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fiasKO [112]
3 years ago
10

Please help, max points and brainliest answer given as reward!

Mathematics
2 answers:
Sergeeva-Olga [200]3 years ago
7 0

Answer:

4. width= (-2+ sqrt(204)) /2

length=  (2+ sqrt(204)) /2

5.  length= 3.65

   width= 1.65

Step-by-step explanation:

Goshia [24]3 years ago
4 0

Answer:  

4. width= (-2+ sqrt(204)) /2

 length=  (2+ sqrt(204)) /2

5.  length= 3.65

    width= 1.65

Step-by-step explanation:

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Simplify, using the distributive property and then combining like terms. 2(x+y)+(2x−2y)
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Answer:

4x

Step-by-step explanation:

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The altitude of a triangle is increasing at a rate of 3 3 centimeters/minute while the area of the triangle is increasing at a r
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3 years ago
Y= 1/2x -3 on a graph<br> I’m confused on how to graph 1/2x - 3
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Put your equation into this!

Step-by-step explanation:

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3 0
4 years ago
Read 2 more answers
Initially, there were equal amount of roses and tulips at a store. Each bouquet was made with 3 roses and 4 tulips. After the bo
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Answer:There were 10 bouquets of roses made and 4.5 bouquets of tulips made

Step-by-step explanation:

8 0
3 years ago
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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