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inysia [295]
3 years ago
14

Which values of P and Q does the following equation have infinitely many solutions?

Mathematics
1 answer:
vladimir1956 [14]3 years ago
4 0

Any value as long as P = Q

For the equation to have infinitely many solutions, we require both sides of the equation to have exactly the same terms.

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A motor racing circuit has length 5 5/6 miles. A straight section of the circuit has 1 1/4 miles. What fraction of the circuit i
aleksklad [387]

Answer:

3/14

Step-by-step explanation:

1 1/4 can be written as 5/4

5 5/6 can be written as 35/6

so divide (5/4)/(35/6)

= (5/4)*(6/35)

=(1/4)*(6/7)

=6/28

simplified = 3/14

4 0
2 years ago
Twelve packs of paper cost $7.20. Of each pack costs the same, how much does one pack cost?
White raven [17]
Each pack costs $1.66 dollars
5 0
3 years ago
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A line has a slope of -6 over 7. What is the slope of the line parallel to it? And what is the slope of the line perpendicular t
olga_2 [115]
The slope of the parallel line is -6/7
the slope of the perpendicular line is 7/6
the slope of the line = -6/7
the gradient of two parallel lines are equal
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: the gradient m1*m2 = -1
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3 0
3 years ago
WHAT WOULD THISSS BEEEEE ANYONEEE!
Eva8 [605]

Answer:

12

Step-by-step explanation:

12×12=144

12 squared equals 144 so square root is 12.

7 0
2 years ago
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Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications,
posledela

Answer:

  • number of multiplies is n!
  • n=10, 3.6 ms
  • n=15, 21.8 min
  • n=20, 77.09 yr
  • n=25, 4.9×10^8 yr

Step-by-step explanation:

Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...

  mpy[n] = n·mp[n-1]

  mpy[2] = 2

So, ...

  mpy[n] = n! . . . n ≥ 2

__

If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...

  10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10

Then the larger matrices take ...

  n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min

  n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years

  n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years

_____

For the shorter time periods (less than 100 years), we use 365.25 days per year.

For the longer time periods (more than 400 years), we use 365.2425 days per year.

8 0
3 years ago
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