to prove that a2 + b2 = c2
a2 + b2 = c2
Explanation:
In the above-given question,
given that,
In the pythagorean theorem.
a2 + b2 = c2
the length of the hypotenuse ie equal to the two side lengths.
a2 + b2 = c2
Question 3.
How can you prove the Pythagorean Theorem?
a2 + b2 = c2
Explanation:
In the above-given question,
given that,
In the pythagorean theorem,
the length of the hypotenuse is equal to the length of the other two sides.
hypotenuse = c.
length = a.
a2 + b2 = c2
Question 4.
Use the Internet or sonic other resource to find a way to prove the Pythagorean Theorem that is different from Explorations 1 and 2.
### Lesson 9.1 The Pythagorean Theorem
Monitoring Progress
Find the value of x. Then tell whether the side lengths form a Pythagorean triple.
Question 1.
x = √52.
Explanation:
In the above-given question,
given that,
the side lengths are 6 and 4.
a2 + b2 = c2
6 x 6 + 4 x 4 = c2
36 + 16 = c2
52 = c2.
c = √52.
Question 2.
x = 4.
Explanation:
In the above-given question,
given that,
the side lengths are 3 and 5.
x2 + 3 x 3 = 5 x 5.
x2 + 9 = 25.
x2 = 25 – 9.
x2 = 16.
x = 4.
Question 3.
An anemometer is a device used to measure wind speed. The anemometer shown is attached to the top of a pole. Support wires are attached to the pole 5 feet above the ground. Each support wire is 6 feet long. How far from the base of the pole is each wire attached to the ground?
x = √11.
Explanation:
In the above-given question,
given that,
An anemometer is a device used to measure wind speed.
support wires are attached to the pole 5 feet above the ground.
Each support wire is 6 feet long.
d2 + 6 x 6 = 5 x 5.
d2 + 36 = 25.
d2 = 25 – 36.
d2 = 11.
d
= √11.
Tell whether the triangle is a right triangle.
Question 4.
Yes the triangle is a right triangle.
Explanation:
In the above-given question,
given that,
the hypotenuse = 3 √34.
one side = 15.
the other side = 9.
so the triangle is a right triangle.
Question 5.
Yes, the triangle is a actute triangle.
Explanation:
In the above-given question,
given that,
the hypotenuse = 22.
one side = 26.
the other side = 14.
so the triangle is a acute triangle.
Question 6.
Verify that segments with lengths of 3, 4, and 6 form a triangle. Is the triangle acute, right, or obtuse?
Yes the lengths of the triangle form a acute triangle.
Explanation:
In the above-given question,
given that,
the side lenths of 3, 4, and 6 form a triangle.
6 x 6 = 3 x 3 + 4 x 4.
36 = 9 + 16.
36 = 25.
so the length of the triangle forms a acute triangle.
Question 7.
Verify that segments with lengths of 2, 1, 2, 8, and 3.5 form a triangle. Is the triangle acute, right, or obtuse?
### Exercise 9.1 The Pythagorean Theorem
Vocabulary and Core Concept Check
Question 1.
VOCABULARY
What is a Pythagorean triple?
Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Find the length of the longest side.
The length of the longest side = 5.
Explanation:
In the above-given question,
given that,
the side lengths are 3 and 4.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 4 x 4 + 3 x 3.
X x X = 16 + 9.
X x X = 25.
X = 5.
Find the length of the hypotenuse
The length of the hypotenuse = 5.
Explanation:
In the above-given question,
given that,
the side lengths are 3 and 4.
in the Pythagorean theorem,
the longest side is equal to the side lengths.
X = 4 x 4 + 3 x 3.
X x X = 16 + 9.
X x X = 25.
X = 5.
Find the length of the longest leg.
The length of the longest leg = 5.
Explanation:
In the above-given question,
given that,
the side lengths are 3 and 4.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 4 x 4 + 3 x 3.
X x X = 16 + 9.
X x X = 25.
X = 5.
Find the length of the side opposite the right angle.
The length of the side opposite to the right angle = 5.
Explanation:
In the above-given question,
given that,
the side lengths are 3 and 4.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 4 x 4 + 3 x 3.
X x X = 16 + 9.
X x X = 25.
X = 5.
Monitoring progress and Modeling with Mathematics
In Exercises 3-6, find the value of x. Then tell whether the side lengths form a Pythagorean triple.
Question 3.
Question 4.
The length of the x = 34.
Explanation:
In the above-given question,
given that,
the side lengths are 30 and 16.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 16 x 16 + 30 x 30.
X x X = 256 + 900.
X x X = 1156.
X = 34.
Question 5.
Question 6.
The length of the x = 7.2.
Explanation:
In the above-given question,
given that,
the side lengths are 6 and 4.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 4 x 4 + 6 x 6.
X x X = 16 + 36.
X x X = 52.
X = 7.2.
In Exercises 7 – 10, find the value of x. Then tell whether the side lengths form a Pythagorean triple.
Question 7.
Question 8.
The length of the X = 25.6.
Explanation:
In the above-given question,
given that,
the side lengths are 24 and 9.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 24 x 24 + 9 x 9.
X x X = 576 + 81.
X x X = 657.
X = 25.6.
Question 9.
Question 10.
The length of the x = 11.4.
Explanation:
In the above-given question,
given that,
the side lengths are 7 and 9.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 7 x 7 + 9 x 9.
X x X = 49 + 81.
X x X = 130.
X = 11.4.
ERROR ANALYSIS
In Exercises 11 and 12, describe and correct the error in using the Pythagorean Theorem (Theorem 9.1).
Question 11.
Question 12.
x = 24.
Explanation:
In the above-given question,
given that,
the side lengths are 26 and 10.
26 x 26 = a x a + 10 x 10.
676 = a x a + 100.
676 – 100 = a x a.
576 = a x a.
a = 24.
x = 24.
Question 13.
MODELING WITH MATHEMATICS
The fire escape forms a right triangle, as shown. Use the Pythagorean Theorem (Theorem 9. 1) to approximate the distance between the two platforms. (See Example 3.)
Question 14.
MODELING WITH MATHEMATICS
The backboard of the basketball hoop forms a right triangle with the supporting rods, as shown. Use the Pythagorean Theorem (Theorem 9.1) to approximate the distance between the rods where the meet the backboard.
x = 9.1.
Explanation:
In the above-given question,
given that,
the side lengths are 13.4 and 9.8.
13.4 x 13.4 = X x X + 9.8 x 9.8.
179.56 = X x X + 96.04.
179.56 – 96.04 = X x X.
83.52 = X x X.
X = 9.1.
In Exercises 15 – 20, tell whether the triangle is a right triangle.
Question 15.
Question 16.
No, the triangle is not a right triangle.
Explanation:
In the above-given question,
given that,
the side lengths are 23 and 11.4.
the hypotenuse = 21.2.
21.2 x 21.2 = 23 x 23 + 11.4 x 11.4.
449.44 = 529 + 129.96.
449.44 = 658.96.
449 is not equal to 658.96.
so the triangle is not a right triangle.
Question 17.
Question 18.
No, the triangle is not a right triangle.
Explanation:
In the above-given question,
given that,
the side lengths are 5 and 1.
the hypotenuse = √26.
26 x 26 = 5 x 5 + 1 x 1.
676 = 25 + 1.
676 = 26.
676 is not equal to 26.
so the triangle is not a right triangle.
Question 19.
Question 20.
Yes, the triangle forms a right triangle.
Explanation:
In the above-given question,
given that,
the side lengths are 80 and 39.
the hypotenuse = 89.
89 x 89 = 80 x 80 + 39 x 39.
7921 = 6400 + 1521.
7921 = 7921.
7921 is equal to 7921.
so the triangle forms a right triangle.
In Exercises 21 – 28, verify that the segment lengths form a triangle. Is the triangle acute, right, or obtuse?
Question 21.
10, 11, and 14
Question 22.
6, 8, and 10
Yes, the triangle is forming a right triangle.
Explanation:
In the above-given question,
given that,
the side lengths are 8 and 6.
the hypotenuse = 10.
10 x 10 = 8 x 8 + 6 x 6.
100 = 64 + 36.
100 = 100.
100 is equal to 100.
so the triangle is forming a right triangle.
Question 23.
12, 16, and 20
Question 24.
15, 20, and 36
Yes, the triangle is obtuse triangle.
Explanation:
In the above-given question,
given that,
the side lengths are 15 and 20.
the hypotenuse = 36.
36 x 36 = 20 x 20 + 15 x 15.
1296 = 400 + 225.
1296 > 625.
1296 is greater than 625.
so the triangle is not a obtuse triangle.
Question 25.
5.3, 6.7, and 7.8
Question 26.
4.1, 8.2, and 12.2
No, the triangle is obtuse triangle.
Explanation:
In the above-given question,
given that,
the side lengths are 4.1 and 8.2.
the hypotenuse = 12.2.
12.2 x 12.2 = 4.1 x 4.1 + 8.2 x 8.2`.
148.84 = 16.81 + 67.24.
148.84 > 84.05.
148.84 is greater than 84.05.
so the triangle is obtuse triangle.
Question 27.
24, 30, and 6√43
Question 28.
10, 15 and 5√13
Yes, the triangle is an acute triangle.
Explanation:
In the above-given question,
given that,
the side lengths are 10 and 5√13.
the hypotenuse = 15.
15 x 15 = 10 x 10 + 5√13 x 5√13.
225 = 100 + 34.81.
225 < 134.81.
225 is less than 134.81.
so the triangle is acute triangle.
Question 29.
MODELING WITH MATHEMATICS
In baseball, the lengths of the paths between consecutive bases are 90 feet, and the paths form right angles. The player on first base tries to steal second base. How far does the ball need to travel from home plate to second base to get the player out?
Question 30.
REASONING
You are making a canvas frame for a painting using stretcher bars. The rectangular painting will be 10 inches long and 8 inches wide. Using a ruler, how can you be certain that the corners of the frame are 90°
x = 12.8.
Explanation:
In the above-given question,
given that,
the side lengths are 10 and 8.
the hypotenuse = x.
X x X = 10 x 10 + 8 x 8.
X = 100 + 64.
X = 12.8.
In Exercises 31 – 34, find the area of the isosceles triangle.
Question 31.
Question 32.
The area of the Isosceles triangle = 12 ft.
Explanation:
In the above-given question,
given that,
base = 32 ft.
hypotenuse = 20ft.
a2 + b2 = c2
h x h + 16 x 16 = 20 x 20.
h x h + 256 = 400.
h x h = 400 – 256.
h x h = 144.
h = 12 ft.
so the area of the isosceles triangle = 12 ft.
Question 33.
Question 34.
The area of the Isosceles triangle = 48 m.
Explanation:
In the above-given question,
given that,
base = 28 m.
hypotenuse = 50 m.
a2 + b2 = c2
h x h + 14 x 14 = 50 x 50.
h x h + 196 = 2500.
h x h = 2500 – 196.
h x h = 2304.
h = 48 m.
so the area of the isosceles triangle = 48 m.
Question 35.
ANALYZING RELATIONSHIPS
Justify the Distance Formula using the Pythagorean Theorem (Thin. 9. 1).
Question 36.
HOW DO YOU SEE IT?
How do you know ∠C is a right angle without using the Pythagorean Theorem (Theorem 9.1) ?
Yes, the triangle is forming a right triangle.
Explanation:
In the above-given question,
given that,
the side lengths are 8 and 6.
the hypotenuse = 10.
10 x 10 = 8 x 8 + 6 x 6.
100 = 64 + 36.
100 = 100.
100 is equal to 100.
so the triangle is forming a right triangle.
Question 37.
PROBLEM SOLVING
You are making a kite and need to figure out how much binding to buy. You need the binding for the perimeter of the kite. The binding Comes in packages of two yards. How many packages should you buy?
Question 38.
PROVING A THEOREM
Use the Pythagorean Theorem (Theorem 9. 1) to prove the Hypotenuse-Leg (HL) Congruence Theorem (Theorem 5.9).
Question 39.
PROVING A THEOREM
Prove the Converse of the Pythagorean Theorem (Theorem 9.2). (Hint: Draw ∆ABC with side lengths a, b, and c, where c is the length of the longest side. Then draw a right triangle with side lengths a, b, and x, where x is the length of the hypotenuse. Compare lengths c and x.)
Question 40.
THOUGHT PROVOKING
Consider two integers m and n. where m > n. Do the following expressions produce a Pythagorean triple? If yes, prove your answer. If no, give a counterexample.
2mn, m2 – n2, m2 + n2
Question 41.
MAKING AN ARGUMENT
Your friend claims 72 and 75 Cannot be part of a pythagorean triple because 722 + 752 does not equal a positive integer squared. Is your friend correct? Explain your reasoning.
Question 42.
PROVING A THEOREM
Copy and complete the proof of the pythagorean Inequalities Theorem (Theorem 9.3) when c2 < a2 + b2.
Given In ∆ABC, c2 < a2 + b2 where c is the length
of the longest side.
∆PQR has side lengths a, b, and x, where x is the length of the hypotenuse, and ∠R is a right angle.
Prove ∆ABC is an acute triangle.
Statements Reasons 1. In ∆ABC, C2 < (12 + h2, where c is the length of the longest side. ∆PQR has side lengths a, b, and x, where x is the length of the hypotenuse, and ∠R is a right angle. 1. _____________________________ 2. a2 + b2 = x2 2. _____________________________ 3. c2 < r2 3. _____________________________ 4. c < x 4. Take the positive square root of each side. 5. m ∠ R = 90° 5. _____________________________ 6. m ∠ C < m ∠ R 6. Converse of the Hinge Theorem (Theorem 6.13) 7. m ∠ C < 90° 7. _____________________________ 8. ∠C is an acute angle. 8. _____________________________ 9. ∆ABC is an acute triangle. 9. _____________________________
Question 43.
PROVING A THEOREM
Prove the Pythagorean Inequalities Theorem (Theorem 9.3) when c2 > a2 + b2. (Hint: Look back at Exercise 42.)
Maintaining Mathematical Proficiency
Simplify the expression by rationalizing the denominator.
Question 44.
$$\frac{7}{\sqrt{2}}$$
7/√2 = 7√2 /2.
Explanation:
In the above-given question,
given that,
$$\frac{7}{\sqrt{2}}$$ = 7/√2.
7/√2 = 7/√2 x √2 /√2 .
7 √2 /√4.
7√2 /2.
Question 45.
$$\frac{14}{\sqrt{3}}$$
Question 46.
$$\frac{8}{\sqrt{2}}$$
8/√2 = 8√2 /2.
Explanation:
In the above-given question,
given that,
$$\frac{8}{\sqrt{2}}$$ = 8/√2.
8/√2 = 8/√2 x √2 /√2 .
8 √2 /√4.
8√2 /2.
Question 47.
$$\frac{12}{\sqrt{3}}$$
### 9.2 Special Right Triangles
Exploration 1
Side Ratios of an Isosceles Right Triangle
Work with a partner:
a. Use dynamic geometry software to construct an isosceles right triangle with a leg length of 4 units.
b. Find the acute angle measures. Explain why this triangle is called a 45° – 45° – 90° triangle.
c. Find the exact ratios of the side lenghts (using square roots).
$$\frac{A B}{A C}$$ = ____________
$$\frac{A B}{B C}$$ = ____________
$$\frac{A B}{B C}$$ = ____________
ATTENDING TO PRECISION
To be proficient in math, you need to express numerical answers with a degree of precision appropriate for the problem context.
d. Repeat parts (a) and (c) for several other isosceles right triangles. Use your results to write a conjecture about the ratios of the side lengths of an isosceles right triangle.
Exploration 2
Work with a partner.
a. Use dynamic geometry software to construct a right triangle with acute angle measures of 30° and 60° (a 30° – 60° – 90° triangle), where the shorter leg length is 3 units.
b. Find the exact ratios of the side lengths (using square roots).
$$\frac{A B}{A C}$$ = ____________
$$\frac{A B}{B C}$$ = ____________
$$\frac{A B}{B C}$$ = ____________
C. Repeat parts (a) and (b) for several other 30° – 60° – 90° triangles. Use your results to write a conjecture about the ratios of the side lengths of a 30° – 60° – 90° triangle.
Question 3.
What is the relationship among the side lengths of 45°- 45° – 90° triangles? 30° – 60° – 90° triangles?
### Lesson 9.2 Special Right Triangles
Monitoring Progress
Find the value of the variable. Write your answer in simplest form.
Question 1.
x = 4
Explanation:
(2√2)² = x² + x²
8 = 2x²
x² = 4
x = 4
Question 2.
y = 2
Explanation:
y² = 2 + 2
y² = 4
y = 2
Question 3.
x = 3, y = 2√3
Explanation:
longer leg = shorter leg • √3
x = √3 • √3
x = 3
hypotenuse = shorter leg • 2
= √3 • 2 = 2√3
Question 4
h = 2√3
Explanation:
longer leg = shorter leg • √3
h = 2√3
Question 5.
The logo on a recycling bin resembles an equilateral triangle with side lengths of 6 centimeters. Approximate the area of the logo.
Area is $$\frac { 1 }{ 3√3 }$$
Explanation:
Area = $$\frac { √3 }{ 4 }$$ a²
= $$\frac { √3 }{ 4 }$$(6)²
= $$\frac { 1 }{ 3√3 }$$
Question 6.
The body of a dump truck is raised to empty a load of sand. How high is the 14-foot-long body from the frame when it is tipped upward by a 60° angle?
28/3 ft high is the 14-foot-long body from the frame when it is tipped upward by a 60° angle.
Explanation:
Height of body at 90 degrees = 14 ft
Height of body at 1 degree = 14/90
Height of body at 60 degrees = 14 x 60/90
= 14 x 2/3
= 28/3 ft
### Exercise 9.2 Special Right Triangles
Vocabulary and Core Concept Check
Question 1.
VOCABULARY
Name two special right triangles by their angle measures.
Question 2.
WRITING
Explain why the acute angles in an isosceles right triangle always measure 45°.
Because the acute angles of a right isosceles triangle must be congruent by the base angles theorem and complementary, their measures must be 90°/2 = 45°.
Monitoring Progress and Modeling with Mathematics
In Exercises 3 – 6, find the value of x. Write your answer in simplest form.
Question 3.
Question 4.
x = 10
Explanation:
hypotenuse = leg • √2
x = 5√2 • √2
x = 10
Question 5.
Question 6.
x = $$\frac { 9 }{ √2 }$$
Explanation:
hypotenuse = leg • √2
9 = x • √2
x = $$\frac { 9 }{ √2 }$$
In Exercises 7 – 10, find the values of x and y. Write your answers in simplest form.
Question 7.
Question 8.
x = 3, y = 6
Explanation:
hypotenuse = 2 • shorter leg
y = 2 • 3
y = 6
longer leg = √3 • shorter leg
3√3 = √3x
x = 3
Question 9.
Question 10.
x = 18, y = 6√3
Explanation:
hypotenuse = 2 • shorter leg
12√3 = 2y
y = 6√3
longer leg = √3 • shorter leg
x = √3 . 6√3
x = 18
ERROR ANALYSIS
In Exercises 11 and 12, describe and correct the error in finding the length of the hypotenuse.
Question 11.
Question 12.
hypotenuse = leg • √2 = √5 . √2 = √10
In Exercises 13 and 14. sketch the figure that is described. Find the indicated length. Round decimal answers to the nearest tenth.
Question 13.
The side length of an equilateral triangle is 5 centimeters. Find the length of an altitude.
Question 14.
The perimeter of a square is 36 inches. Find the length of a diagonal.
The length of a diagonal is 9√2
Explanation:
Side of the square = 36/4 = 9
square diagonal = √2a = √2(9) = 9√2
In Exercises 15 and 16, find the area of the figure. Round decimal answers to the nearest tenth.
Question 15.
Question 16.
Area is 40√(1/3) sq m
Explanation:
longer leg = √3 • shorter leg
4 = √3 • shorter leg
shorter leg = 4/√3
h² = 16/3 + 16
h² = 16(4/3)
h = 8√(1/3)
Area of the parallelogram = 5(8√(1/3)) = 40√(1/3) sq m
Question 17.
PROBLEM SOLVING
Each half of the drawbridge is about 284 feet long. How high does the drawbridge rise when x is 30°? 45°? 60°?
Question 18.
MODELING WITH MATHEMATICS
A nut is shaped like a regular hexagon with side lengths of 1 centimeter. Find the value of x. (Hint: A regular hexagon can be divided into six congruent triangles.)
Side length = 1cm
A regular hexagon has six equal the side length. A line drawn from the centre to any vertex will have the same length as any side.
This implies the radius is equal to the side length.
As a result, when lines are drawn from the centre to each of the vertexes, a
regular hexagon is said to be made of six equilateral triangles.
From the diagram, x = 2× apothem
Apothem is the distance from the centre of a regular polygon to the midpoint of side.
Using Pythagoras theorem, we would get the apothem
1² = apothem² + (½)²
Apothem = √(1² -(½)²)
= √(1-¼) = √¾
Apothem = ½√3
x = 2× Apothem = 2 × ½√3
x = √3
Question 19.
PROVING A THEOREM
Write a paragraph proof of the 45°- 45°- 90° Triangle Theorem (Theorem 9.4).
Given ∆DEF is a 45° – 45° – 90° triangle.
Prove The hypotenuse is √2 times as long as each leg.
Question 20.
HOW DO YOU SEE IT?
The diagram shows part of the wheel of Theodorus.
a. Which triangles, if any, are 45° – 45° – 90° triangles?
b. Which triangles, if any, are 30° – 60° – 90° triangles?
Question 21.
PROVING A THEOREM
Write a paragraph proof of the 30° – 60° – 90° Triangle Theorem (Theorem 9.5).
(Hint: Construct ∆JML congruent to ∆JKL.)
Given ∆JKL is a 30° 60° 9o° triangle.
Prove The hypotenuse is twice as long as the shorter leg, and the longer leg is √3 times as long as the shorter leg.
Question 22.
THOUGHT PROVOKING
A special right triangle is a right triangle that has rational angle measures and each side length contains at most one square root. There are only three special right triangles. The diagram below is called the Ailles rectangle. Label the sides and angles in the diagram. Describe all three special right triangles.
Question 23.
WRITING
Describe two ways to show that all isosceles right triangles are similar to each other.
Question 24.
MAKING AN ARGUMENT
Each triangle in the diagram is a 45° – 45° – 90° triangle. At Stage 0, the legs of the triangle are each 1 unit long. Your brother claims the lengths of the legs of the triangles added are halved at each stage. So, the length of a leg of a triangle added in Stage 8 will be $$\frac{1}{256}$$ unit. Is your brother correct? Explain your reasoning.
Question 25.
USING STRUCTURE
ΔTUV is a 30° – 60° – 90° triangle. where two vertices are U(3, – 1) and V( – 3, – 1), $$\overline{U V}$$ is the hypotenuse. and point T is in Quadrant I. Find the coordinates of T.
Maintaining Mathematical Proficiency
Find the Value of x.
Question 26.
ΔDEF ~ ΔLMN
x = 18
Explanation:
$$\frac { DE }{ LM }$$ = $$\frac { DF }{ LN }$$
$$\frac { 12 }{ x }$$ = $$\frac { 20 }{ 30 }$$
$$\frac { 12 }{ x }$$ = $$\frac { 2 }{ 3 }$$
x = 12($$\frac { 3 }{ 2 }$$) = 18
Question 27.
ΔABC ~ ΔQRS
### 9.3 Similar Right Triangles
Exploration 1
Writing a Conjecture
a. Use dynamic geometry software to construct right ∆ABC, as shown. Draw $$\overline{C D}$$ so that it is an altitude from the right angle to the hypotenuse of ∆ABC.
b. The geometric mean of two positive numbers a and b is the positive number x that satisfies
$$\frac{a}{x}=\frac{x}{b}$$
x is the geometric mean of a and b.
Write a proportion involving the side lengths of ∆CBD and ∆ACD so that CD is the geometric mean of two of the other side lengths. Use similar triangles to justify your steps.
c. Use the proportion you wrote in part (b) to find CD.
d. Generalize the proportion you wrote in part (b). Then write a conjecture about how the geometric mean is related to the altitude from the right angle to the hypotenuse of a right triangle.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results in constructing arguments.
Exploration 2
Comparing Geometric and Arithmetic Means
Work with a partner:
Use a spreadsheet to find the arithmetic mean and the geometric mean of several pairs of positive numbers. Compare the two means. What do you notice?
Question 3.
How are altitudes and geometric means of right triangles related?
### Lesson 9.3 Similar Right Triangles
Monitoring progress
Identify the similar triangles.
Question 1.
△QRS ~ △ QST
Question 2.
△EFG ~ △ EHG
Question 3.
△EGH ~ △EFG
$$\frac { EF }{ EG }$$ = $$\frac { GF }{ GH }$$
$$\frac { 5 }{ 3 }$$ = $$\frac { 4 }{ x }$$
x = 2.4
Question 4.
△JLM ~ △LMK
$$\frac { JL }{ LM }$$ = $$\frac { JM }{ KM }$$
$$\frac { 13 }{ 5 }$$ = $$\frac { 12 }{ x }$$
x = 4.615
Find the geometric mean of the two numbers.
Question 5.
12 and 27
x = √(12 x 27)
x = √324 = 18
Question 6.
18 and 54
x = √(18 x 54) = √(972)
x = 31.17
Question 7.
16 and 18
x = √(16 x 18) = √(288)
x = 16.970
Question 8.
Find the value of x in the triangle at the left.
x = √(9 x 4)
x = 6
Question 9.
WHAT IF?
In Example 5, the vertical distance from the ground to your eye is 5.5 feet and the distance from you to the gym wall is 9 feet. Approximate the height of the gym wall.
9² = 5.5 x w
81 = 5.5 x w
w = 14.72
The height of the wall = 14.72 + 5.5 = 20.22
### Exercise 9.3 Similar Right Triangles
Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and ____________ .
Question 2.
WRITING
In your own words, explain geometric mean.
The geometric mean is the average value or mean that signifies the central tendency of set of numbers by finding the product of their values.
Monitoring progress and Modeling with Mathematics
In Exercises 3 and 4, identify the similar triangles.
Question 3.
Question 4.
△LKM ~ △LMN ~ △MKN
In Exercises 5 – 10, find the value of x.
Question 5.
Question 6.
x = 9.6
Explanation:
$$\frac { QR }{ SR }$$ = $$\frac { SR }{ TS }$$
$$\frac { 20 }{ 16 }$$ = $$\frac { 12 }{ x }$$
1.25 = $$\frac { 12 }{ x }$$
x = 9.6
Question 7.
Question 8.
x = 14.11
Explanation:
$$\frac { AB }{ AC }$$ = $$\frac { BD }{ BC }$$
$$\frac { 16 }{ 34 }$$ = $$\frac { x }{ 30 }$$
x = 14.11
Question 9.
Question 10.
x = 2.77
Explanation:
$$\frac { 5.8 }{ 3.5 }$$ = $$\frac { 4.6 }{ x }$$
x = 2.77
In Exercises 11 – 18, find the geometric mean of the two numbers.
Question 11.
8 and 32
Question 12.
9 and 16
x = √(9 x 16)
x = 12
Question 13.
14 and 20
Question 14.
25 and 35
x = √(25 x 35)
x = 29.5
Question 15.
16 and 25
Question 16.
8 and 28
x = √(8 x 28)
x = 14.96
Question 17.
17 and 36
Question 18.
24 and 45
x = √(24 x 45)
x = 32.86
In Exercises 19 – 26. find the value of the variable.
Question 19.
Question 20.
y = √(5 x 8)
y = √40
y = 2√10
Question 21.
Question 22.
10 • 10 = 25 • x
100 = 25x
x = 4
Question 23.
Question 24.
b² = 16(16 + 6)
b² = 16(22) = 352
b = 18.76
Question 25.
Question 26.
x² = 8(8 + 2)
x² = 8(10) = 80
x = 8.9
ERROR ANALYSIS
In Exercises 27 and 28, describe and correct the error in writing an equation for the given diagram.
Question 27.
Question 28.
d² = g • e
MODELING WITH MATHEMATICS
In Exercises 29 and 30, use the diagram.
Question 29.
You want to determine the height of a monument at a local park. You use a cardboard square to line up the top and bottom of the monument, as shown at the above left. Your friend measures the vertical distance from the ground to your eye and the horizontal distance from you to the monument. Approximate the height of the monument.
Question 30.
Your classmate is standing on the other side of the monument. She has a piece of rope staked at the base of the monument. She extends the rope to the cardboard square she is holding lined up to the top and bottom of the monument. Use the information in the diagram above to approximate the height of the monument. Do you get the same answer as in Exercise 29? Explain your reasoning.
MATHEMATICAL CONNECTIONS
In Exercises 31 – 34. find the value(s) of the variable(s).
Question 31.
Question 32.
$$\frac { 6 }{ b + 3 }$$ = $$\frac { 8 }{ 6 }$$
36 = 8(b + 3)
36 = 8b + 24
8b = 12
b = $$\frac { 3 }{ 2 }$$
Question 33.
Question 34.
x = 42.66, y = 40, z = 53
Explanation:
$$\frac { 24 }{ 32 }$$ = $$\frac { 32 }{ x }$$
0.75 = $$\frac { 32 }{ x }$$
x = 42.66
y = √24² + 32²
y = √576 + 1024 = 40
z = √42.66² + 32² = √1819.87 + 1024 = 53
Question 35.
REASONING
Use the diagram. Decide which proportions are true. Select all that apply.
(A) $$\frac{D B}{D C}=\frac{D A}{D B}$$
(B) $$\frac{B A}{C B}=\frac{C B}{B D}$$
(C) $$\frac{C A}{B \Lambda}=\frac{B A}{C A}$$
(D) $$\frac{D B}{B C}=\frac{D A}{B A}$$
Question 36.
ANALYZING RELATIONSHIPS
You are designing a diamond-shaped kite. You know that AD = 44.8 centimeters, DC = 72 centimeters, and AC = 84.8 centimeters. You Want to use a straight crossbar $$\overline{B D}$$. About how long should it be? Explain your reasoning.
BD = 76.12
Explanation:
AD = 44.8 cm, DC = 72 cm, and AC = 84.8 cm
Two disjoint pairs of consecutive sides are congruent.
So, AD = AB = 44.8 cm
DC = BC = 72 cm
The diagonals are perpendicular.
So, AC ⊥ BD
AC = AO + OC
AX = x + y = 84.8 — (i)
Perpendicular bisects the diagonal BD into equal parts let it be z.
BD = BO + OD
BD = z + z
Using pythagorean theorem
44.8² = x² + z² —- (ii)
72² = y² + z² —– (iii)
Subtract (ii) and (iii)
72² – 44.8² = y²+ z² – x² – z²
5184 – 2007.04 = (x + y) (x – y)
3176.96 = (84.8)(x – y)
37.464 = x – y —- (iv)
x + y + x – y = 84.8 + 37.464
2x = 122.264
x = 61.132
x + y = 84.8
61.132 + y = 84.8
y = 23.668
44.8² = x² + z²
z = 38.06
BD = z + z
BD = 76.12
Question 37.
ANALYZING RELATIONSHIPS
Use the Geometric Mean Theorems (Theorems 9.7 and 9.8) to find AC and BD.
Question 38.
HOW DO YOU SEE IT?
In which of the following triangles does the Geometric Mean (Altitude) Theorem (Theorem 9.7) apply?
(A)
(B)
(C)
(D)
Question 39.
PROVING A THEOREM
Use the diagram of ∆ABC. Copy and complete the proof of the Pythagorean Theorem (Theorem 9. 1).
Given In ∆ABC, ∆BCA is a right angle.
Prove c2 = a2 + b2
Statements Reasons 1. In ∆ABC, ∠BCA is a right angle. 1. ________________________________ 2. Draw a perpendicular segment (altitude) from C to $$\overline{A B}$$. 2. Perpendicular Postulate (Postulate 3.2) 3. ce = a2 and cf = b2 3. ________________________________ 4. ce + b2 = ___ + b2 4. Addition Property of Equality 5. ce + cf = a2 + b2 5. ________________________________ 6. c(e + f) a2 + b2 6. ________________________________ 7. e + f = ________ 7. Segment Addition Postulate (Postulate 1.2) 8. c • c = a2 + b2 8. ________________________________ 9. c2 = a2 + b2 9. Simplify.
Question 40.
MAKING AN ARGUMENT
Your friend claims the geometric mean of 4 and 9 is 6. and then labels the triangle, as shown. Is your friend correct? Explain your reasoning.
G.M = √(4 x 9) = √36 = 6
My friend is correct.
In Exercises 41 and 42, use the given statements to prove the theorem.
Gien ∆ABC is a right triangle.
Altitude $$\overline{C D}$$ is dravn to hypotenuse $$\overline{A B}$$.
Question 41.
PROVING A THEOREM
Prove the Geometric Mean (Altitude) Theorem (Theorem 9.7) b showing that CD2 = AD • BD.
Question 42.
PROVING A THEOREM
Prove the Geometric Mean ( Leg) Theorem (Theorem 9.8) b showing that CB2 = DB • AB and AC2 = AD • AB.
Question 43.
CRITICAL THINKING
Draw a right isosceles triangle and label the two leg lengths x. Then draw the altitude to the hypotenuse and label its length y. Now, use the Right Triangle Similarity Theorem (Theorem 9.6) to draw the three similar triangles from the image and label an side length that is equal to either x or y. What can you conclude about the relationship between the two smaller triangles? Explain your reasoning.
Question 44.
THOUGHT PROVOKING
The arithmetic mean and geometric mean of two nonnegative numbers x and y are shown.
arithmetic mean = $$\frac{x+y}{2}$$
geometric mean = $$\sqrt{x y}$$
Question 45.
PROVING A THEOREM
Prove the Right Triangle Similarity Theorem (Theorem 9.6) by proving three similarity statements.
Given ∆ABC is a right triangle.
Altitude $$\overline{C D}$$ is drawn to hvpotenuse $$\overline{A B}$$.
Prove ∆CBD ~ ∆ABC, ∆ACD ~ ∆ABC,
∆CBD ~ ∆ACD
Maintaining Mathematical proficiency
Solve the equation for x.
Question 46.
13 = $$\frac{x}{5}$$
13 = $$\frac{x}{5}$$
x = 65
Question 47.
29 = $$\frac{x}{4}$$
Question 48.
9 = $$\frac{78}{x}$$
9 = $$\frac{78}{x}$$
9x = 78
x = 8.6
Question 49.
30 = $$\frac{115}{x}$$
### 9.1 to 9.3 Quiz
Find the value of x. Tell whether the side lengths form a Pythagorean triple.
Question 1.
x = 15
Explanation:
x² = 9² + 12²
x² = 81 + 144
x² = 225
x = 15
Question 2.
x = 10.63
Explanation:
x² = 7² + 8² = 49 + 64
x = √113
x = 10.63
Question 3.
x = 4√3
Explanation:
8² = x² + 4²
64 = x² + 16
x² = 48
x = 4√3
Verify that the segment lengths form a triangle. Is the triangle acute, right, or obtuse?
(Section 9.1)
Question 4.
24, 32, and 40
Triangle is a right angle trinagle.
Explanation:
40² = 1600
24² + 32² = 576 + 1024 = 1600
40² = 24² + 32²
So, the triangle is a right angle trinagle.
Question 5.
7, 9, and 13
Triangle is an obtuse trinagle.
Explanation:
13² = 169
7² + 9² = 49 + 81 = 130
13² > 7² + 9²
So, the triangle is an obtuse trinagle.
Question 6.
12, 15, and 10√3
Triangle is an acute trinagle.
Explanation:
15² = 225
12² + (10√3)² = 144 + 300 = 444
15² < 12² + (10√3)²
So, the triangle is an acute trinagle.
Find the values of x and y. Write your answers in the simplest form.
Question 7.
x = 6, y = 6√2
Explanation:
x = 6
hypotenuse = leg • √2
y = 6√2
Question 8.
y = 8√3, x = 16
Explanation:
longer leg = shorter leg • √3
y = 8√3
x² = 8² + (8√3)² = 64 + 192
x = 16
Question 9.
x = 5√2, y = 5√6
Explanation:
longer leg = shorter leg • √3
y = x√3
y = 5√6
hypotenuse = shorter leg • 2
10√2 = 2x
x = 5√2
Find the geometric mean of the two numbers.
Question 10.
6 and 12
G.M = √(6 • 12) = 6√2
Question 11.
15 and 20
G.M = √(15 • 20) = 10√3
Question 12.
18 and 26
G.M = √(18 • 26) = 6√13
Identify the similar right triangles. Then find the value of the variable.
Question 13.
x = √(8 • 4)
x = 4√2
Question 14.
y = √(9 • 6) = 3√6
Question 15.
Question 16.
Television sizes are measured by the length of their diagonal. You want to purchase a television that is at least 40 inches. Should you purchase the television shown? Explain your reasoning.
x² = 20.25² + 36²
x² = 410.0625 + 1296 = 1706.0625
x = 41.30
Yes, i will purchase the television.
Question 17.
Each triangle shown below is a right triangle.
a. Are any of the triangles special right triangles? Explain your reasoning.
A is a similar triangle.
b. List all similar triangles. if any.
B, C and D, E are similar triangles.
c. Find the lengths of the altitudes of triangles B and C.
B altitude = √(9 + 27) = 6
C altitude = √(36 + 72) = 6√3
### 9.4 The Tangent Ratio
Exploration 1
Calculating a Tangent Ratio
Work with a partner
a. Construct ∆ABC, as shown. Construct segments perpendicular to $$\overline{A C}$$ to form right triangles that share vertex A and arc similar to ∆ABC with vertices, as shown.
b. Calculate each given ratio to complete the table for the decimal value of tan A for each right triangle. What can you Conclude?
Exploration 2
Using a calculator
Work with a partner: Use a calculator that has a tangent key to calculate the tangent of 36.87°. Do you get the same result as in Exploration 1? Explain.
ATTENDING TO PRECISION
To be proficient in math, you need to express numerical answers with a degree of precision appropriate for the problem context.
Question 3.
Repeat Exploration 1 for ∆ABC with vertices A(0, 0), B(8, 5), and C(8, 0). Construct the seven perpendicular segments so that not all of them intersect $$\overline{A C}$$ at integer values of x. Discuss your results.
Question 4.
How is a right triangle used to find the tangent of an acute angle? Is there a unique right triangle that must be used?
### Lesson 9.4 The Tangent Ratio
Monitoring progress
Find tan J and tan K. Write each answer as a fraction and as a decimal rounded to four places.
Question 1.
tan K = $$\frac { opposite side }{ adjacent side }$$ = $$\frac { JL }{ KL }$$
= $$\frac { 32 }{ 24 }$$ = $$\frac { 4 }{ 3 }$$ = 1.33
tan J = $$\frac { KL }{ JL }$$ = $$\frac { 24 }{ 32 }$$ = $$\frac { 3 }{ 4 }$$ = 0.75
Question 2.
tan K = $$\frac { LJ }{ LK }$$ = $$\frac { 15 }{ 8 }$$
tan J = $$\frac { LK }{ LJ }$$ = $$\frac { 8 }{ 15 }$$
Find the value of x. Round your answer to the nearest tenth.
Question 3.
Tan 61 = $$\frac { 22 }{ x }$$
1.804 = $$\frac { 22 }{ x }$$
x = 12.1951
Question 4.
tan 56 = $$\frac { x }{ 13 }$$
1.482 = $$\frac { x }{ 13 }$$
x = 19.266
Question 5.
WHAT IF?
In Example 3, the length of the shorter leg is 5 instead of 1. Show that the tangent of 60° is still equal to √3.
longer leg = shorter leg • √3
= 5√3
tan 60 = $$\frac { 5√3 }{ 5 }$$
= √3
Question 6.
You are measuring the height of a lamppost. You stand 40 inches from the base of the lamppost. You measure the angle ot elevation from the ground to the top of the lamppost to be 70°. Find the height h of the lamppost to the nearest inch.
tan 70 = $$\frac { h }{ 40 }$$
2.7474 = $$\frac { h }{ 40 }$$
h = 109.896 in
### Exercise 9.4 The Tangent Ratio
Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
The tangent ratio compares the length of _________ to the length of ___________ .
Question 2.
WRITING
Explain how you know the tangent ratio is constant for a given angle measure.
When two triangles are similar, the corresponding sides are proportional which makes the ratio constant for a given acute angle measurement.
Monitoring Progress and Modeling with Mathematics
In Exercises 3 – 6, find the tangents of the acute angles in the right triangle. Write each answer as a fraction and as a decimal rounded to four decimal places.
Question 3.
Question 4.
tan F = $$\frac { DE }{ EF }$$ = $$\frac { 24 }{ 7 }$$
tan D = $$\frac { EF }{ DE }$$ = $$\frac { 7 }{ 24 }$$
Question 5.
Question 6.
tan K = $$\frac { JL }{ LK }$$ = $$\frac { 3 }{ 5 }$$
tan J = $$\frac { LK }{ JL }$$ = $$\frac { 5 }{ 3 }$$
In Exercises 7 – 10, find the value of x. Round your answer to the nearest tenth.
Question 7.
Question 8.
tan 27 = $$\frac { x }{ 15 }$$
0.509 = $$\frac { x }{ 15 }$$
x = 7.635
Question 9.
Question 10.
tan 37 = $$\frac { 6 }{ x }$$
0.753 = $$\frac { 6 }{ x }$$
x = 7.968
ERROR ANALYSIS
In Exercises 11 and 12, describe the error in the statement of the tangent ratio. Correct the error if possible. Otherwise, write not possible.
Question 11.
Question 12.
In Exercises 13 and 14, use a special right triangle to find the tangent of the given angle measure.
Question 13.
45°
Question 14.
30°
tan 30° = $$\frac { 1 }{ √3 }$$
Question 15.
MODELING WITH MATHEMATICS
A surveyor is standing 118 Feet from the base of the Washington Monument. The surveyor measures the angle of elevation from the ground to the top of the monument to be 78°. Find the height h of the Washington Monument to the nearest foot.
Question 16.
MODELING WITH MATHEMATICS
Scientists can measure the depths of craters on the moon h looking at photos of shadows. The length of the shadow cast by the edge of a crater is 500 meters. The angle of elevation of the rays of the Sun is 55°. Estimate the depth d of the crater.
tan 55 = $$\frac { d }{ 500 }$$
1.428 = $$\frac { d }{ 500 }$$
d = 714 m
The depth of the crater is 714 m
Question 17.
USING STRUCTURE
Find the tangent of the smaller acute angle in a right triangle with side lengths 5, 12, and 13.
Question 18.
USING STRUCTURE
Find the tangent 0f the larger acute angle in a right triangle with side lengths 3, 4, and 5.
tan x = $$\frac { 4 }{ 3 }$$
Question 19.
REASONING
How does the tangent of an acute angle in a right triangle change as the angle measure increases? Justify your answer.
Question 20.
CRITICAL THINKING
For what angle measure(s) is the tangent of an acute angle in a right triangle equal to 1? greater than 1? less than 1? Justify your answer.
In order for the tangent of an angle to equal 1, the opposite and adjacent sides of a right triangle must be the same. This means the right triangle is an isosceles right triangle so the angles are 45 – 45 – 90. The acute angle must be 1. In order for the tangent to be greater than 1, the opposite side must be greater than the adjacent side. This means the angle must be between 45 and 90 degrees. If the tangent is less than 1, this means the opposite side must be smaller than the adjacent side. The acute angle must be between 0 and 45.
Question 21.
MAKING AN ARGUMENT
Your family room has a sliding-glass door. You want to buy an awning for the door that will be just long enough to keep the Sun out when it is at its highest point in the sky. The angle of elevation of the rays of the Sun at this points is 70°, and the height of the door is 8 feet. Your sister claims you can determine how far the overhang should extend by multiplying 8 by tan 70°. Is your sister correct? Explain.
Question 22.
HOW DO YOU SEE IT?
Write expressions for the tangent of each acute angle in the right triangle. Explain how the tangent of one acute angle is related to the tangent of the other acute angle. What kind of angle pair is ∠A and ∠B?
tan A = $$\frac { BC }{ AC }$$ = $$\frac { a }{ b }$$
tan B = $$\frac { AC }{ BC }$$ = $$\frac { b }{ a }$$
Question 23.
REASONING
Explain why it is not possible to find the tangent of a right angle or an obtuse angle.
Question 24.
THOUGHT PROVOKING
To create the diagram below. you begin with an isosceles right triangle with legs 1 unit long. Then the hypotenuse of the first triangle becomes the leg of a second triangle, whose remaining leg is 1 unit long. Continue the diagram Until you have constructed an angle whose tangent is $$\frac{1}{\sqrt{6}}$$. Approximate the measure of this angle.
Question 25.
PROBLEM SOLVING
Your class is having a class picture taken on the lawn. The photographer is positioned 14 feet away from the center of the class. The photographer turns 50° to look at either end of the class.
a. What is the distance between the ends of the class?
b. The photographer turns another 10° either way to see the end of the camera range. If each student needs 2 feet of space. about how many more students can fit at the end of each row? Explain.
Question 26.
PROBLEM SOLVING
Find the perimeter of the figure. where AC = 26, AD = BF, and D is the midpoint of $$\overline{A C}$$.
Maintaining Mathematical proficiency
Find the value of x.
Question 27.
Question 28.
longer side = shorter side • √3
7 = x√3
x = $$\frac { 7 }{ √3 }$$
x = 4.04
Question 29.
### 9.5 The Sine and Cosine Ratios
Exploration 1
Work with a partner: Use dynamic geometry software.
a. Construct ∆ABC, as shown. Construct segments perpendicular to $$\overline{A C}$$ to form right triangles that share vertex A arid are similar to ∆ABC with vertices, as shown.
b. Calculate each given ratio to complete the table for the decimal values of sin A and cos A for each right triangle. What can you conclude?
Question 2.
How is a right triangle used to find the sine and cosine of an acute angle? Is there a unique right triangle that must be used?
Question 3.
In Exploration 1, what is the relationship between ∠A and ∠B in terms of their measures’? Find sin B and cos B. How are these two values related to sin A and cos A? Explain why these relationships exist.
LOOKING FOR STRUCTURE
To be proficient in math, you need to look closely to discern a pattern or structure.
### Lesson 9.5 The Sine and Cosine Ratios
Monitoring Progress
Question 1.
Find sin D, sin F, cos D, and cos F. Write each answer as a fraction and as a decimal rounded to four places.
sin D = $$\frac { 7 }{ 25 }$$
sin F = $$\frac { 24 }{ 25 }$$
cos D = $$\frac { 24 }{ 25 }$$
cos F = $$\frac { 7 }{ 25 }$$
Explanation:
sin D = $$\frac { EF }{ DF }$$ = $$\frac { 7 }{ 25 }$$
sin F = $$\frac { DE }{ DF }$$ = $$\frac { 24 }{ 25 }$$
cos D = $$\frac { DE }{ DF }$$ = $$\frac { 24 }{ 25 }$$
cos F = $$\frac { EF }{ DF }$$ = $$\frac { 7 }{ 25 }$$
Question 2.
Write cos 23° in terms of sine.
cos X = sin(90 – X)
cos 23° = sin (90 – 23)
= sin(67)
So, cos 23° = sin 67°
Question 3.
Find the values of u and t using sine and cosine. Round your answers to the nearest tenth.
t = 7.2, u = 3.3
Explanation:
sin 65 = $$\frac { t }{ 8 }$$
0.906 = $$\frac { t }{ 8 }$$
t = 7.2
cos 65 = $$\frac { u }{ 8 }$$
0.422 x 8 = u
u = 3.3
Question 4.
Find the sine and cosine of a 60° angle.
sin 60° = $$\frac { √3 }{ 2 }$$
cos 60° = $$\frac { 1 }{ 2 }$$
Question 5.
WHAT IF?
In Example 6, the angle of depression is 28°. Find the distance x you ski down the mountain to the nearest foot.
sin 28° = $$\frac { 1200 }{ x }$$
x = $$\frac { 1200 }{ 0.469 }$$
x = 2558.6
### Exercise 9.5 The Sine and Cosine Ratios
Vocabulary and Core Concept Check
Question 1.
VOCABULARY
The sine raio compares the length of ______________ to the length of _____________
Question 2.
WHICH ONE DOESN’T BELONG?
Which ratio does not belong with the other three? Explain your reasoning.
sin B
sin B = $$\frac { AC }{ BC }$$
cos C
cos C = $$\frac { AC }{ BC }$$
tan B
tan B = $$\frac { AC }{ AB }$$
$$\frac{A C}{B C}$$
$$\frac{A C}{B C}$$ = sin B
Monitoring Progress and Modeling with Mathematics
In Exercises 3 – 8, find sin D, sin E, cos D, and cos E. Write each answer as a Fraction and as a decimal rounded to four places.
Question 3.
Question 4.
sin D = $$\frac { 35 }{ 37 }$$
sin E = $$\frac { 12 }{ 37 }$$
cos D = $$\frac { 12 }{ 37 }$$
cos E = $$\frac { 35 }{ 37 }$$
Question 5.
Question 6.
sin D = $$\frac { 36 }{ 45 }$$
sin E = $$\frac { 27 }{ 45 }$$
cos D = $$\frac { 27 }{ 45 }$$
cos E = $$\frac { 36 }{ 45 }$$
Question 7.
Question 8.
sin D = $$\frac { 8 }{ 17 }$$
sin E = $$\frac { 15 }{ 17 }$$
cos D = $$\frac { 15 }{ 17 }$$
cos E = $$\frac { 8 }{ 17 }$$
In Exercises 9 – 12. write the expression in terms of cosine.
Question 9.
sin 37°
Question 10.
sin 81°
sin 81° = cos(90° – 81°) = cos9°
Question 11.
sin 29°
Question 12.
sin 64°
sin 64° = cos(90° – 64°) = cos 26°
In Exercise 13 – 16, write the expression in terms of sine.
Question 13.
cos 59°
Question 14.
cos 42°
cos 42° = sin(90° – 42°) = sin 48°
Question 15.
cos 73°
Question 16.
cos 18°
cos 18° = sin(90° – 18°) = sin 72°
In Exercises 17 – 22, find the value of each variable using sine and cosine. Round your answers to the nearest tenth.
Question 17.
Question 18.
p = 30.5, q = 14.8
Explanation:
sin 64° = $$\frac { p }{34 }$$
p = 0.898 x 34
p = 30.5
cos 64° = $$\frac { q }{ 34 }$$
q = 0.4383 x 34
q = 14.8
Question 19.
Question 20.
s = 17.7, r = 19
Explanation:
sin 43° = $$\frac { s }{26 }$$
s = 0.681 x 26
s = 17.7
cos 43° = $$\frac { r }{ 26 }$$
r = 0.731 x 26
r = 19
Question 21.
Question 22.
m = 6.7, n = 10.44
Explanation:
sin 50° = $$\frac { 8 }{n }$$
0.766 = $$\frac { 8 }{n }$$
n = 10.44
cos 50° = $$\frac { m }{ n }$$
0.642 = $$\frac { m }{ 10.44 }$$
m = 6.7
Question 23.
REASONING
Which ratios are equal? Select all that apply.
sin X
cos X
sin Z
cos Z
Question 24.
REASONING
Which ratios arc equal to $$\frac{1}{2}$$ Select all
sin L
sin L = $$\frac { 2 }{ 4 }$$ = $$\frac { 1 }{ 2 }$$
cos L
cos L = $$\frac { 2√3 }{ 4 }$$ = $$\frac { √3 }{ 2 }$$
sin J
sin J = $$\frac { 2√3 }{ 4 }$$ = $$\frac { √3 }{ 2 }$$
cos J
cos J = $$\frac { 2 }{ 4 }$$ = $$\frac { 1 }{ 2 }$$
Question 25.
ERROR ANALYSIS
Describe and correct the error in finding sin A.
Question 26.
WRITING
Explain how to tell which side of a right triangle is adjacent to an angle and which side is the hypotenuse.
Question 27.
MODELING WITH MATHEMATICS
The top of the slide is 12 feet from the ground and has an angle of depression of 53°. What is the length of the slide?
Question 28.
MODELING WITH MATHEMATICS
Find the horizontal distance x the escalator covers.
cos 41 = $$\frac { x }{ 26 }$$
0.754 = $$\frac { x }{ 26 }$$
x = 19.6 ft
Question 29.
PROBLEM SOLVING
You are flying a kite with 20 feet of string extended. The angle of elevation from the spool of string to the kite is 67°.
a. Draw and label a diagram that represents the situation.
b. How far off the ground is the kite if you hold the spool 5 feet off the ground? Describe how the height where you hold the spool affects the height of the kite.
Question 30.
MODELING WITH MATHEMATICS
Planes that fly at high speeds and low elevations have radar s sterns that can determine the range of an obstacle and the angle of elevation to the top of the obstacle. The radar of a plane flying at an altitude of 20,000 feet detects a tower that is 25,000 feet away. with an angle of elevation of 1°
a. How many feet must the plane rise to pass over the tower?
sin 1 = $$\frac { h }{ 25000 }$$
0.017 = $$\frac { h }{ 25000 }$$
h = 425 ft
425 ft the plane rise to pass over the tower
b. PIanes Caillot come closer than 1000 feet vertically to any object. At what altitude must the plane fly in order to pass over the tower?
Question 31.
MAKING AN ARGUMENT
Your friend uses che equation sin 49° = $$\frac{x}{16}$$ to find BC. Your cousin uses the equation cos 41° = $$\frac{x}{16}$$ to find BC. Who is correct? Explain your reasoning.
Question 32.
WRITING
Describe what you must know about a triangle in order to use the sine ratio and what you must know about a triangle in order to use the cosine ratio
sin = $$\frac { opposite side }{ hypotenuse }$$
cos = $$\frac { adjacent side }{ hypotenuse }$$
Question 33.
MATHEMATICAL CONNECTIONS
If ∆EQU is equilateral and ∆RGT is a right triangle with RG = 2, RT = 1. and m ∠ T = 90°, show that sin E = cos G.
Question 34.
MODELING WITH MATHEMATICS
Submarines use sonar systems, which are similar to radar systems, to detect obstacles, Sonar systems use sound to detect objects under water.
a. You are traveling underwater in a submarine. The sonar system detects an iceberg 4000 meters a head, with an angle of depression of 34° to the bottom of the iceberg. How many meters must the submarine lower to pass under the iceberg?
tan 34 = $$\frac { x }{ 4000 }$$
.674 = $$\frac { x }{ 4000 }$$
x = 2696
b. The sonar system then detects a sunken ship 1500 meters ahead. with an angle of elevation of 19° to the highest part of the sunken ship. How many meters must the submarine rise to pass over the sunken ship?
tan 19 = $$\frac { x }{ 1500 }$$
0.344 = $$\frac { x }{ 1500 }$$
x = 516 m
Question 35.
ABSTRACT REASONING
Make a conjecture about how you could use trigonometric ratios to find angle measures in a triangle.
Question 36.
HOW DO YOU SEE IT?
Using only the given information, would you use a sine ratio or a cosine ratio to find the length of the hypotenuse? Explain your reasoning.
sin 29 = $$\frac { 9 }{ x }$$
0.48 = $$\frac { 9 }{ x }$$
x = 18.75
The length of hypotenuse is 18.75
Question 37.
MULTIPLE REPRESENTATIONS
You are standing on a cliff above an ocean. You see a sailboat from your vantage point 30 feet above the ocean.
a. Draw and label a diagram of the situation.
b. Make a table showing the angle of depression and the length of your line of sight. Use the angles 40°, 50°, 60°, 70°, and 80°.
c. Graph the values you found in part (b), with the angle measures on the x-axis.
d. Predict the length of the line of sight when the angle of depression is 30°.
Question 38.
THOUGHT PROVOKING
One of the following infinite series represents sin x and the other one represents cos x (where x is measured in radians). Which is which? Justify your answer. Then use each series to approximate the sine and cosine of $$\frac{\pi}{6}$$.
(Hints: π = 180°; 5! = 5 • 4 • 3 • 2 • 1; Find the values that the sine and cosine ratios approach as the angle measure approaches zero).
a.
For x = 0
0 – $$\frac { 0³ }{ 3! }$$ + $$\frac { 0⁵ }{ 5! }$$ – $$\frac { 0⁷ }{ 7! }$$ + . . . = 0
sin x = x – $$\frac { x³ }{ 3! }$$ + $$\frac { x⁵ }{ 5! }$$ – $$\frac { x⁷ }{ 7! }$$ + . . .
sin $$\frac { π }{ 6 }$$ = 0.5
b.
1 – $$\frac { 1² }{ 2! }$$ + $$\frac { 1⁴ }{ 4! }$$ – $$\frac { 1⁶ }{ 6! }$$ + . . = 1
cos x =x1 – $$\frac { x² }{ 2! }$$ + $$\frac { x⁴ }{ 4! }$$ – $$\frac { x⁶ }{ 6! }$$ + . .
cos $$\frac { π }{ 6 }$$ = 0.86
Question 39.
CRITICAL THINKING
Let A be any acute angle of a right triangle. Show that
(a) tan A = $$\frac{\sin A}{\cos A}$$ and
(b) (sin A)2 + (cos A)2 = 1.
Question 40.
CRITICAL THINKING
Explain why the area ∆ ABC in the diagram can be found using the formula Area = $$\frac{1}{2}$$ ab sin C. Then calculate the area when a = 4, b = 7, and m∠C = 40°:
Area = $$\frac{1}{2}$$ ab sin C
= $$\frac{1}{2}$$ (4 x 7) sin 40°
= 14 x 0.642
= 8.988
Maintaining Mathematical Proficiency
Find the value of x. Tell whether the side lengths form a Pythagorean triple.
Question 41.
Question 42.
x = 12√2
Explanation:
c² = a² + b²
x² = 12² + 12²
x² = 144 + 144
x² = 288
x = 12√2
Question 43.
Question 44.
x = 6√2
Explanation:
c² = a² + b²
9² = x² + 3²
81 = x² + 9
x² = 81 – 9
x = 6√2
### 9.6 Solving Right Triangles
Exploration 1
Solving Special Right Triangles
Work with a partner. Use the figures to find the values of the sine and cosine of ∠A and ∠B. Use these values to find the measures of ∠A and ∠B. Use dynamic geometry software to verify your answers.
a.
b.
Exploration 2
Solving Right Triangles
Work with a partner: You can use a calculator to find the measure of an angle when you know the value of the sine, cosine, or tangent of the rule. Use the inverse sine, inverse cosine, 0r inverse tangent feature of your calculator to approximate the measures of ∠A and ∠B to the nearest tenth of a degree. Then use dynamic geometry software to verify your answers.
ATTENDING TO PRECISION
To be proficient in math, you need to calculate accurately and efficiently, expressing numerical answers with a degree of precision appropriate for the problem context.
a.
b.
Question 3.
When you know the lengths of the sides of a right triangle, how can you find the measures of the two acute angles?
Question 4.
A ladder leaning against a building forms a right triangle with the building and the ground. The legs of the right triangle (in meters) form a 5-12-13 Pythagorean triple. Find the measures of the two acute angles to the nearest tenth of a degree.
### Lesson 9.6 Solving Right Triangles
Determine which of the two acute angles has the given trigonometric ratio.
Question 1.
The sine of the angle is $$\frac{12}{13}$$.
Sin E = $$\frac{12}{13}$$
m∠E = sin-1($$\frac{12}{13}$$) = 67.3°
Question 2.
The tangent of the angle is $$\frac{5}{12}$$
tan F = $$\frac{5}{12}$$
m∠F = tan-1($$\frac{5}{12}$$) = 22.6°
Let ∠G, ∠H, and ∠K be acute angles. Use a calculator to approximate the measures of ∠G, ∠H, and ∠K to the nearest tenth of a degree.
Question 3.
tan G = 0.43
∠G = inverse tan of 0.43 = 23.3°
Question 4.
sin H = 0.68
∠H = inverse sin of 0.68 = 42.8°
Question 5.
cos K = 0.94
∠K = inverse cos of 0.94 = 19.9°
Solve the right triangle. Round decimal answers to the nearest tenth.
Question 6.
DE = 29, ∠D = 46.05°, ∠E = 42.84°
Explanation:
c² = a² + b²
x² = 20² + 21²
x² = 400 + 441
x² = 841
x = 29
sin D = $$\frac { 21 }{ 29 }$$
∠D = 46.05
sin E = $$\frac { 20 }{ 29 }$$
∠E = 42.84
Question 7.
GJ = 60, ∠G = 56.09°, ∠H = 33.3°
Explanation:
c² = a² + b²
109² = 91² + x²
x² = 11881 – 8281
x² = 3600
x = 60
sin G = $$\frac { 91 }{ 109 }$$
∠G = 56.09
sin H = $$\frac { 60 }{ 109 }$$
∠H = 33.3
Question 8.
Solve the right triangle. Round decimal answers to the nearest tenth.
XY = 13.82, YZ = 6.69, ∠Y = 37.5
Explanation:
cos 52 = $$\frac { 8.5 }{ XY }$$
0.615 = $$\frac { 8.5 }{ XY }$$
XY = 13.82
sin 52 = $$\frac { YZ }{ XY }$$
0.788 = $$\frac { YZ }{ 8.5 }$$
YZ = 6.69
sin Y = $$\frac { 8.5 }{ 13.82 }$$
∠Y = 37.5
Question 9.
WHAT IF?
In Example 5, suppose another raked stage is 20 feet long from front to back with a total rise of 2 feet. Is the raked stage within your desired range?
x = inverse sine of $$\frac { 2 }{ 20 }$$
x = 5.7°
### Exercise 9.6 Solving Right Triangles
Question 1.
COMPLETE THE SENTENCE
To solve a right triangle means to find the measures of all its ________ and _______ .
Question 2.
WRITING
Explain when you can use a trigonometric ratio to find a side length of a right triangle and when you can use the Pythagorean Theorem (Theorem 9.1 ).
Monitoring Progress and Modeling with Mathematics
In Exercises 3 – 6. determine which of the two acute angles has the given trigonometric ratio.
Question 3.
The cosine of the angle is $$\frac{4}{5}$$
Question 4.
The sine of the angle is $$\frac{5}{11}$$
Sin(angle) = $$\frac { opposite }{ hypo }$$
sin A = $$\frac{5}{11}$$
The acute angle that has a sine of the angle is $$\frac{5}{11}$$ is ∠A.
Question 5.
The sine of the angle is 0.95.
Question 6.
The tangent of the angle is 1.5.
tan(angle) = $$\frac { opposite }{ adjacent }$$
1.5 = $$\frac { 18 }{ 12 }$$
tan C = 1.5
The acute angle that has a tangent of the angle is 1.5 ∠C.
In Exercises 7 – 12, let ∠D be an acute angle. Use a calculator to approximate the measure of ∠D to the nearest tenth of a degree.
Question 7.
sin D = 0.75
Question 8.
sin D = 0.19
sin D = 0.19
∠D = inverse sine of 0.19
∠D = 10.9°
Question 9.
cos D = 0.33
Question 10.
cos D = 0.64
cos D = 0.64
∠D = inverse cos of 0.64
∠D = 50.2°
Question 11.
tan D = 0.28
Question 12.
tan D = 0.72
tan D = 0.72
∠D = inverse tan of 0.72
∠D = 35.8°
In Exercises 13 – 18. solve the right triangle. Round decimal answers to the nearest tenth.
Question 13.
Question 14.
ED = 2√65, ∠E = 59.3, ∠D = 29.7
Explanation:
c² = 8² + 14²
x² = 64 + 196
x² = 260
x = 2√65
sin E = $$\frac { 14 }{ 2√65 }$$
∠E = 59.3
sin D = $$\frac { 8 }{ 2√65 }$$
∠D = 29.7
Question 15.
Question 16.
HJ = 2√15, ∠G = 28.9, ∠J = 61
Explanation:
c² = a² + b²
16² = 14² + x²
x² = 256 – 196
x² = 60
x = 2√15
sin G = $$\frac { 2√15 }{ 16 }$$
∠G = 28.9
sin J = $$\frac { 14 }{ 16 }$$
∠J = 61
Question 17.
Question 18.
RT = 17.8, RS = 9.68, ∠T = 32.8
Explanation:
sin 57 = $$\frac { 15 }{ x }$$
0.838 = $$\frac { 15 }{ x }$$
x = 17.899
RT = 17.8
cos 57 = $$\frac { x }{ 17.8 }$$
0.544 = $$\frac { x }{ 17.8 }$$
x = 9.68
RS = 9.68
sin T = $$\frac { 9.68 }{ 17.8 }$$
∠T = 32.8
Question 19.
ERROR ANALYSIS
Describe and correct the error in using an inverse trigonometric ratio.
Question 20.
PROBLEM SOLVING
In order to unload clay easily. the body of a dump truck must be elevated to at least 45° The body of a dump truck that is 14 feet long has been raised 8 feet. Will the clay pour out easily? Explain your reasoning.
Angle of elevation: sin x = $$\frac { 8 }{ 14 }$$
x = inverse sine of $$\frac { 8 }{ 14 }$$ = 34.9
The clay will not pour out easily.
Question 21.
PROBLEM SOLVING
You are standing on a footbridge that is 12 feet above a lake. You look down and see a duck in the water. The duck is 7 feet away from the footbridge. What is the angle of elevation from the duck to you
Question 22.
HOW DO YOU SEE IT?
Write three expressions that can be used to approximate the measure of ∠A. Which expression would you choose? Explain your choice.
Three expressions are ∠A = inverse tan of ($$\frac { 15 }{ 22 }$$) = 34.2°
∠A = inverse sine of ($$\frac { 15 }{ BA }$$)
∠A = inverse cos of ($$\frac { 22 }{ BA }$$)
Question 23.
MODELING WITH MATHEMATICS
The Uniform Federal Accessibility Standards specify that awheel chair ramp may not have an incline greater than 4.76. You want to build a ramp with a vertical rise of 8 inches. you want to minimize the horizontal distance taken up by the ramp. Draw a diagram showing the approximate dimensions of your ramp.
Question 24.
MODELING WITH MATHEMATICS
The horizontal part of a step is called the tread. The vertical part is called the riser. The recommended riser – to – tread ratio is 7 inches : 11 inches.
a. Find the value of x for stairs built using the recommended riser-to-tread ratio.
b. you want to build stairs that are less steep than the stairs in part (a). Give an example of a riser – to – tread ratio that you could use. Find the value of x for your stairs.
Question 25.
USING TOOLS
Find the measure of ∠R without using a protractor. Justify your technique.
Question 26.
MAKING AN ARGUMENT
Your friend claims that tan-1x = $$\frac{1}{\tan x}$$. Is your friend correct? Explain your reasoning.
No
For example
tan-1(√3) = 60
$$\frac{1}{\tan √3}$$ = 33.1
USING STRUCTURE
In Exercises 27 and 28, solve each triangle.
Question 27.
∆JKM and ∆LKM
Question 28.
∆TUS and ∆VTW
TS = 8.2, UT = 7.3, ∠T = 28.6
TV = 13.2, TW = 9.6, ∠V = 46, ∠T = 42.84
Explanation:
tan 64 = $$\frac { TS }{ 4 }$$
TS = 2.05 x 4
TS = 8.2
sin 64 = $$\frac { UT }{ 8.2 }$$
0.898= $$\frac { UT }{ 8.2 }$$
UT = 7.3
sin T = $$\frac { 4 }{ 8.2 }$$
∠T = 28.6
TV = TS + SV
TV = 8.2 + 5 = 13.2
13.2² = TW² + 9²
TW² = 174.24 – 81
TW = 9.6
sin V = $$\frac { 9.6 }{ 13.2 }$$
∠V = 46
sin T = $$\frac { 9 }{ 13.2 }$$
∠T = 42.84
Question 29.
MATHEMATICAL CONNECTIONS
Write an expression that can be used to find the measure of the acute angle formed by each line and the x-axis. Then approximate the angle measure to the nearest tenth of a degree.
a. y = 3x
b. y = $$\frac{4}{3}$$x + 4
Question 30.
THOUGHT PROVOKING
a. sin-1 (sin x)
sin-1 (sin x) = x
b. tan(tan-1 y)
tan(tan-1 y) = y
C. cos(cos-1 z)
cos(cos-1 z) = z
Question 31.
REASONING
Explain why the expression sin-1 (1.2) does not make sense.
Question 32.
USING STRUCTURE
The perimeter of the rectangle ABCD is 16 centimeters. and the ratio of its width to its length is 1 : 3. Segment BD divides the rectangle into two congruent triangles. Find the side lengths and angle measures of these two triangles.
The perimeter of the rectangle ABCD is 16 centimeters
2(l + b) = 16
l + b = 8
b : l = 1 : 3
4x = 8
x = 2
l = 6, b = 2
BD = √(6² + 2²) = √40 = 2√10
sin B = $$\frac { AD }{ BD }$$ = $$\frac { 2 }{ 2√10 }$$
∠ABD = 18.4
∠CBD = 71.6
sin D = $$\frac { AB }{ BD }$$ = $$\frac { 6 }{ 2√10 }$$
∠CDB = 19
Maintaining Mathematical Proficiency
Solve the equation
Question 33.
$$\frac{12}{x}=\frac{3}{2}$$
Question 34.
$$\frac{13}{9}=\frac{x}{18}$$
$$\frac{13}{9}=\frac{x}{18}$$
x = 1.44 x 18
x = 26
Question 35.
$$\frac{x}{2.1}=\frac{4.1}{3.5}$$
Question 36.
$$\frac{5.6}{12.7}=\frac{4.9}{x}$$
$$\frac{5.6}{12.7}=\frac{4.9}{x}$$
0.44 = 4.9/x
x = 11.13
### 9.7 Law of Sines and Law of Cosines
Exploration 1
Discovering the Law of Sines
Work with a partner.
a. Copy and complete the table for the triangle shown. What can you conclude?
b. Use dynamic geometry software to draw two other triangles. Copy and complete the table in part (a) for each triangle. Use your results to write a conjecture about the relationship between the sines of the angles and the lengths of the sides of a triangle.
USING TOOLS STRATEGICALLY
To be proficient in math, you need to use technology to compare predictions with data.
Exploration 2
Discovering the Law of Cosines
Work with a partner:
a. Copy and complete the table for the triangle in Exploration 1 (a). What can you conclude?
b. Use dynamic geometry software to draw two other triangles. Copy and complete the table in part (a) for each triangle. Use your results to write a conjecture about what you observe in the completed tables.
Question 3.
What are the Law of Sines and the Law of Cosines?
Question 4.
When would you use the Law of Sines to solve a triangle? When would you use the Law of Cosines to solve a triangle?
### Lesson 9.7 Law of Sines and Law of Cosines
Monitoring Progress
Use a calculator to find the trigonometric ratio. Round your answer to four decimal places.
Question 1.
tan 110°
tan 110° = -2.7474
Question 2.
sin 97°
sin 97° = 0.9925
Question 3.
cos 165°
cos 165° = -0.9659
Find the area of ∆ABC with the given side lengths and included angle. Round your answer to the nearest tenth.
Question 4.
m ∠ B = 60°, a = 19, c = 14
Area = 155.18
Explanation:
Area = $$\frac { 1 }{ 2 }$$ ac sin B
= $$\frac { 1 }{ 2 }$$ (19 x 14) sin 60°
= 133 x 0.866
= 155.18
Question 5.
m ∠ C = 29°, a = 38, b = 31
Area = 282.72
Explanation:
Area = $$\frac { 1 }{ 2 }$$ ab sin C
= $$\frac { 1 }{ 2 }$$ (38 x 31) sin 29
= 598 x 0.48
= 282.72
Solve the triangle. Round decimal answers to the nearest tenth.
Question 6.
∠C = 46.6, ∠B = 82.4, AC = 23.57
Explanation:
Using law of sines
$$\frac { c }{ sin C }$$ = $$\frac { a }{ sin A }$$
$$\frac { 17 }{ sin C }$$ = $$\frac { 18 }{ sin 51 }$$
$$\frac { 17 }{ sin C }$$ = $$\frac { 18 }{ 0.77 }$$
sin C = 0.7274
∠C = 46.6
∠A + ∠B + ∠C = 180
51 + 46.6 + ∠B = 180
∠B = 82.4
$$\frac { sin B }{ b }$$ = $$\frac { sin A }{ a }$$
$$\frac { 0.99 }{ b }$$ = $$\frac { 0.77 }{ 18 }$$
b = 23.57
Question 7.
∠B = 31.3, ∠C = 108.7, c = 23.6
Explanation:
$$\frac { sin B }{ b }$$ = $$\frac { sin A }{ a }$$
$$\frac { sin B }{ 13 }$$ = $$\frac { sin 40 }{ 16 }$$
sin B = $$\frac { .64 }{ 16 }$$ x 13
sin B = 0.52
∠B = 31.3
∠A + ∠B + ∠C = 180
40 + 31.3 + ∠C = 180
∠C = 108.7
$$\frac { c }{ sin C }$$ = $$\frac { a }{ sin A }$$
$$\frac { c }{ sin 108.7 }$$ = $$\frac {16 }{ sin 40 }$$
c = $$\frac {16 }{ .64 }$$ x 0.947
c = 23.6
Solve the triangle. Round decimal answers to the nearest tenth.
Question 8.
∠C = 66, a = 4.36, c = 8.27
Explanation:
∠A + ∠B + ∠C = 180
29 + 85 + ∠C = 180
∠C = 66
$$\frac { a }{ sin A }$$ = $$\frac { b }{ sin B }$$ = $$\frac { c }{ sin C }$$
$$\frac { a }{ sin 29 }$$ = $$\frac { 9 }{ sin 85 }$$
$$\frac { a }{ 0.48 }$$ = $$\frac { 9 }{ 0.99 }$$
a = 4.36
$$\frac { b }{ sin B }$$ = $$\frac { c }{ sin C }$$
$$\frac { 9 }{ sin 85 }$$ = $$\frac { c }{ sin 66 }$$
$$\frac { 9 }{ 0.99 }$$ = $$\frac { c }{ 0.91 }$$
c = 8.27
Question 9.
∠A = 29, b = 19.37, c = 20.41
Explanation:
∠A + ∠B + ∠C = 180
∠A + 70 + 81 = 180
∠A = 29
$$\frac { a }{ sin A }$$ = $$\frac { b }{ sin B }$$ = $$\frac { c }{ sin C }$$
$$\frac { 10 }{ sin 29 }$$ = $$\frac { b }{ sin 70 }$$
$$\frac { 10 }{ 0.48 }$$ = $$\frac { b }{ 0.93 }$$
b = 19.37
$$\frac { a }{ sin A }$$ = $$\frac { c }{ sin C }$$
$$\frac { 10 }{ sin 29 }$$ = $$\frac { c }{ sin 81 }$$
$$\frac { 10 }{ 0.48 }$$ = $$\frac { c }{ 0.98 }$$
c = 20.41
Question 10.
WHAT IF?
In Example 5, what would be the length of a bridge from the South Picnic Area to the East Picnic Area?
The length of a bridge from the South Picnic Area to the East Picnic Area is 188 m.
Explanation:
$$\frac { a }{ sin A }$$ = $$\frac { b }{ sin B }$$ = $$\frac { c }{ sin C }$$
$$\frac { a }{ sin 71 }$$ = $$\frac { 150 }{ sin 49 }$$
a = 188
Solve the triangle. Round decimal answers to the nearest tenth.
Question 11.
b = 61.3, ∠A = 46, ∠C = 46
Explanation:
b² = a² + c² − 2ac cos B
b² = 45² + 43² – 2(45)(43) cos 88
b² = 2025 + 1849 – 3870 x 0.03 = 3757.9
b = 61.3
$$\frac { sin A }{ a }$$ = $$\frac { sin B }{ b }$$
$$\frac { sin A }{ 45 }$$ = $$\frac { sin 88 }{ 61.3 }$$
sin A = 0.72
∠A = 46
∠A + ∠B + ∠C = 180
46 + 88 + ∠C = 180
∠C = 46
Question 12.
a = 41.1, ∠C = 35.6, ∠B = 30.4
Explanation:
a² = b² + c² − 2bc cos A
a² = 23² + 26² – 2(23)(26) cos 114
a² = 529 + 676 – 1196 x -0.406
a² = 1690.5
a = 41.1
$$\frac { sin 114 }{ 41.1 }$$ = $$\frac { sin B }{ 23 }$$
0.02 = $$\frac { sin B }{ 23 }$$
sin B = 0.507
∠B = 30.4
∠A + ∠B + ∠C = 180
114 + 30.4 + ∠C = 180
∠C = 35.6
Question 13.
∠A = 41.4, ∠B = 81.8, ∠C = 56.8
Explanation:
a² = b² + c² − 2bc cos A
4² = 6² + 5² – 2(6)(5) cos A
16 = 36 + 25 – 60 cos A
-45 = – 60 cos A
cos A = 0.75
∠A = 41.4
$$\frac { sin 41.4 }{ 4 }$$ = $$\frac { sin B }{ 6 }$$
0.165 = $$\frac { sin B }{ 6 }$$
sin B = 0.99
∠B = 81.8
∠A + ∠B + ∠C = 180
41.4 + 81.8 + ∠C = 180
∠C = 56.8
Question 14.
∠B = 81.8, ∠A = 58.6, ∠C = 39.6
Explanation:
a² = b² + c² − 2bc cos A
23² = 27² + 16² – 2(27)(16) cos A
529 = 729 + 256 – 864 cos A
456 = 864 cos A
cos A = 0.52
∠A = 58.6
$$\frac { sin 58.6 }{ 23 }$$ = $$\frac { sin B }{ 27 }$$
0.03 = $$\frac { sin B }{ 27 }$$
sin B = 0.99
∠B = 81.8
∠A + ∠B + ∠C = 180
58.6 + 81.8 + ∠C = 180
∠C = 39.6
### Exercise 9.7 Law of Sines and Law of Cosines
Vocabulary and Core Concept Check
Question 1.
WRITING
What type of triangle would you use the Law of Sines or the Law of Cosines to solve?
Question 2.
VOCABULARY
What information do you need to use the Law of Sines?
Monitoring progress and Modeling with Mathematics
In Exercises 3 – 8, use a calculator to find the trigonometric ratio, Round your answer to four decimal places.
Question 3.
sin 127°
Question 4.
sin 98°
sin 98° = 0.9902
Question 5.
cos 139°
Question 6.
cos 108°
cos 108° = -0.309
Question 7.
tan 165°
Question 8.
tan 116°
tan 116° = -2.0503
In Exercises 9 – 12, find the area of the triangle. Round your answer to the nearest tenth.
Question 9.
Question 10.
Area = $$\frac{1}{2}$$bc sin A
Area = $$\frac{1}{2}$$(28)(24) sin83
Area = 332.64
Question 11.
Question 12.
Area = $$\frac{1}{2}$$ab sin C
Area = $$\frac{1}{2}$$(15)(7) sin 96
Area = 51.9
In Exercises 13 – 18. solve the triangle. Round decimal answers to the nearest tenth.
Question 13.
Question 14.
∠B = 38.3, ∠A = 37.7, a = 15.7
Explanation:
$$\frac { sin B }{16 }$$ = $$\frac { sin 104 }{ 25 }$$
sin B = 0.62
∠B = 38.3
∠A + ∠B + ∠C = 180
∠A + 38.3 + 104 = 180
∠A = 37.7
$$\frac { sin 37.7 }{ a }$$ = $$\frac { sin 104 }{ 25 }$$
$$\frac { 0.61 }{ a }$$ = 0.0388
a = 15.7
Question 15.
Question 16.
∠B = 65, b = 33.55, a = 24.4
Explanation:
∠A + ∠B + ∠C = 180
42 + 73 + ∠B = 180
∠B = 65
$$\frac { sin B }{ b }$$ = $$\frac { sin C }{ c }$$
$$\frac { sin 65 }{ b }$$ = $$\frac { sin 73 }{ 34 }$$
b = 33.55
$$\frac { sin A }{ a }$$ = $$\frac { sin C }{ c }$$
$$\frac { sin 42 }{ a }$$ = $$\frac { sin 73 }{ 34 }$$
a = 24.4
Question 17.
Question 18.
∠C = 90, b = 39.56, a = 17.6
Explanation:
∠A + ∠B + ∠C = 180
24 + 66 + ∠C = 180
∠C = 90
$$\frac { sin B }{ b }$$ = $$\frac { sin C }{ c }$$
$$\frac { sin 66 }{ b }$$ = $$\frac { sin 90 }{ 43 }$$
$$\frac { .91 }{ b }$$ = 0.023
b = 39.56
$$\frac { sin A }{ a }$$ = $$\frac { sin C }{ c }$$
$$\frac { sin 24 }{a }$$ = $$\frac { sin 90 }{ 43 }$$
$$\frac { 0.406 }{a }$$ = 0.023
a = 17.6
In Exercises 19 – 24, solve the triangle. Round decimal answers to the nearest tenth.
Question 19.
Question 20.
b = 29.9, ∠A = 26.1, ∠C = 15.07
Explanation:
b² = a² + c² – 2ac cos B
b² = 20² + 12² – 2(12)(20) cos 138
b² = 400 + 144 – 480 (-0.74)
b² = 899.2
b = 29.9
$$\frac { sin B }{ b }$$ = $$\frac { sin A }{ a }$$
$$\frac { sin 138 }{ 29.9 }$$ = $$\frac { sin A }{ 20 }$$
$$\frac { 0.66 }{ 29.9 }$$ = $$\frac { sin A }{ 20 }$$
sin A = 0.44
∠A = 26.1
$$\frac { sin B }{ b }$$ = $$\frac { sin C }{ c }$$
$$\frac { sin 138 }{ 29.9 }$$ = $$\frac { sin C }{ 12 }$$
sin C = 0.26
∠C = 15.07
Question 21.
Question 22.
∠A = 107.3, ∠B = 51.6, ∠C = 21.1
Explanation:
b² = a² + c² – 2ac cos B
28² = 18² + 13² – 2(18)(13) cos B
784 = 324 + 169 – 468 cos B
291 = 468 cos B
cos B = 0.62
∠B = 51.6
$$\frac { sin C }{ c }$$ = $$\frac { sin B }{ b }$$
$$\frac { sin C }{ 13 }$$ = $$\frac { sin 51.6 }{ 28 }$$
sin C = 0.36
∠C = 21.1
∠A + ∠B + ∠C = 180
51.6 + 21.1 + ∠A = 180
∠A = 107.3
Question 23.
Question 24.
∠A = 23, ∠B = 132.1, ∠C = 24.9
Explanation:
b² = a² + c² – 2ac cos B
5² = 12² + 13² – 2(12)(13) cos B
25 = 144 + 169 – 312 cos B
288 = 312 cos B
cos B = 0.92
∠B = 23
$$\frac { sin C }{ c }$$ = $$\frac { sin B }{ b }$$
$$\frac { sin C }{ 13 }$$ = $$\frac { sin 23 }{ 5 }$$
sin C = 1.014
∠C = 24.9
∠A + ∠B + ∠C = 180
23 + 24.9 + ∠B = 180
∠B = 132.1
Question 25.
ERROR ANALYSIS
Describe and correct the error in finding m ∠ C.
Question 26.
ERROR ANALYSIS
Describe and correct the error in finding m ∠ A in ∆ABC when a = 19, b = 21, and c = 11.
a² = b² + c² – 2bc cos A
19² = 21² + 11² – 2(21)(11) cos A
361 = 441 + 121 – 462 cosA
201 = 462 cosA
cos A = 0.43
∠A = 64.5
COMPARING METHODS
In Exercise 27 – 32. tell whether you would use the Law of Sines, the Law of Cosines. or the Pythagorean Theorem (Theorem 9.1) and trigonometric ratios to solve the triangle with the given information. Explain your reasoning. Then solve the triangle.
Question 27.
m ∠ A = 72°, m ∠ B = 44°, b = 14
Question 28.
m ∠ B = 98°, m ∠ C = 37°, a = 18
∠A = 45, b = 25.38, c = 15.38
Explanation:
∠A + ∠B + ∠C = 180
∠A + 98 + 37 = 180
∠A = 45
$$\frac { sin B }{ b }$$ = $$\frac { sin A }{ a }$$
$$\frac { sin 98 }{ b }$$ = $$\frac { sin 45 }{ 18 }$$
$$\frac { 0.99 }{ b }$$ = 0.039
b = 25.38
$$\frac { sin A }{ a }$$ = $$\frac { sin C }{ c }$$
$$\frac { sin 45 }{ 18 }$$ = $$\frac { sin 37 }{ c }$$
0.039 = $$\frac { sin 37 }{ c }$$
c = 15.38
Question 29.
m ∠ C = 65°, a = 12, b = 21
Question 30.
m ∠ B = 90°, a = 15, c = 6
b = 3√29, ∠A = 66.9, ∠C = 23.1
Explanation:
b² = a² + c²- 2ac cos B
b² = 15² + 6² – 2(15)(6) cos 90
= 225 + 36 – 180(0)
b² = 261
b = 3√29
$$\frac { sin B }{ b }$$ = $$\frac { sin A }{ a }$$
$$\frac { sin 90 }{ 3√29 }$$ = $$\frac { sin A }{ 15 }$$
sin A = 0.92
∠A = 66.9
∠A + ∠B + ∠C = 180
66.9 + 90 + ∠C = 180
∠C = 23.1
Question 31.
m ∠ C = 40°, b = 27, c = 36
Question 32.
a = 34, b = 19, c = 27
∠B = 33.9, ∠A = 78.5, ∠C = 67.6
Explanation:
b² = a² + c²- 2ac cos B
19² = 34² + 27²- 2(34)(27) cos B
361 = 1156 + 729 – 1836 cos B
cos B = 0.83
∠B = 33.9
$$\frac { sin 33.9 }{ 19 }$$ = $$\frac { sin A }{ 34 }$$
sin A = 0.98
∠A = 78.5
∠A + ∠B + ∠C = 180
78.5 + 33.9 + ∠C = 180
∠C = 67.6
Question 33.
MODELING WITH MATHEMATICS
You and your friend are standing on the baseline of a basketball court. You bounce a basketball to your friend, as shown in the diagram. What is the distance between you and your friend?
Question 34.
MODELING WITH MATHEMATICS
A zip line is constructed across a valley, as shown in the diagram. What is the width w of the valley?
w = 92.5 ft
Explanation:
w² = 25² + 84² – 2(25)(84) cos 102
w² = 7681 – 4200 cos 102
w = 92.5 ft
Question 35.
MODELING WITH MATHEMATICS
You are on the observation deck of the Empire State Building looking at the Chrysler Building. When you turn 145° clockwise, you see the Statue of Liberty. You know that the Chrysler Building and the Empire Slate Building arc about 0.6 mile apart and that the Chrysler Building and the Statue of Liberty are about 5.6 miles apart. Estimate the distance between the Empire State Building and the Statue of Liberty.
Question 36.
MODELING WITH MATHEMATICS
The Leaning Tower of Pisa in Italy has a height of 183 feet and is 4° off vertical. Find the horizontal distance d that the top of the tower is off vertical.
Question 37.
MAKING AN ARGUMENT
Your friend says that the Law of Sines can be used to find JK. Your cousin says that the Law of Cosines can be used to find JK. Who is correct’? Explain your reasoning.
Question 38.
REASONING
Use ∆XYZ
a. Can you use the Law of Sines to solve ∆XYZ ? Explain your reasoning.
b. Can you use another method to solve ∆XYZ ? Explain your reasoning.
Question 39.
MAKING AN ARGUMENT
Your friend calculates the area of the triangle using the formula A = $$\frac{1}{2}$$qr sin S and says that the area is approximately 208.6 square units. Is your friend correct? Explain your reasoning.
Question 40.
MODELING WITH MATHEMATICS
You are fertilizing a triangular garden. One side of the garden is 62 feet long, and another side is 54 feet long. The angle opposite the 62-foot side is 58°.
a. Draw a diagram to represent this situation.
b. Use the Law of Sines to solve the triangle from part (a).
c. One bag of fertilizer covers an area of 200 square feet. How many bags of fertilizer will you need to cover the entire garden?
C = 47.6, A = 74.4, a = 70.4
9 bags of fertilizer.
Question 41.
MODELING WITH MATHEMATICS
A golfer hits a drive 260 yards on a hole that is 400 yards long. The shot is 15° off target.
a. What is the distance x from the golfer’s ball to the hole?
b. Assume the golfer is able to hit the ball precisely the distance found in part (a). What is the maximum angle θ (theta) by which the ball can be off target in order to land no more than 10 yards fr0m the hole?
Question 42.
COMPARING METHODS
A building is constructed on top of a cliff that is 300 meters high. A person standing on level ground below the cliff observes that the angle of elevation to the top of the building is 72° and the angle of elevation to the top of the cliff is 63°.
a. How far away is the person from the base of the cliff?
b. Describe two different methods you can use to find the height of the building. Use one of these methods to find the building’s height.
Consider △SYZ and evaluate d using tangent function
tan SYZ = $$\frac { 300 }{ d }$$
d = $$\frac { 300 }{ tan 63 }$$
d = 152.86
The person is standing 152.86 m away from the base of the cliff.
Consider △XYS and evaluate h + 300
tan XYZ = $$\frac { h + 300 }{ d }$$
h = 152.86 x tan 72 – 300
h = 170.45
The building is 170.45 m high.
Question 43.
MATHEMATICAL CONNECTIONS
Find the values of x and y.
Question 44.
HOW DO YOU SEE IT?
Would you use the Law of Sines or the Law of Cosines to solve the triangle?
Question 45.
REWRITING A FORMULA
A Simplify the Law of Cosines for when the given angle is a right angle.
Question 46.
THOUGHT PROVOKING
Consider any triangle with side lengths of a, b, and c. Calculate the value of s, which is half the perimeter of the triangle. What measurement of the triangle is represented by $$\sqrt{s(s-a)(s-b)(s-c)} ?$$
Question 47.
ANALYZING RELATIONSHIPS
The ambiguous case of the Law of Sines occurs when you are given the measure of one acute angle. the length of one adjacent side, and the length of the side opposite that angle, which is less than the length of the adjacent side. This results in two possible triangles. Using the given information, find two possible solutions for ∆ABC
Draw a diagram for each triangle.
(Hint: The inverse sine function gives only acute angle measures. so consider the acute angle and its supplement for ∠B.)
a. m ∠ A = 40°, a = 13, b = 16
b. m ∠ A = 21°, a = 17, b = 32
Question 48.
ABSTRACT REASONING
Use the Law of Cosines to show that the measure of each angle of an equilateral triangle is 60°. Explain your reasoning.
a² = b² + c²- 2bc cos A
a² = a² + a² – 2 aa cos A
a² = 2a² coas A
cos A = 1/2
∠A = 60
Question 49.
CRITICAL THINKING
An airplane flies 55° east of north from City A to City B. a distance of 470 miles. Another airplane flies 7° north of east from City A to City C. a distance of 890 miles. What is the distance between Cities B and C?
Question 50.
REWRITING A FORMULA
Follow the steps to derive the formula for the area of a triangle.
Area = $$\frac{1}{2}$$ab sin C.
a. Draw the altitude from vertex B to $$\overline{A C}$$. Label the altitude as h. Write a formula for the area of the triangle using h.
b. Write an equation for sin C
c. Use the results of parts (a) and (b) to write a formula for the area of a triangle that does not include h.
Question 51.
PROVING A THEOREM
Follow the steps to use the formula for the area of a triangle to prove the Law of Sines (Theorem 9.9).
a. Use the derivation in Exercise 50 to explain how to derive the three related formulas for the area of a triangle.
Area = $$\frac{1}{2}$$bc sin A,
Area = $$\frac{1}{2}$$ac sin B,
Area = $$\frac{1}{2}$$ab sin C
b. why can you use the formulas in part (a) to write the following statement?
$$\frac{1}{2}$$bc sin A = $$\frac{1}{2}$$ac sin B = $$\frac{1}{2}$$ab sin C
c. Show how to rewrite the statement in part (b) to prove the Law of Sines. Justify each step.
Question 52.
PROVING A THEOREM
Use the given information to complete the two – column proof of the Law of Cosines (Theorem 9.10).
Given $$\overline{B D}$$ is an altitude of ∆ABC.
Prove a2 = b2 + c2 – 2bc cos A
Statements Reasons 1. $$\overline{B D}$$ is an altitude of ∆ABC. 1. Given 2. ∆ADB and ∆CDB are right triangles. 2. _______________________ 3. a2 = (b – x)2 + h2 3. _______________________ 4. _______________________ 4. Expand binomial. 5. x2 + h2 = c2 5. _______________________ 6. _______________________ 6. Substitution Property of Equality 7. cos A = $$\frac{x}{c}$$ 7. _______________________ 8. x = c cos A 8. _______________________ 9. a2 = b2 + c2 – 2bc Cos A 9. _______________________
Maintaining Mathematical Proficiency
Find the radius and diameter of the circle.
Question 53.
Question 54.
The radius is 10 in and the diameter is 20 in.
Question 55.
Question 56.
The radius is 50 in and the diameter is 100 in.
### 9.1 The Pythagorean Theorem
Find the value of x. Then tell whether the side lengths form a Pythagorean triple.
Question 1.
x = 2√34
The sides will not form a Pythagorean triple.
Explanation:
x² = 6² + 10²
x²= 36 + 100
x = 2√34
Question 2.
x = 12
The sides form a Pythagorean triple.
Explanation:
20² = 16²+ x²
400 = 256 + x²
x² = 144
x = 12
Question 3.
x = 2√30
The sides will not form a Pythagorean triple.
Explanation:
13² = 7² + x²
169 = 49 + x²
x = 2√30
Verify that the segments lengths form a triangle. Is the triangle acute, right, or obtuse?
Question 4.
6, 8, and 9
9² = 81
6² + 8² = 36 + 64 = 100
9² < 6² + 8²
So, the triangle is acute
Question 5.
10, 2√2, and 6√3
10² = 100
(2√2)² + (6√3)² = 8 + 108 = 116
So, the triangle is acute.
Question 6.
13, 18 and 3√55
18² = 324
13² + (3√55)² = 169 + 495 = 664
So, the triangle is acute.
### 9.2 Special Right Triangles
Question 7.
hypotenuse = leg • √2
x = 6√2
Question 8.
longer leg = shorter leg • √3
14 = x • √3
x = 8.08
Question 9.
longer leg = shorter leg • √3
x = 8√3 • √3
x = 24
### 9.3 Similar Right Triangles
Identify the similar triangles. Then find the value of x.
Question 10.
9 = √(6x)
81 = 6x
x = $$\frac { 27 }{ 2 }$$
Question 11.
x = √(6 x 4)
x = 2√6
Question 12.
$$\frac { RP }{ RQ }$$ = $$\frac { SP }{ RS }$$
$$\frac { 9 }{ x }$$ = $$\frac { 6 }{ 3 }$$
x = 3.5
Question 13.
$$\frac { SU }{ ST }$$ = $$\frac { SV }{ VU }$$
$$\frac { 16 }{ 20 }$$ = $$\frac { x }{ (x – 16) }$$
4(x – 16) = 5x
x = 16
Find the geometric mean of the two numbers.
Question 14.
9 and 25
mean = √(9 x 25)
= 15
Question 15.
36 and 48
mean = √(36 x 48)
= 24√3
Question 16.
12 and 42
mean = √(12 x 42)
= 6√14
#### 9.4 The Tangent Ratio
Find the tangents of the acute angles in the right triangle. Write each answer as a fraction and as a decimal rounded to four decimal places.
Question 17.
tan J = $$\frac { Opposite side }{ Adjacent side }$$
tan J = $$\frac { LK }{ JK }$$ = $$\frac { 11 }{ 60 }$$
tan L = $$\frac { JK }{ LK }$$ = $$\frac { 60 }{ 11 }$$
Question 18.
tan P = $$\frac { MN }{ MP }$$ = $$\frac { 35 }{ 12 }$$
tan N = $$\frac { MP }{ MN }$$ = $$\frac { 12 }{ 35 }$$
Question 19.
tan A = $$\frac { BC }{ AC }$$ = $$\frac { 7 }{ 4√2 }$$
tan B = $$\frac { AC }{ BC }$$ = $$\frac { 4√2 }{ 7 }$$
Find the value of x. Round your answer to the nearest tenth.
Question 20.
tan 54 = $$\frac { x }{ 32 }$$
1.37 = $$\frac { x }{ 32 }$$
x = 43.8
Question 21.
tan 25 = $$\frac { x }{ 20 }$$
0.46 x 20 = x
x = 9.2
Question 22.
tan 38 = $$\frac { 10 }{ x }$$
x = 12.82
Question 23.
The angle between the bottom of a fence and the top of a tree is 75°. The tree is 4 let from the fence. How tall is the tree? Round your answer to the nearest foot.
tan 75 = $$\frac { x }{ 4 }$$
x = 14.92
### 9.5 The Sine and Cosine Ratios
Find sin X, sin Z, cos X, and cos Z. Write each answer as a fraction and as a decimal rounded to four decimal places.
Question 24.
sin X = $$\frac { 3 }{ 5 }$$
sin Z = $$\frac { 4 }{ 5 }$$
cos X = $$\frac { 4 }{ 5 }$$
cos Z = $$\frac { 3 }{ 5 }$$
Question 25.
sin X = $$\frac { 7 }{ √149 }$$
sin Z = $$\frac { 10 }{ √149 }$$
cos X = $$\frac { 10 }{ √149 }$$
cos Z = $$\frac { 7 }{ √149 }$$
Question 26.
sin X = $$\frac { 55 }{ 73 }$$
sin Z = $$\frac { 48 }{ 73 }$$
cos X = $$\frac { 48 }{ 73 }$$
cos Z = $$\frac { 55 }{ 73 }$$
Find the value of each variable using sine and cosine. Round your answers to the nearest tenth.
Question 27.
sin 23 = $$\frac { t }{ 34 }$$
t = 13.26
cos 23 = $$\frac { s }{ 34 }$$
s = 31.28
Question 28.
sin 36 = $$\frac { s }{ 5 }$$
s = 2.9
cos 36 = $$\frac { r }{ 5 }$$
r = 4
Question 29.
sin 70 = $$\frac { v }{ 10 }$$
v = 9.39
cos 70 = $$\frac { w }{ 10 }$$
w = 3.42
Question 30.
Write sin 72° in terms of cosine.
sin 72 = cos(90 – 72)
= cos 18 = 0.95
Question 31.
Write cos 29° in terms of sine.
sin 29 = cos(90 – 29)
= cos 61 = 0.48
### 9.6 Solving Right Triangles
Let ∠Q be an acute angle. Use a calculator to approximate the measure of ∠Q to the nearest tenth of a degree.
Question 32.
cos Q = 0.32
cos Q = 0.32
∠Q = inverse cos of .32
∠Q = 71.3
Question 33.
sin Q = 0.91
sin Q = 0.91
∠Q = inverse sin of 0.91
∠Q = 65.5
Question 34.
tan Q = 0.04
tan Q = 0.04
∠Q = inverse tan of 0.04
∠Q = 2.29
Solve the right triangle. Round decimal answers to the nearest tenth.
Question 35.
a = 5√5, ∠A = 47.7, ∠B = 42.3
Explanation:
c² = a² + b²
15² = a² + 10²
a² = 125
a = 5√5
sin A = $$\frac { 5√5 }{ 15 }$$ = 0.74
∠A = 47.7
∠A + ∠B + ∠C = 180
47.7 + ∠B + 90 = 180
∠B = 42.3
Question 36.
NL = 7.59, ∠L = 53, ML = 4.55
Explanation:
cos 37 = $$\frac { 6 }{ NL }$$
NL = 7.59
∠N + ∠M + ∠L = 180
37 + 90 + ∠L = 180
∠L = 53
sin 37 = $$\frac { ML }{ 7.59 }$$
ML = 4.55
Question 37.
XY = 17.34, ∠X = 46, ∠Z = 44
Explanation:
c² = a² + b²
25² = 18²+ b²
b² = 301
b = 17.34
sin X = $$\frac { 18 }{ 25 }$$
∠X = 46
sum of angles = 180
46 + 90 + ∠Z = 180
∠Z = 44
### 9.7 Law of Sines and Law of Cosines
Find the area of ∆ABC with the given side lengths and included angle.
Question 38.
m ∠ B = 124°, a = 9, c = 11
Area = $$\frac { 1 }{ 2 }$$ ac sin B
= $$\frac { 1 }{ 2 }$$ (9 x 11) sin 124
= 40.59
Question 39.
m ∠ A = 68°, b = 13, c = 7
Area = $$\frac { 1 }{ 2 }$$ bc sin A
= $$\frac { 1 }{ 2 }$$ (13 x 7) sin 68
= 41.86
Question 40.
m ∠ C = 79°, a = 25 b = 17
Area = $$\frac { 1 }{ 2 }$$ ab sin C
= $$\frac { 1 }{ 2 }$$ (25 x 17) sin 79
= 208.25
Solve ∆ABC. Round decimal answers to the nearest tenth.
Question 41.
m ∠ A = 112°, a = 9, b = 4
∠B = 24, ∠C = 44, c = 6.76
Explanation:
$$\frac { sin B }{ b }$$ = $$\frac { sin A }{ a }$$
$$\frac { sin B }{ 4 }$$ = $$\frac { sin 112 }{ 9 }$$
sin B = 0.408
∠B = 24
∠A + ∠B + ∠C = 180
112 + 24 + ∠C = 180
∠C = 44
$$\frac { sin 112 }{ 9 }$$ = $$\frac { sin 44 }{ c }$$
c = 6.76
Question 42.
m ∠ 4 = 28°, m ∠ B = 64°, c = 55
∠C = 88, b = 49.4, a = 25.5
Explanation:
∠A + ∠B + ∠C = 180
28 + 64 + ∠C = 180
∠C = 88
$$\frac { sin B }{ b }$$ = $$\frac { sin C }{ c }$$
$$\frac { sin 64 }{ b }$$ = $$\frac { sin 88 }{ 55 }$$
$$\frac { sin 64 }{ b }$$ = 0.018
b = 49.4
$$\frac { sin 28 }{ a }$$ = $$\frac { sin 88 }{ 55 }$$
a = 25.5
Question 43.
m ∠ C = 48°, b = 20, c = 28
∠B = 31.3, ∠A = 100.7, a = 37.6
Explanation:
$$\frac { sin B }{ b }$$ = $$\frac { sin C }{ c }$$
$$\frac { sin B }{ 20 }$$ = $$\frac { sin 48 }{ 28 }$$
$$\frac { sin B }{ 20 }$$ = 0.026
sin B = 0.52
∠B = 31.3
∠A + ∠B + ∠C = 180
∠A + 31.3 + 48 = 180
∠A = 100.7
$$\frac { sin 100.7 }{ a }$$ = $$\frac { sin 48 }{ 28 }$$
a = 37.6
Question 44.
m ∠ B = 25°, a = 8, c = 3
b = 5.45, ∠A = 37.5, ∠C = 117.5
Explanation:
b² = a² + c²- 2ac cos B
b² = 8² + 3² – 2(8 x 3) cos 25 = 73 – 43.2 = 29.8
b = 5.45
$$\frac { sin A }{ 8 }$$ = $$\frac { sin 25 }{ 5.45 }$$
sin A = 0.61
∠A = 37.5
∠A + ∠B + ∠C = 180
37.5 + 25 + ∠C = 180
∠C = 117.5
Question 45.
m ∠ B = 102°, m ∠ C = 43°, b = 21
∠A = 35, c = 14.72, a = 12.3
Explanation:
∠A + ∠B + ∠C = 180
∠A + 102 + 43 = 180
∠A = 35
$$\frac { sin 102 }{ 21 }$$ = $$\frac { sin 43 }{ c }$$
0.046 = $$\frac { sin 43 }{ c }$$
c = 14.72
$$\frac { sin 102 }{ 21 }$$ = $$\frac { sin 35 }{ a }$$
0.046 = $$\frac { sin 35 }{ a }$$
a = 12.3
Question 46.
a = 10, b = 3, c = 12
∠B = 11.7, ∠C = 125.19, ∠A = 43.11
Explanation:
b² = a² + c²- 2ac cos B
3² = 10² + 12²- 2(10 x 12) cos B
9 = 100 + 144 – 240 cos B
cos B = 0.979
∠B = 11.7
a² = b² + c²- 2bc cos A
100 = 9 + 144 – 72 cos A
cos A = 0.73
∠A = 43.11
∠A + ∠B + ∠C = 180
43.11 + 11.7 + ∠C = 180
∠C = 125.19
### Right Triangles and Trigonometry Test
Find the value of each variable. Round your answers to the nearest tenth.
Question 1.
sin 25 = $$\frac { t }{ 18 }$$
t = 7.5
cos 25 = $$\frac { s }{ 18 }$$
s = 16.2
Question 2.
sin 22 = $$\frac { 6 }{ x }$$
x = 16.21
cos 22 = $$\frac { y }{ 16.21 }$$
y = 14.91
Question 3.
tan 40 = $$\frac { k }{ 10 }$$
k = 8.3
cos 40 = $$\frac { 10 }{ j }$$
j = 13.15
Verity that the segment lengths form a triangle. Is the triangle acute, right, or obtuse?
Question 4.
16, 30, and 34
34²= 16² + 30²
So, the triangle is a right-angled triangle.
Question 5.
4, √67, and 9
9² = 81
4² + (√67)² = 83
So the triangle is acute
Question 6.
√5. 5. and 5.5
5.5² = 30.25
√5² + 5² = 30
So the triangle is obtuse
Solve ∆ABC. Round decimal answers to the nearest tenth.
Question 7.
c = 12.08, ∠A = 24.22, ∠C = 65.78
Explanation:
tan A = $$\frac { 5 }{ 11 }$$
∠A = 24.22
c² = 11²+ 5²
c = 12.08
24.22 + 90 + ∠C = 180
∠C = 65.78
Question 8.
∠B = 35.4, ∠C = 71.6, c = 17.9
Explanation:
$$\frac { sin 73 }{ 18 }$$ = $$\frac { sin B }{ 11 }$$
sin B = 0.58
∠B = 35.4
73 + 35.4 + ∠C = 180
∠C = 71.6
$$\frac { sin 73 }{ 18 }$$ = $$\frac { sin 71.6 }{ c }$$
c = 17.9
Question 9.
BC = 4.54, ∠B = 59.3, ∠A = 30.7
Explanation:
9.2² = 8² + x²
x = 4.54
$$\frac { sin 90 }{ 9.2 }$$ = $$\frac { sin B }{ 8 }$$
sin B = 0.86
∠B = 59.3
∠A + 59.3 + 90 = 180
∠A = 30.7
Question 10.
m ∠ A = 103°, b = 12, c = 24
a = 29, ∠B = 53.5
Explanation:
a² = b² + c²- 2bc cos A
a² = 144 + 24² – 2(12 x 24) cos 103
a = 29
$$\frac { sin B }{ 12 }$$ = $$\frac { sin 103 }{ 29 }$$
∠B = 23.5
∠C = 180 – (103 + 23.5) = 53.5
Question 11.
m ∠ A = 26°, m ∠ C = 35°, b = 13
∠B = 119, a = 6.42, c = 8.5
Explanation:
∠B + 26 + 35 = 180
∠B = 119
$$\frac { sin 119 }{ 13 }$$ = $$\frac { sin 26 }{ a }$$
a = 6.42
$$\frac { sin 119 }{ 13 }$$ = $$\frac { sin 35 }{ c }$$
c = 8.5
Question 12.
a = 38, b = 31, c = 35
∠B=50.2, ∠C = 59.8, ∠A = 70
Explanation:
b² = a² + c²- 2ac cos B
31² = 38²+ 35²- 2(35 x 38) cos B
cos B = 0.64
∠B=50.2
a² = b² + c²- 2bc cos A
38² = 31²+ 35²- 2(31 x 35) cos A
cos A = 0.341
∠A = 70
∠C = 59.8
Question 13.
Write cos 53° in terms of sine.
cos 53° = sin (90 – 53) = sin 37
Find the value of each variable. Write your answers in simplest form.
Question 14.
sin 45 = $$\frac { 16 }{ q }$$
q = 22.6
cos 45 = $$\frac { r }{ q }$$
r = 16
Question 15.
Question 16.
sin 30 = $$\frac { f }{ 9.2 }$$
f = 4.6
cos 30 = $$\frac { 8 }{ h }$$
h = 9.2
Question 17.
In ∆QRS, m ∠ R = 57°, q = 9, and s = 5. Find the area of ∆QRS.
Area = $$\frac { 1 }{ 2 }$$ qs sin R
= $$\frac { 1 }{ 2 }$$ (9 x 5) sin 57 = 18.675
Question 18.
You are given the measures of both acute angles of a right triangle. Can you determine the side lengths? Explain.
No.
Question 19.
You are at a parade looking up at a large balloon floating directly above the street. You are 60 feet from a point on the street directly beneath the balloon. To see the top of the balloon, you look up at an angle of 53°. To see the bottom of the balloon, you look up at an angle of 29°. Estimate the height h of the balloon.
Question 20.
You warn to take a picture of a statue on Easter Island, called a moai. The moai is about 13 feet tall. Your camera is on a tripod that is 5 feet tall. The vertical viewing angle of your camera is set at 90°. How far from the moai should you stand so that the entire height of the moai is perfectly framed in the photo?
### Right Triangles and Trigonometry Cummulative Assessment
Question 1.
The size of a laptop screen is measured by the length of its diagonal. you Want to purchase a laptop with the largest screen possible. Which laptop should you buy?
(A)
(B)
(C)
(D)
(B)
Explanation:
(a) d = √9² + 12² = 15
(b) d = √11.25² + 20² = 22.94
(c) d = √12² + 6.75² = 13.76
(d) d = √8² + 6² = 10
Question 2.
In ∆PQR and ∆SQT, S is between P and Q, T is between R and Q, and What must be true about $$\overline{S T}$$ and $$\overline{P R}$$? Select all that apply.
$$\overline{S T}$$ ⊥ $$\overline{P R}$$ $$\overline{S T}$$ || $$\overline{P R}$$ ST = PR ST = $$\frac{1}{2}$$PR
Question 3.
In the diagram, ∆JKL ~ ∆QRS. Choose the symbol that makes each statement true.
< = >
sin J ___________ sin Q sin L ___________ cos J cos L ___________ tan Q
cos S ___________ cos J cos J ___________ sin S tan J ___________ tan Q
tan L ___________ tan Q tan S ___________ cos Q sin Q ___________ cos L
sin J = sin Q sin L = cos J cos L = tan Q
cos S > cos J cos J > sin S tan J = tan Q
tan L < tan Q tan S > cos Q sin Q = cos L
Question 4.
A surveyor makes the measurements shown. What is the width of the river.
tan 34 = $$\frac { AB }{ 84 }$$
AB = 56.28
Question 5.
Create as many true equations as possible.
____________ = ______________
sin X cos X tan x $$\frac{X Y}{X Z}$$ $$\frac{Y Z}{X Z}$$
Sin Z cos Z tan Z $$\frac{X Y}{Y Z}$$ $$\frac{Y Z}{X Y}$$
sin X = $$\frac{Y Z}{X Z}$$ = cos Z
cos X = $$\frac{X Y}{X Z}$$ = sin Z
tan x = $$\frac{Y Z}{X Y}$$
tan Z = $$\frac{X Y}{Y Z}$$
Question 6.
Prove that quadrilateral DEFG is a kite.
Given $$\overline{H E} \cong \overline{H G}$$, $$\overline{E G}$$ ⊥ $$\overline{D F}$$
Prove $$\overline{F E} \cong \overline{F G}$$, $$\overline{D E} \cong \overline{D G}$$
Question 7.
What are the coordinates of the vertices of the image of ∆QRS after the composition of transformations show?
(A) Q’ (1, 2), R'(5, 4), S'(4, -1)
(B) Q'(- 1, – 2), R’ (- 5, – 4), S’ (- 4, 1)
(C) Q'(3, – 2), R’ (- 1, – 4), S’ (0, 1)
(D) Q’ (-2, 1), R'(- 4, 5), S'(1, 4)
Question 8.
The Red Pyramid in Egypt has a square base. Each side of the base measures 722 feet. The height of the pyramid is 343 fee.
a. Use the side length of the base, the height of the pyramid, and the Pythagorean Theorem to find the slant height, AB, of the pyramid.
tan 1 = $$\frac { 722 }{ 635.3 }$$
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