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Tresset [83]
3 years ago
7

Find the distance between 6,-2 and 1,-2​

Mathematics
2 answers:
lesantik [10]3 years ago
8 0

The distance between (6,-2) and (1,-2) is 5.

Anika [276]3 years ago
3 0

Answer:

The distance:

d = 5

Step-by-step explanation:

To find the distance between points (6,-2) and (1,-2), you need the <u>distance formula</u>:

d = \sqrt{(x_{2} - x_{2})^{2} + (y_{2} - y_{1})^{2}}

-Use the given points (6,-2) and (1.-2) for the distance formula:

d = \sqrt{(1 - 6)^{2} + (-2 + 2)^{2}}

-Then, solve the formula:

d = \sqrt{(1 - 6)^{2} + (-2 + 2)^{2}}

d = \sqrt{(-5)^{2} + (0)^{2}}

d = \sqrt{25 + 0^{2}}

d = \sqrt{25 + 0}

d = \sqrt{25}

d = 5

So, the distance is 5.

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SIZIF [17.4K]

52 weeks in a year

27600/52 = 530.769

so 1 weeks pay = 530.77

 she shouldn't pay more than that on rent


(Round answer as needed)

 

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3 years ago
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What is the slope of the line 7x + 2y = 5
blondinia [14]

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-7/2

Step-by-step explanation:

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At the beginning of a basketball season, the panthers won 40 games out of 100 games. At this rate, how many games will they win
denis23 [38]

At the rate of winning, they should win 44 games in  normal 110-game season.

7 0
2 years ago
Solve for the values of x and y in the diagram
qaws [65]

Answer: y = 0.5, x = 50

Step-by-step explanation: Both triangles in the picture are isosceles, telling us that the 2 angles at the bottom are congruent. With this, we can find y by doing the following:

a triangle has 180 degrees so we subtract the given 50 which gives us 130

2(2y + 64) = 130

4y + 128 = 130

y = .5

This means that the bottom 2 angles are both 65. Since the top angle of the second triangle is supplementary to the bottom angle of the first one, the top angle of the second triangle is 115. So, we find x by:

2(45 - x/4) = 65

x = 50

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7 0
2 years ago
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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
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