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krok68 [10]
2 years ago
10

A wheel has a radious of 50cm. How many times would the wheel go around if it rolled for 10km

Mathematics
1 answer:
Zinaida [17]2 years ago
6 0
First you must make sure all measurements are in the same value: You can choose either cm or km

Distance = 1,000,000 cm or 10 km

Radius = 50 cm or 0.0005 km

Distance/(2πR)

1,000,000/(2π*50)

1,000,000/314.059…….

Approx = 3183.1 Rotations
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Write in expanded form six hundred twenty six and fourteen thousandths
True [87]

Answer:

626,140

or

600,000+20,000+6,000+100+40

Step-by-step explanation:

8 0
3 years ago
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For any triangle ABC note down the sine and cos theorems ( sinA/a= sinB/b etc..)
SCORPION-xisa [38]

Answer:

Step-by-step explanation:

Law of sines is:

(sin A) / a = (sin B) / b = (sin C) / c

Law of cosines is:

c² = a² + b² − 2ab cos C

Note that a, b, and c are interchangeable, so long as the angle in the cosine corresponds to the side on the left of the equation (for example, angle C is opposite of side c).

Also, angles of a triangle add up to 180° or π.

(i) sin(B−C) / sin(B+C)

Since A+B+C = π, B+C = π−A:

sin(B−C) / sin(π−A)

Using angle shift property:

sin(B−C) / sin A

Using angle sum/difference identity:

(sin B cos C − cos B sin C) / sin A

Distribute:

(sin B cos C) / sin A − (cos B sin C) / sin A

From law of sines, sin B / sin A = b / a, and sin C / sin A = c / a.

(b/a) cos C − (c/a) cos B

From law of cosines:

c² = a² + b² − 2ab cos C

(c/a)² = 1 + (b/a)² − 2(b/a) cos C

2(b/a) cos C = 1 + (b/a)² − (c/a)²

(b/a) cos C = ½ + ½ (b/a)² − ½ (c/a)²

Similarly:

b² = a² + c² − 2ac cos B

(b/a)² = 1 + (c/a)² − 2(c/a) cos B

2(c/a) cos B = 1 + (c/a)² − (b/a)²

(c/a) cos B = ½ + ½ (c/a)² − ½ (b/a)²

Substituting:

[ ½ + ½ (b/a)² − ½ (c/a)² ] − [ ½ + ½ (c/a)² − ½ (b/a)² ]

½ + ½ (b/a)² − ½ (c/a)² − ½ − ½ (c/a)² + ½ (b/a)²

(b/a)² − (c/a)²

(b² − c²) / a²

(ii) a (cos B + cos C)

a cos B + a cos C

From law of cosines, we know:

b² = a² + c² − 2ac cos B

2ac cos B = a² + c² − b²

a cos B = 1/(2c) (a² + c² − b²)

Similarly:

c² = a² + b² − 2ab cos C

2ab cos C = a² + b² − c²

a cos C = 1/(2b) (a² + b² − c²)

Substituting:

1/(2c) (a² + c² − b²) + 1/(2b) (a² + b² − c²)

Common denominator:

1/(2bc) (a²b + bc² − b³) + 1/(2bc) (a²c + b²c − c³)

1/(2bc) (a²b + bc² − b³ + a²c + b²c − c³)

Rearrange:

1/(2bc) [a²b + a²c + bc² + b²c − (b³ + c³)]

Factor (use sum of cubes):

1/(2bc) [a² (b + c) + bc (b + c) − (b + c)(b² − bc + c²)]

(b + c)/(2bc) [a² + bc − (b² − bc + c²)]

(b + c)/(2bc) (a² + bc − b² + bc − c²)

(b + c)/(2bc) (2bc + a² − b² − c²)

Distribute:

(b + c)/(2bc) (2bc) + (b + c)/(2bc) (a² − b² − c²)

(b + c) + (b + c)/(2bc) (a² − b² − c²)

From law of cosines, we know:

a² = b² + c² − 2bc cos A

2bc cos A = b² + c² − a²

cos A = (b² + c² − a²) / (2bc)

-cos A = (a² − b² − c²) / (2bc)

Substituting:

(b + c) + (b + c)(-cos A)

(b + c)(1 − cos A)

From half angle formula, we can rewrite this as:

2(b + c) sin²(A/2)

(iii) (b + c) cos A + (a + c) cos B + (a + b) cos C

From law of cosines, we know:

cos A = (b² + c² − a²) / (2bc)

cos B = (a² + c² − b²) / (2ac)

cos C = (a² + b² − c²) / (2ab)

Substituting:

(b + c) (b² + c² − a²) / (2bc) + (a + c) (a² + c² − b²) / (2ac) + (a + b) (a² + b² − c²) / (2ab)

Common denominator:

(ab + ac) (b² + c² − a²) / (2abc) + (ab + bc) (a² + c² − b²) / (2abc) + (ac + bc) (a² + b² − c²) / (2abc)

[(ab + ac) (b² + c² − a²) + (ab + bc) (a² + c² − b²) + (ac + bc) (a² + b² − c²)] / (2abc)

We have to distribute, which is messy.  To keep things neat, let's do this one at a time.  First, let's look at the a² terms.

-a² (ab + ac) + a² (ab + bc) + a² (ac + bc)

a² (-ab − ac + ab + bc + ac + bc)

2a²bc

Repeating for the b² terms:

b² (ab + ac) − b² (ab + bc) + b² (ac + bc)

b² (ab + ac − ab − bc + ac + bc)

2ab²c

And the c² terms:

c² (ab + ac) + c² (ab + bc) − c² (ac + bc)

c² (ab + ac + ab + bc − ac − bc)

2abc²

Substituting:

(2a²bc + 2ab²c + 2abc²) / (2abc)

2abc (a + b + c) / (2abc)

a + b + c

8 0
3 years ago
The sides of a cube are measured in whole centimeters. Which of the following could not be the cube's volume?
laila [671]
A: 100 cm³
This is because the others can be multiplied by something 3 times to get it, but A can not.

6 0
3 years ago
Read 2 more answers
Given f(x)=x3 and g(x)= 1-5x2, fine (fog)(x) and it’s Domain
andre [41]

Answer:

Option B. f(g(x)) = (1-5x ^ 2) ^ 3  all real numbers

Step-by-step explanation:

We have

f(x) = x ^ 3 and g(x) = 1-5x ^ 2

They ask us to find

(fog)(x) and it's Domain

To solve this problem we must introduce the function g(x) within the function f(x)

That is, we must do f(g(x)).

So, we have:

f(x) = x ^ 3

g(x) = 1-5x ^ 2

Then:

f(g(x)) = (1-5x ^ 2) ^ 3

The domain of the function f(g(x)) is the range of the function g(x) = 1-5x ^ 2.

Since the domain and range of g(x) are all real numbers then the domain of f(g(x)) are all real numbers

Therefore the correct answer is the option b: f(g(x)) = (1-5x ^ 2) ^ 3

And his domain is all real.

5 0
3 years ago
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Rename 1/2 into its proper fraction
AveGali [126]
It is into its proper fraction and cannot be simplified down any further. Therefore the proper fraction is 1/2
3 0
3 years ago
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