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2 years ago

11

1. What is the domainof the relation {(3,4), (1,2), (-3, -5), (0, -8)}?

1 answer:

Marizza181 [45]2 years ago

3 0

Wouldn’t it be D. since the domain is x. (x,y)

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Solve S = 2πrh for r

Elodia [21]

Divide both sides by 2<span>πh:-

r = S / 2</span><span>πh Answer</span>

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2 years ago

What is the midpoint of the line segment with endpoints (-1,3) and (-5,5)?

andrew11 [14]

Answer:

(4, 5)

Step-by-step explanation:

Calculate the midpoint using the midpoint formula

midpoint = [( + ), ( + ) ]

with (, ) = (1, 2) and (, ) = (7, 8)

midpoint = [(1 + 7), (2 + 8) ]

= (4, 5)

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2 years ago

Show that (1 + 2V3)2 = 13 +4V3

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2 years ago

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Rasek [7]

Answer:

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2 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 $x^{5}$

8 0

2 years ago