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taurus [48]
3 years ago
12

Help me..it has to be rounded to the tenths

Mathematics
2 answers:
klio [65]3 years ago
7 0

Answer:

10.2 cm²

Step-by-step explanation:

The area (A) of a regular hexagon is

A = \frac{1}{2} × perimeter × apothem

Perimeter = 6 × 2 = 12 cm ( hexagon has 6 sides ), hence

A = 0.5 × 12 × 1.7 = 6 × 1.7 = 10.2 cm²

DaniilM [7]3 years ago
3 0

Answer:

10.2cm^2

Step-by-step explanation:

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3 years ago
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PLEASEEEEEE HELPPPPP!!
DaniilM [7]

Answer:

y-b+am/m

Step-by-step explanation:

y-b=m(x-a)

y-b=MX-ma

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x=y-b+ma/m

4 0
3 years ago
WHAT IS X³-27 SIMPLIFIED
Eduardwww [97]

Answer:

<u>It</u><u> </u><u>is</u><u> </u><u>(</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>)</u><u>³</u><u> </u><u>-</u><u> </u><u>9</u><u>x</u><u>(</u><u>3</u><u> </u><u>-</u><u> </u><u>x</u><u>)</u>

Step-by-step explanation:

Express 27 in terms of cubes, 27 = 3³:

=  {x}^{3}  -  {3}^{3}

From trinomial expansion:

{(x - y)}^{3}  = (x - y)(x - y)(x - y) \\

open first two brackets to get a quadratic equation:

{(x - y)}^{3}  = ( {x}^{2}  - 2xy +  {y}^{2} )(x - y)

expand further:

{(x - y)}^{3}  =  {x}^{3}   - y {x}^{2}  - 2y {x}^{2}  + 2x {y}^{2}  + x {y}^{2}  -  {y}^{3}  \\  {(x - y)}^{3}  =  {x}^{3}  -  {y}^{3}  + 3x {y}^{2}  - 3y {x}^{2}  \\  {(x - y)}^{3}  =  {x}^{3}  -  {y}^{3}  + 3xy(y - x) \\  \\ { \boxed{( {x}^{3} -  {y}^{3} ) =  {(x - y)}^{3}   - 3xy(y - x)}}

take y to be 3, then substitute:

( {x}^{3}  - 3^3) =  {(x - 3)}^{3}  - 9x(3 - x)

5 0
3 years ago
Simplify the expression. A) 8^4•8^9
OlgaM077 [116]

Answer:

{8}^{13}

Step-by-step explanation:

{8}^{4}  \times  {8}^{9}  \\  {8}^{4 + 9}  \\  {8}^{13}  \\  = 5.497558139 \times 10 ^{11}

Hope this helps you.

Let me know if you have any other questions :-)

7 0
3 years ago
What is m∠Q ?
ddd [48]
Unlike the previous problem, this one requires application of the Law of Cosines.  You want to find angle Q when you know the lengths of all 3 sides of the triangle.

Law of Cosines:  a^2 = b^2 + c^2 - 2bc cos A

Applying that here:

                           40^2 = 32^2 + 64^2 - 2(32)(64)cos Q

Do the math.  Solve for cos Q, and then find Q in degrees and Q in radians.

7 0
3 years ago
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