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z. Math > Week 5 > Flashcards

Week 5 Flashcards

1
Q

5x+ 4 = 19

A

5x+ 4 = 19

5x+ 4−4 = 19−4

5x= 15

5x/5=15/5

x= 3

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2
Q
A
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3
Q

3x+ 9 =−3

A
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4
Q

−2x+ 6 = 26

A
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5
Q

x/5 + 2 = 7

A
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6
Q

2x/3 − 4 = 0

A
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7
Q

6n+ 8 = 44

A
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8
Q

5(3m−1) = 70

A
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9
Q

−18−3h= 5

A
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10
Q

4x+ 7 = 8−x

A
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11
Q

2(4y+ 1) = 5(3−y)

A
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12
Q

7(5−c) = 3(c+ 1)−10

A
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13
Q
A
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14
Q
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15
Q
A
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16
Q

The kinetic energy, E, of an object is given by E=1/2mv2 where m is the mass of the object in kg and v is its velocity in m/s. Kinetic energy is measured in Joules (J).

Find the kinetic energy, E, of an object that has a mass of 8kg and is travelling at 6m/s.

A
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17
Q

The kinetic energy, E, of an object is given by E=1/2mv2 where m is the mass of the object in kg and v is its velocity in m/s. Kinetic energy is measured in Joules (J).

An object has a kinetic energy of 180J and is travelling at 8m/s. Find the mass of the object.

A
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18
Q

The kinetic energy, E, of an object is given by E=1/2mv2 where m is the mass of the object in kg and v is its velocity in m/s. Kinetic energy is measured in Joules (J).

Find the velocity of an object if it has a mass of 11.5kg and has a kinetic energy of 81.6J.

A
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19
Q

The surface area, S, of a box with length L, width W and height H is given by

S= 2(L×W) + 2(L×H) + 2(H×W).

Find the surface area of a box with a length of 5cm, width of 3cm and a height of 2.5cm.

A
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20
Q

The surface area, S, of a box with length L, width W and height H is given by

S= 2(L×W) + 2(L×H) + 2(H×W).

Find the height of a box with a surface area of 154.78cm3, a length of 8.5cm and a width of 4.4cm.

A
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21
Q

n2−4n= 0

A

n2−4n= 0

n(n−4) = 0

n= 0, n= 4

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22
Q

3h2−27 = 0

A

3h2−27 = 0

3(h2−9) = 0

3(h−3)(h+ 3) = 0

h= 3, h=−3

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23
Q

k2−13k+ 42 = 0

A

k2−13k+ 42 = 0

(k−6)(k−7) = 0

k= 6, k= 7

24
Q

3x2+x−2 = 0

A

3x2+x−2 = 0

(3x−2)(x+ 1) = 0

x=2/3, x=−1

25
Q

2a2= 5a+ 3

A

2a2= 5a+ 3

2a2−5a−3 = 0

(2a+ 1)(a−3) = 0

a=−1/2, a= 3

26
Q

9y2= 4

A

9y2= 4

y2=4/9

y=±√4/9

y=±2/3

or

y=−2/3, y=2/3

27
Q

2x2+ 4x−6 = 0

A

2x2+ 4x−6 = 0

2(x2+ 2x−3) = 0

x2+ 2x−3 = 0

(x+ 3)(x−1) = 0

x=−3, x= 1

28
Q

p2= 7p−4

A

p2= 7p−4

p2−7p+ 4 = 0

This quadratic does not factorise easily. Therefore it is necessary to use the quadratic formula.

p=7±√(−7)2−4×1×4/2×1

=7±√33 /2

29
Q

2x2−9x+ 5 = 0

A
30
Q

5x2−3x−1 = 0

A
31
Q

Twice a number is added to 3 times its square. The result is 16. What was the original number?

A
32
Q

A soccer ball is kicked such that its height, h, above the ground t seconds after it is kicked is given by h=−t(t−6). How long after it is kicked does it hit the ground?

A

The equation for the height (h) of the ball t seconds after it has been kicked is given by

h=−t(t−6)

To find when the ball hits the ground, let h= 0. This gives

0 =−t(t−6)

The solutions to this quadratic are t= 0 and t= 6. If we assume that the ball was on the ground at t= 0 and then kicked into the air, the ball will hit the ground six seconds after it has been kicked.

33
Q

log2(x) = 4

A

log2(x) = 4

x= 24

x= 16

34
Q

log5(x) = 3

A

log5(x) = 3

x= 53

x= 125

35
Q

6 log4(x) = 12

A

6 log4(x) = 12

log4(x) = 2

x= 42

x= 16

36
Q

log3(x) + 7 = 10

A

log3(x) + 7 = 10

log3(x) = 3

x= 33

x= 27

37
Q

log10(5x) = 2

A

log10(5x) = 2

5x= 102

5x= 100

x= 20

38
Q

4 log7(9x) = 8

A

4 log7(9x) = 8

log7(9x) = 2

9x= 72

9x= 49

x=49/9

39
Q

8 log2(4x) = 24

A

8 log2(4x) = 24

log2(4x) = 3

4x= 23

4x= 8

x= 2

40
Q

5 log3(6x−1) = 10

A

5 log3(6x−1) = 10

log3(6x−1) = 2

6x−1 = 32

6x−1 = 9

6x= 10

x=10/6

x=5/3

41
Q

9 log5(3x+ 2) + 7 = 25

A

9 log5(3x+ 2) + 7 = 25

9 log5(3x+ 2) = 18

log5(3x+ 2) = 2

3x+ 2 = 52

3x+ 2 = 25

3x= 23

x=23/3

42
Q

3 log4(5x+ 8)−4 = 2

A

3 log4(5x+ 8)−4 = 2

3 log4(5x+ 8) = 6

log4(5x+ 8) = 2

5x+ 8 = 42

5x+ 8 = 16

5x= 8

x=8/5

43
Q

2x= 12

A

This question could be solved with logarithms of any base. The obvious choices are base 2, base 10 and base e. The solutions for base 2 are shown below.

Using base 2 logarithms:

2x = 12

log2(2x) = log2(12)

x * log2(2) = log2(12)

x * 1 = log2(12)

x = log2(12)

However, the question asked us to round our answer to two decimal places. To do this,we will need to use the change of base rule.

x= log2(12)

=log10(12)/log10(2)

= 3.58

44
Q

5x= 30

A

5x= 30

log10(5x) = log10(30)

x × log10(5) = log10(30)

x=log10(30)/log10(5)

x= 2.11

45
Q

9x= 58

A

9x= 58

log10(9x) = log10(58)

x×log10(9) = log10(58)

x=log10(58)/log10(9)

x= 1.85

46
Q

ex= 4

A

ex= 4

ln(ex) = ln(4)

x×ln(e) = ln(4)

x= 1.39

47
Q

2ex= 12

A

2ex= 12

ex= 6

ln(ex) = ln(6)

x×ln(e) = ln(6)

x= 1.79

48
Q

7ex= 21

A

7ex= 21

ex= 3

ln(ex) = ln(3)

x×ln(e) = ln(3)

x= 1.10

49
Q

5ex+ 4 = 14

A

5ex+ 4 = 14

5ex= 10

ex= 2

ln(ex) = ln(2)

x×ln(e) = ln(2)

x= 0.69

50
Q

3e2x+1= 15

A

3e2x+1= 15

e2x+1= 5

ln(e2x+1) = ln(5)

(2x+ 1)×ln(e) = ln(5)

2x+ 1 = ln(5)

2x= ln(5)−1

x=ln(5)−1/2

x= 0.30

51
Q

8e5x−4= 32

A

8e5x−4= 32

e5x−4= 4

ln(e5x−4) = ln(4)

(5x−4)×ln(e) = ln(4)

5x−4 = ln(4)

5x= ln(4) + 4

x=ln(4)+4/5

x= 1.08

52
Q

6e9x+7+ 2 = 20

A

6e9x+7+ 2 = 20

6e9x+7= 18

e9x+7= 3

ln(e9x+7) = ln(3)

(9x+ 7)×ln(e) = ln(3)

9x+ 7 = ln(3)

9x= ln(3)−7

x=ln(3)−7/9

x=−0.66

53
Q

The mass, m, of a baby animal (in kg) t weeks after it is born can be modelled by

m= 3 log10(8t+ 10)

Find the mass of the animal when it is born.

A

m= 3 log10(8t+ 10)

To find the mass of the animal at birth, substitute t= 0 into the equation form. (This is because it has been zero weeks since its birth).

m= 3 log10(8×0 + 10)

= 3 log10(10)

= 3 kg

54
Q

The mass, m, of a baby animal (in kg) t weeks after it is born can be modelled by

m= 3 log10(8t+ 10)

Find the mass of the animal 10 weeks after it was born

A

m= 3 log10(8t+ 10)

m= 3 log10(8×10 + 10)

= 3 log10(90)

≈5.86 kg

55
Q

The mass, m, of a baby animal (in kg) t weeks after it is born can be modelled by

m= 3 log10(8t+ 10)

How long until the animal is 8kg?

A
56
Q

32x−4×3x+ 4 = 0

A

z. Math (14 decks)

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