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kvasek [131]
3 years ago
13

Given that a triangle has sides of length 4.1 cm and 7 cm. What is the range of possible length of the third side?

Mathematics
1 answer:
Citrus2011 [14]3 years ago
5 0

Answer:

82.8

Step-by-step explanation:

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If g(x-1) = (2-g(x))/8 and g(1) = 3, find the value of g(2).
aniked [119]

Answer:

-22

Step-by-step explanation:

g(2-1)=g(1)=(2-g(2))/8=3

2-g(2)=24

-g(2)=22

g(2)=-22

5 0
3 years ago
The table shows the percentage of students in each of three grade levels who list soccer as their favorite sport. Soccer Sophomo
Olegator [25]
The answer should be roughly 35.4%. 

You can obtain this answer by looking at the percentage of each subgrouping. For instance, 33% of the class in juniors and 45% of them list soccer as their favorites. Thus showing that 14.85% of the entire school is made up of juniors that enjoy soccer. 

If you do the totaling for all soccer lovers, you get a total of 41.95% of the school. By dividing the two numbers you get the answer above. 
4 0
3 years ago
Which Expression is equivalent to 8x - 2y + x + x
Ray Of Light [21]
8x -2y + x+x simply combine like terms 8x+ x + x= 10x therefore 10x-2y is equivalent
4 0
3 years ago
Read 2 more answers
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
Write 420 as a product 9f prime numbers​
stich3 [128]

Answer:

2, 2, 3, 5, 7 that is the farthest you can go

Step-by-step explanation:

2,2,3,5,7 prime factorization of 420 but impossible to get 9 factors

5 0
3 years ago
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