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JulsSmile [24]
3 years ago
11

What is the solution for 2y=x+2 x-3y=-5

Mathematics
2 answers:
Mars2501 [29]3 years ago
8 0

Answer:

x = 4

y = 3

Step-by-step explanation:

<u>Given </u><u>equations </u><u>:</u><u>-</u>

  • 2y = x + 2
  • x - 3y = -5

<u>Second</u><u> </u><u>equation</u><u> </u><u>can </u><u>be</u><u> written</u><u> as</u><u> </u><u>,</u><u> </u>

  • x - 3y = -5
  • -3y = -x - 5

<u>Adding</u><u> </u><u>them </u><u>:</u><u>-</u><u> </u>

  • -3y + 2y = 2 -5
  • -y = -3
  • y = 3

<u>Put </u><u>this </u><u>in </u><u>(</u><u>ii)</u><u> </u><u>:</u><u>-</u><u> </u>

  • x = 3y - 5
  • x = 3*3 - 5
  • x = 9 -5
  • X = 4
Ainat [17]3 years ago
3 0
Answer:
X=4
Y=3
You welcome
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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

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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
Help help help pls :)
kykrilka [37]

Answer:

opposite\approx 70.02

Step-by-step explanation:

The triangle in the given problem is a right triangle, as the tower forms a right angle with the ground. This means that one can use the right angle trigonometric ratios to solve this problem. The right angle trigonometric ratios are as follows;

sin(\theta)=\frac{opposite}{hypotenuse}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\\\\tan(\theta)=\frac{opposite}{adjacent}

Please note that the names (opposite) and (adjacent) are subjective and change depending on the angle one uses in the ratio. However the name (hypotenuse) refers to the side opposite the right angle, and thus it doesn't change depending on the reference angle.

In this problem, one is given an angle with the measure of (35) degrees, and the length of the side adjacent to this angle. One is asked to find the length of the side opposite the (35) degree angle. To achieve this, one can use the tangent (tan) ratio.

tan(\theta)=\frac{opposite}{adjacent}

Substitute,

tan(35)=\frac{opposite}{100}

Inverse operations,

tan(35)=\frac{opposite}{100}

100(tan(35))=opposite

Simplify,

100(tan(35))=opposite

70.02\approx opposite

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When this work has been done, we get P(x) = 160x - x^2 - 100x - 500, or

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