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fenix001
15 days ago
5

the length of the swimming pool is 8 m longer than its width and the area is 105 m squared write in quadratic equation ​

Mathematics
1 answer:
babunello [3.6K]14 days ago
5 0

Response:

Detailed explanation:

The length of the pool is longer than its width by 8 meters.

If we designate L as length and W as width, we can express this as:

L = 8 + W

We also know that the area amounts to 105 m squared.

Note:

Area of a Rectangle = Length x Width

Thus, 105 = (8 + W) x W

105 = 8w + w^2\\

To adjust the equation, subtract 105 from both sides:

w^2 + 8w - 105 = 0

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7 days ago
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In triangle RST, m∠R > m∠S + m∠T. Which must be true of triangle RST? Check all that apply.
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Answer:

∠ R > 90°

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Explicación paso a paso:

Dado que en el triángulo RST

\angle R> (\angle s +\angle T)

Ahora, según la condición, un ángulo es mayor que la suma de los otros dos ángulos.

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Por lo tanto, si un ángulo mide  90°, la suma de los otros dos debe ser igual a 90°

Y si uno de los ángulos es de 90°, solo los otros dos pueden ser de 45° cada uno.

Aquí suponiendo que el ángulo s = ángulo T = 30°, entonces con esta condición el ángulo r sería de 120°, que es mayor que 90°.

De lo anterior, se concluye que

Para, \angle R> (\angle s +\angle T)

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11 days ago
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Consider the vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + u = 8 co
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Answer:

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The characteristic equation is k² + 1 = 0, which leads to k² = -1, resulting in k = ±i.

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Applying the method of undetermined coefficients, we have

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Calculating the derivatives gives us Uc’(t) = -Pwsin wt  + Qwcos wt

And differentiating again yields Uc’’(t) = -Pw^2cos wt  - Qw^2sin wt

With the equation U’’ + u = 8cos wt, we substitute:

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This simplifies to (-Pw^2 + P) cos wt   + (-Qw^2 + Q) sin wt = 8cos wt.

From -Pw^2 + P = 8, we find P= 8  /(1- w^2).

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U’(t) = -C1 sin t + c2 cos t - 8 w sin wt /(1- w^2).

Evaluating at t = 0 gives us:

U’(0) = -C1 sin (0) + c2 cos (0) - 8 w sin (0) /(1- w^2) = 7.

Thus, c2 = 7.

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