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Umnica [9.8K]
3 years ago
8

La altura de un trapecio isósceles es de 4 cm, la suma de las medidas de las bases es de 14 cm y los lados oblicuos miden 5 cm.

Averigua las medidas de las bases del trapecio.
Por favor, necesito ayuda.
Mathematics
1 answer:
Temka [501]3 years ago
3 0
Teorema de Pitagoras a^2 + b^2 = c^2
4^2 + b^2 = 5^2
16 + b^2 = 25
- 16. -16
b^2 =. 9
b = 3

Si el fondo base mide mas que el superior base, miden 14 (fondo) y 14 - 3 (izquierda) -3 (derecho) = 8
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Anastasy [175]

Answer:

Exercise (a)

The work done in pulling the rope to the top of the building is 750 lb·ft

Exercise (b)

The work done in pulling half the rope to the top of the building is 562.5 lb·ft

Step-by-step explanation:

Exercise (a)

The given parameters of the rope are;

The length of the rope = 50 ft.

The weight of the rope = 0.6 lb/ft.

The height of the building = 120 ft.

We have;

The work done in pulling a piece of the upper portion, ΔW₁ is given as follows;

ΔW₁ = 0.6Δx·x

The work done for the second half, ΔW₂, is given as follows;

ΔW₂ = 0.6Δx·x + 25×0.6 × 25 =  0.6Δx·x + 375

The total work done, W = W₁ + W₂ = 0.6Δx·x + 0.6Δx·x + 375

∴ We have;

W = 2 \times \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= 2 \times \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 750

The work done in pulling the rope to the top of the building, W = 750 lb·ft

Exercise (b)

The work done in pulling half the rope is given by W₂ as follows;

W_2 =  \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 562.5

The work done in pulling half the rope, W₂ = 562.5 lb·ft

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Which quadrilateral makes this statement true?<br>Quadrilaterat ABCD≈____​
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FGHE

Step-by-step explanation:

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Irina18 [472]

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Step-by-step explanation:

Let the ends of the given segment are A and B.

Coordinates of A → (8, 6)

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(x, y) → (kx, ky)

If k = \frac{2}{3}

(x, y) → (\frac{2}{3}x, \frac{2}{3}y)

By this rule coordinates of the image points of A and B will be,

A(8, 6) → A'(\frac{2\times 8}{3},\frac{2\times 6}{3})

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3 years ago
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Answer:

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Step-by-step explanation:

The given data set is

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Formula for mean:

Mean=\frac{\sum x}{n}

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Mean of the data set is

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\sigma=\sqrt{\frac{\sum (x-\overline{x})^2}{n}}

Formula for standard deviation for sample:

\sigma=\sqrt{\frac{\sum (x-\overline{x})^2}{n-1}}

\sigma=\sqrt{\frac{0.231792}{50-1}}

\sigma=\sqrt{0.004730449}

\sigma=0.06877826

\sigma\approx 0.069

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3 years ago
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