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Sergio [31]
4 years ago
8

A circular lid has a diameter of 12 inches. What is the exact area of the lid? A. 6π square inches B. 12π square inches C. 24π s

quare inches D. 36π square inches
Mathematics
1 answer:
kaheart [24]4 years ago
6 0
The area of any circle is equal to \pi r^2, where r is the radius.

We know that the diameter is 12. The radius is always half of the diameter, so the radius of our circle must be 6.

Substitute 6 for r in our equation and simplify.

π6² = π6×6 = 36π in²
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3 0
3 years ago
Factorize 2a^2+7a-15​
loris [4]

\huge\bf  \pink {\underline{Solution :-}}

2 {a}^{2}  + 7a - 15

= 2 {a}^{2}  + 10a - 3a - 15

= 2a(a + 5) - 3(a + 5)

= (a + 5)(2a - 3)

\sf \red{Hence, Answer \:  is  \: (a + 5)(2a - 3).}

\\

\bf \purple{ \underline{ Important  \: Formulas  \: for  \: Factorization :-}}

• \:  {a}^{2}  + 2ab + {b}^{2} =  {(a + b)}^{2}

• \: (a + b)(a - b) =  {a}^{2}  -  {b}^{2}

• \:  {a}^{2} - 2 ab  +  {b}^{2}  =  {(a - b)}^{2}

• \:  {a}^{3}  + 3 {a}^{2} b + 3a {b}^{2}  +  {b}^{3} =  {(a + b)}^{3}

• \:  {a}^{3}   -  3 {a}^{2} b + 3a {b}^{2}   -  {b}^{3} =  {(a  - b)}^{3}

•  \:  {(a + b)}^{2}  +  {(a - b)}^{2}  = 2( {a}^{2}  +  {b}^{2} )

• \: {(a + b)}^{2}   -   {(a - b)}^{2}  = 4ab

• \: (a + b)( {a}^{2}   -ab   +  {b}^{2}  ) =  {a}^{3}  +  {b}^{3}

• \: (a  -  b)( {a}^{2}    + ab   +  {b}^{2}  ) =  {a}^{3}   -  {b}^{3}

• \:   {( \frac{a + b}{2} )}^{2}  - ( {\frac{a - b}{2} } )^{2}  = ab

5 0
3 years ago
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Please help me find x. I really don't get it.
inysia [295]

Answer:

  x = 11

Step-by-step explanation:

The relationship between the sine and cosine functions can be written as ...

  sin(x) = cos(90 -x)

  sin(A) = cos(90 -A) = cos(B) . . . . substituting the given values

Equating arguments of the cosine function, we have ...

 90 -(3x+4) = 8x -35

  86 -3x = 8x -35

  86 +35 = 8x +3x . . . . . add 3x+35 to both sides

  121 = 11x . . . . . . . . . . . . collect terms

  121/11 = x = 11 . . . . . . . . divide by 11

_____

<em>Comment on the solution</em>

There are other applicable relationships between sine and cosine as well. The result is that there are many solutions to this equation. One set is ...

  11 +(32 8/11)k . . . for any integer k

Another set is ...

  61.8 +72k . . . . . for any integer k

3 0
3 years ago
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