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attashe74 [19]
1 year ago
6

Please help me! in need of help due tomorrow!​

Mathematics
1 answer:
julsineya [31]1 year ago
3 0

Answer:

Step-by-step explanation:

(1). A = π r²

A = 4 π

Area of shaded region is \frac{4\pi *260}{360} = \frac{26}{9} \pi or \frac{26\pi }{9}

(2). C = 4 π

In this case, the length of the bigger arc and the area of shaded region happen to be the same.

The length of the arc ADB is  \frac{26}{9} \pi or \frac{26\pi }{9}

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I may not be correct on this but I believe this is either Linear Function or other

Step-by-step explanation:

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Calculate the area of triangle ABC with altitude CD, given A(6,-2) B(1,3) C(5,5) D(2,2)
Vikki [24]
The area will be half the product of the lengths of AB and CD.

|AB| = √((6-1)² +(-2-3)²) = 5√2
|CD| = √((5-2)² +(5-2)²) = 3√2

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6 0
3 years ago
Given: sin(18m-12)=cos(7m+2), find the value of m.
V125BC [204]

Answer:

m = -2/7 + π/14 + (π n_1)/7 for n_1 element Z

or m = 2/3 - (π n_2)/18 for n_2 element Z

Step-by-step explanation:

Solve for m:

-cos(7 m + 2) sin(12 - 18 m) = 0

Multiply both sides by -1:

cos(7 m + 2) sin(12 - 18 m) = 0

Split into two equations:

cos(7 m + 2) = 0 or sin(12 - 18 m) = 0

Take the inverse cosine of both sides:

7 m + 2 = π n_1 + π/2 for n_1 element Z

or sin(12 - 18 m) = 0

Subtract 2 from both sides:

7 m = -2 + π/2 + π n_1 for n_1 element Z

or sin(12 - 18 m) = 0

Divide both sides by 7:

m = -2/7 + π/14 + (π n_1)/7 for n_1 element Z

or sin(12 - 18 m) = 0

Take the inverse sine of both sides:

m = -2/7 + π/14 + (π n_1)/7 for n_1 element Z

or 12 - 18 m = π n_2 for n_2 element Z

Subtract 12 from both sides:

m = -2/7 + π/14 + (π n_1)/7 for n_1 element Z

or -18 m = π n_2 - 12 for n_2 element Z

Divide both sides by -18:

Answer: m = -2/7 + π/14 + (π n_1)/7 for n_1 element Z

or m = 2/3 - (π n_2)/18 for n_2 element Z

3 0
3 years ago
Can someone help me?
Lana71 [14]

Answer:

I'm not sure from my answer but I'm going to try

A =  { a^2  -  4  = 0 }

A = { a^2  = 0+ 4 }

A = { a^2  = 4}

Take root of both sides

A = {a = 2}

A = {2}

B = { -2, -1, 0, 1, 2}

C = { 1, 2, 3, 4, 5, 6,7}

i) A\B

= {2} \ {-2,-1,0,1,2}

= ⌀ ( "null set" )

ii) (A\B) \C

= ( {2} \ {-2,-1,0,1,2} ) \ C

= ⌀ \ C

= ⌀ \ { 1, 2, 3, 4, 5, 6,7}

= ⌀ ( "null set" )

iii) (A ∪ B) \ C

= ( { 2 } ∪ { -2, -1, 0, 1, 2} ) \ C

= { -2, -1, 0, 1, 2} \C

= { -2, -1, 0, 1, 2} \ { 1, 2, 3, 4, 5, 6,7}

= { -2, -1, 0 }

iv) ( A\B) ∩ ( A\C)

= (  { 2 } \ {-2,-1,0,1,2} ) ∩ ( { 2 } \ { { 1, 2, 3, 4, 5, 6,7} )

= ⌀ ∩ ⌀

= ⌀

v) ( A ∩ B ) \ C

= ( { 2 }  ∩  { -2, -1, 0, 1, 2 } ) \ C

= { 2 } \ { 1, 2, 3, 4, 5, 6, 7 }

= ⌀ "null set"

7 0
3 years ago
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