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3 years ago

Px + qy = r 2px - qy = 2r When solving this system of equations, x =

Mathematics

2 answers:

VladimirAG [237]3 years ago

8 0

Px + qy = r
2px - qy = 2r
----------------add
3px = 3r

x = 3r/3p
x = r/p

expeople1 [14]3 years ago

8 0

You just add the equations:
px + qy = r
2px - qy = 2r
3px = 3r
divide by 3
px = r
divide by p
x = (r)/(p)

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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

2 years ago

A sample of 20 joint specimens of a particular type gave a sample mean proportional limit stress of 8.41 mpa and a sample standa

nignag [31]

For this problem, the confidence interval is the one we are looking for. Since the confidence level is not given, we assume that it is 95%.

The formula for the confidence interval is: mean ± t (α/2)(n-1) * s √1 + 1/n

Where:

α= 5%

α/2 = 2.5%

t 0.025, 19 = 2.093 (check t table)

n = 20

df = n – 1 = 20 – 1 = 19

So plugging in our values:

8.41 ± 2.093 * 0.77 √ 1 + 1/20

= 8.41 ± 2.093 * 0.77 (1.0247)

= 8.41 ± 2.093 * 0.789019

= 8.41 ± 1.65141676

3 0

2 years ago

Order these numbers from least to greatest.

PSYCHO15rus [73]

<h3>Answer: 3.1691, 5.8, 5.802, 5.82</h3>

Explanation:

The smallest item is 3.1691 which is listed first. This is because 3 is smaller than 5.

Now to sort the values that start with 5.

Think of 5.82 as 5.820; think of 5.8 as 5.800

We can see that 820 is larger than 800, which means 5.820 is larger than 5.800; in short, 5.82 > 5.8

Through similar logic, we can see that 5.82 > 5.802 and it further means 5.82 is the largest item. The next largest is 5.802

The sub-list {5.802, 5.82, 5.8} sorts to {5.8, 5.802, 5.82}. These values are then written after the 3.1691 mentioned. This will produce the fully sorted list from smallest to largest.

4 0

2 years ago

The time to complete a standardized exam in the BYU-Idaho Testing Center is approximately normal with a mean of 70 minutes and a

leva [86]

Answer:

Step-by-step explanation:

Given that X the time to complete a standardized exam in the BYU-Idaho Testing Center is approximately normal with a mean of 70 minutes and a standard deviation of 10 minutes.

We have 68 rule as 2/3 of total would lie within 1 std deviation, and 95 rule as nearly 95% lie within 2 std deviations from the mean.

We have std deviation = 10

Hence 2 std deviations from the mean

= Mean ±2 std deviations

= $70$ ±20