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3 years ago

5

Circumference of a circle

2 answers:

irakobra [83]3 years ago

5 0

Answer:

Multiply the radius by 2 to get the diameter.

Multiply the result by π, or 3.14 for an estimation.

That's it; you found the circumference of the circle.

Step-by-step explanation:

k0ka [10]3 years ago

4 0

Answer:

Step-by-step explanation:

the straight sides are just 4.5 *2 = 9 cm

the arcs are 2 $\pi$ r * Ф/360

where Ф= 50 for this problem

and r = 8

the inner arc is 2 $\pi$ (3.5) *50/360 = 3.05432

the outer arc is 2 $\pi$ (8)*50/360 = 6.98131

total is 19.0356

to 3 sig figs is 19.036 :)

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3 years ago

Using the binomial theorem , obtain the expansion of :

andrezito [222]

Answer:

see explanation

Step-by-step explanation:

Expand both factors and collect like term

Using Pascal' triangle with n = 6 to obtain the coefficients

1 6 15 20 15 6 1

Decreasing powers of 1 from $1^{6}$ to $1^{0}$

Increasing powers of 3x from $(3x)^{0}$ to $(3x)^{6}$

$1+3x)^{6}$

= 1. $1^{6}$ $(3x)^{0}$ + 6. $1^{5}$ $(3x)^{1}$ + 15. $1^{4}$ $(3x)^{2}$ + 20. $1^{3}$ $(3x)^{3}$ + 15.1² $(3x)^{4}$ + 6. $1^{1}$ $(3x)^{5}$ + 1. $1^{0}$ $(3x)^{6}$

= 1 + 18x + 135x² + 540x³ + 1215 $x^{4}$ + 1458 $x^{5}$ + 729 $x^{6}$

--------------------------------------------------------------------------------------

$(1-3x)^{6}$

= 1. $1^{6}$ $(-3x)^{0}$ + 6. $1^{5}$ $(-3x)^{1}$ + 15. $1^{4}$ $(-3x)^{2}$ + 20. $1^{3}$ $(-3x)^{3}$ + 15.1² $(-3x)^{4}$ + 6. $1^{1}$ $(-3x)^{5}$ + 1. $1^{0}$ $(-3x)^{6}$

= 1 - 18x + 135x² - 540x³ + 1215 $x^{4}$ - 1458 $x^{5}$ + 729 $x^{6}$

----------------------------------------------------------------------------------

Collecting like terms from both expressions

$(1+3x)^{6}$ + $(1-3x)^{6}$

= 2 + 270x² + 2430 $x^{4}$ + 1458 $x^{6}$

----------------------------------------------------

(2)

Using Pascal's triangle with n = 5

1 5 10 10 5 1

Decreasing powers of 1 from $1^{5}$ to $1^{0}$

Increasing powers of 2x from $(2x)^{0}$ to $(2x)^{5}$

$(1+2x)^{5}$

= 1. $1^{5}$ $(2x)^{0}$ + 5. $1^{4}$ $(2x)^{1}$ + 10. $1^{3}$ $(2x)^{2}$ + 10. $1^{2}$ $(2x)^{3}$ + 5. $1^{1}$ $(2x)^{4}$ + 1. $1^{0}$ $(2x)^{5}$

= 1 + 10x + 40x² + 80x³ + 80 $x^{4}$ + 32 $x^{5}$

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3 years ago