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artcher [175]
3 years ago
10

Significant Figures PracticeName________________________ Date_____________ Section________________

Mathematics
2 answers:
Free_Kalibri [48]3 years ago
5 0

Answer:1)2804m^2=4s.f

2)84Km^3=2s.f

3)5.029m=4s.f

4)0.00003068m=4s.f

5)4.6×10^5m=2s.f

6)4.06×10^-5m=3s.f

7)750m=3s.f

8)75m=2s.f

9)75,000.0m=5s.f

10)75.00m=2s.f

11)75,000.00m=5s.f

12)10cm=2s.t

B) to four significant figures

1)3.682

2)860100

3)375.7

4)51150

5)4673

C to 1 decimal places

1)1.4

2)4735.0

3)687524.0

4)7.6

5)235.0

Dto 2d.p

1)22.49

2)25880.00

3)30623.00

4)412541.00

5)866323.00

E)

1)6.201cm+7.4cm+0.68cm+12.0cm=26.281cm

2)1.6km+1.62km+1200km=1203.22km

3)8.264g-7.8g=0.464g

4)10.4168m-6m=4.4168m

5)12.00m+15.001kg=12.00m+15.001kg

6)1.31cm×2.3cm=3.013cm^2

7)5.7621m×6.201m=35.7308m^2

8)20.2cm/7.41s=2.726cm/s

Step-by-step explanation:

S.f means significant figures or Digits.

2)cm +kg is impossible because they are unlike termsi.e they do not have any thing in common, hence 12.00m+15.001kg cannot be added up.

balu736 [363]3 years ago
4 0

Answer:

1. State the number of significant digits in each measurement.

1) 2804 m = 4 significant digits

2) 2.84 km = 3 significant digits

3) 5.029 m = 4 significant digits

4) 0.003068 m = 4 significant digits

5) 4.6 x 10^5 m = 2 significant digits

6) 4.06 x 10^-5 m = 3 significant digits

7) 750 m = 2 significant digits

8) 75 m = 2 significant digits

9) 75,000 m = 2 significant digits

10) 75.00 m = 4 significant digits

11) 75,000.0 m = 6 significant digits

12) 10 cm = 1 significant digit

2. Round the following numbers as indicated:

To four figures:

3.6824172 = 3.682

1.86005137 = 1.860

5.652311 = 5.652

2.5114 = 2.511

5.4673 = 5.467

To one decimal place:

1.3511 = 1.4

2.473 = 2.5

5.687524 = 5.7

7.555 = 7.6

8.235 = 8.2

To two decimal places:

22.494 = 22.49

79.258 = 79.26

80.0306 = 80.03

23.4125 = 23.41

41.866323 = 41.87

Solve the following problems and report answers with appropriate number of significant digits.

1) 6.201 cm + 7.4 cm + 0.68 cm +12.0 cm = 26.281 cm (5 significant digits)

2) 1.6 km + 1.62 km +1200 km = 1203.22 km (6 significant digits)

3) 8.264 g - 7.8 g = 0.464 g (3 significant digits)

4) 10.4168 m - 6.0 m = 6.4168 (5 significant digits)

5) 12.00 kg+15.001 kg= 27.001 kg (5 significant digits)

6) 1.31 cm x 2.3 cm = 3.013 cm² (4 significant digits)

7) 5.7621 m x 6.201 m = 35.7307821 (9 significant digits)

8) 20.2 cm / 7.41 s = 2.726046 (7 significant digits)

Explanation

The significant digits of a number are the digits that have meaning or contribute to the value of the number. Sometimes they are also called significant figures.

Which digits are significant?

There are some basic rules that tell you which digits in a number are significant:

All non-zero digits are significant

Any zeros between significant digits are also significant

Trailing zeros to the right of a decimal point are significant

Which digits aren't significant?

The only digits that aren't significant are zeros that are acting only as place holders in a number. These are:

Trailing zeros to the left of the decimal point (note: these zeros may or may not be significant)

Leading zeros to the right of the decimal point

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3 years ago
Please help me to prove this!<br>I need is no.(c). So, please help me do it.<br>​
zloy xaker [14]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = 90°                  → A + B = 90° - C

                                                     → C = 90° - (A + B)

Use the Double Angle Identity:      cos 2A = 1 - 2 sin² A

                                                       → sin² A = (1 - cos 2A)/2

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]

Use the Product to Sum Identity: cos (A - B) - cos (A + B) = 2 sin A · sin B

Use the Cofunction Identities:      cos (90° - A) = sin A

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<u>Proof LHS → RHS:</u>

LHS:                       sin² A + sin² B + sin² C

\text{Double Angle:}\qquad \dfrac{1-\cos 2A}{2}+\dfrac{1-\cos 2B}{2}+\sin^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2-\cos 2A-\cos 2B\bigg)+\sin^2 C\\\\\\.\qquad \qquad \qquad =1-\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\sin^2 C

\text{Sum to Product:}\quad 1-\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\sin^2 C\\\\\\.\qquad \qquad \qquad =1-\cos (A+B)\cdot \cos (A-B)+\sin^2 C

Given:                1 - cos (90° - C) · cos (A - B) + sin² C

Cofunction:       1 - sin C · cos (A - B) + sin² C

Factor:               1 - sin C [cos (A - B) + sin C]

Given:                1 - sin C[cos (A - B) - sin (90° - (A + B))]

Cofunction:       1 - sin C[cos (A - B) - cos (A + B)]

Sum to Product:       1 - sin C [2 sin A · sin B]

                            = 1 - 2 sin A · sin B · sin C

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6 0
3 years ago
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jonny [76]

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so first, we move the constant to the right.

-3x < 15 + 6

the, we calculate that.

-3x < 21

then, we divide both sides.

solution: x > -7

6 0
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Katarina [22]
Hi there!

Th' Eqn. is :-

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Hence, The required answer is : x = - 7

~ Hope it helps!
3 0
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