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pogonyaev
2 years ago
14

How do I figure out how many faces are glued together?

Mathematics
1 answer:
Anika [276]2 years ago
8 0
16 is the answer. I think
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The number of blogs has grown rapidly. Assuming that two new blogs are created each second,
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The Answer is C, 5.27 x 10^6

Please mark it as brainlest answer:).

5 0
2 years ago
See attachment the problem can be found in there
zloy xaker [14]

Answer:

\frac{ - 2 {x}^{2}  + 19x   + 3 }{3 {x} (4x^{2}     -  9 )}

Step-by-step explanation:

\frac{5}{6 {x}^{2}  + 9x}  +  \frac{1}{2x - 3}  -  \frac{2}{3x}  \\  \\  =  \frac{5}{3x(2 {x}   + 3)}  +  \frac{1}{2x - 3}  -  \frac{2}{3x}  \\  \\ =  \frac{5(2x - 3) + 1 \times 3x(2x + 3)}{3x(2 {x}   + 3)(2x - 3)}  -  \frac{2}{3x}  \\  \\  = \frac{10x - 15 + 6 {x}^{2}  + 9x}{3x((2 {x} ) ^{2}     -  {3}^{2} )}  -  \frac{2}{3x} \\  \\  = \frac{6 {x}^{2}   + 19x - 15}{3x(4x^{2}     -  9 )}  -  \frac{2}{3x} \\  \\  = \frac{3x(6 {x}^{2}   + 19x - 15) - 2 \times 3x(4x^{2}     -  9 )}{3x(4x^{2}     -  9 ) \times 3x}  \\  \\  = \frac{18{x}^{3}   + 57 {x}^{2}  - 45x - 24x^{3}      + 54x }{3x(4x^{2}     -  9 ) \times 3x}   \\  \\ = \frac{ - 6 {x}^{3}  + 57 {x}^{2}   + 9x }{9 {x}^{2} (4x^{2}     -  9 )}  \\  \\ = \frac{ 3x(- 2 {x}^{2}  + 19x   + 3) }{9 {x}^{2} (4x^{2}     -  9 )}  \\  \\  \huge \orange{ \boxed{= \frac{ - 2 {x}^{2}  + 19x   + 3 }{3 {x} (4x^{2}     -  9 )}  }}

7 0
3 years ago
Fr.ee points because you deserve it
Advocard [28]

Answer:

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Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
In a survey, one-fifth youths like cell phone only and 16 did not like cell phone at all. Also 60%
Pepsi [2]

Answer: Let, n(U) be=x only (p)=x/5 only(ca)=16 n(pnc)=?n(c)=60%ofx=0.6x

then,

Only (c)=n(c)-n(cnp)

16=0.6x-n(cnp)

n(cnp)=0.6x-16

again,

n(U)=only(p)+only (c)+n(cnp)+n(pUc)of complement

x=x/5+16+0.6x-16+8

5x=x+3x+40

x=40

n(cnp)=0.6x-16=0.6×40-16=8

Step-by-step explanation:

6 0
3 years ago
A circle with circumfrence 10, has an arc with 160 degree angle
Diano4ka-milaya [45]
Soo what is the question is there more to this
4 0
3 years ago
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