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3 years ago

Karen has a cube that has a side length of 6 in. The net of the cube is shown.

Mathematics

1 answer:

nordsb [41]3 years ago

6 0

Answer:

<h3>C. 180 square inches</h3>

Step-by-step explanation:

<h2>#carry on learning </h2><h2>#mark brainlits</h2>

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......Help Please.....

Ad libitum [116K]

Answer:

A= 4 , B= 3

Step-by-step explanation:

Simply put, use systems of equations to solve for A and B. First make y equal 4, and to make A equal to four, while being multiplied by 1 it has to be 4. Then solve backwards and now you know that A equals four, what multiplied by 4 is equal to 36, 9. What is the square root of 9? 3!

8 0

3 years ago

Si el área de una figura rectangular esta dada por A-bxh,¿cual es la siguiente expresiones representa el área de la puerta

stealth61 [152]

Respuesta:

3x² + 5x - 2

Explicación paso a paso:

Dado que :

Área del rectángulo, A = largo * ancho

Dimensión = 3x - 1; x + 2

Área = (3x - 1) * (x + 2)

Área = 3x² + 6x - x - 2

Área = 3x² + 5x - 2

8 0

3 years ago

What property was used to rewrite the polynomial expression (x+5)+3(a+(6x)y) to 5+(x+3a+(18x)y)

Umnica [9.8K]

<span>3(a+(6x)y) was clearly multiplied out as seen by the 3a and 18xy, so the distributive property was used there. In addition, the commutative and associative properties state that you can rearrange sums, so those were used too </span><span />

4 0

3 years ago

A heavy rope, 50 ft long, weighs 0.6 lb/ft and hangs over the edge of a building 120 ft high. Approximate the required work by a

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

6 0

3 years ago

Write the phrase as an expression. Then evaluate the expression when x = 5.

Nadusha1986 [10]

Answer:

6 + 8x

Step-by-step explanation:

1. Write the expression

6 + 8x

2. Plug in 5 for x

6 + 8(5) → 6 + 40 = 46

6 0

3 years ago