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1 month ago

Suppose the coordinate of a is 0, ar = 5, and at =7. What are the possible coordinates of the midpoint of rt ?

Mathematics

1 answer:

Zina [12.3K]1 month ago

4 0

Here, 'a' relates to 0.

There are two scenarios for 'r' and 't'.

Scenario 1.

Both are positioned on the same side to the right of 'a'.

In this case, 'r' would equal 5, and 't' would equal 7.

The midpoint between 'r' and 't' is $\frac{5+7}{2} =6$ .

Scenario 2.

If both are found to the left of 'a'.

Then 'r' would equal -5, while 't' would equal -7.

The midpoint is $\frac{-7-5}{2} =-6$ .

Scenario 3.

If 'r' is right of 'a' and 't' is left of 'a'.

Thus 'r' equals 5 and 't' equals -7.

The midpoint is $\frac{-7+5}{2}=-1$ .

Scenario 4.

If 'r' is left of 'a' while 't' is right of 'a'.

In this case, 'r' corresponds to -5 and 't' corresponds to 7.

The midpoint is $\frac{-5+7}{2}=1$ .

The potential midpoint coordinates for 'rt' are 6, -6, 1, and -1.

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A product is composed of four parts. In order for the product to function properly in a given situation, each of the parts must

lawyer [12517]

Answer:

P(functioning product) =.99*.99*.96*.96 =.0.903

Step-by-step explanation:

In order for the product to function, all four probabilities must succeed, hence:

P(Part-1)*P(Part-2)*P(Part-3)*P(Part-4)

where:

P(Part-1) = 0.96

P(Part-2) = 0.96

P(Part-3) = 0.99

P(Part-4) = 0.99

Since each part is independent, we apply the formula P(A∩B) = P(A)*P(B)

P (Functioning Product) = P(Part-1)*P(Part-2)*P(Part-3)*P(Part-4)

P (Functioning Product) = 0.96*0.96*0.96*0.99*0.99

P(Functioning Product) = 0.903

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1 month ago

Determine the volume of the solid that lies between planes perpendicular to the x-axis at x=0 and x=4. The cross sections perpen

babunello [11817]

Answer:

Volume = 16 unit^3

Step-by-step explanation:

Given:

- The solid is situated between x = 0 and x = 4.

- Diagonal lines stretch from y = sqrt(x) to y = -sqrt(x)

Find:

Calculate the volume enclosed.

Solution:

- We will begin by finding the area projection of the solid on the plane at x = 0.

A(x) = 0.5*(diagonal)^2

- The diagonal ranges from y = sqrt(x) to y = -sqrt(x), thus

A(x) = 0.5*(sqrt(x) + sqrt(x))^2

A(x) = 0.5*(4x) = 2x

- By utilizing the area function, we will integrate across x from 0 to 4 to determine the solid's volume:

V = integral(A(x)).dx

V = integral(2*x).dx

V = x^2

- Evaluating from 0 to 4:

V= 16 - 0 = 16 unit^3

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1 month ago

Select the correct answer. What is the surface area of the victory podium shown here? Include all surfaces of the podium, includ

lawyer [12517]

Answer:

The correct answer is Option C.

Step-by-step explanation:

We are provided with the dimensions of the podium, including its length, width, and height, and need to determine its surface area.

To find the surface area of a cuboid, the formula used is 2(LW + WH + HL)

Referring to the provided figure, the dimensions of the upper cuboid are

L=2.5 ft., W=1.5 ft, H=1.5 ft

$\text{Surface area=}2(2.5(1.5)+1.5(1.5)+1.5(2.5))=19.5ft^2$

$\text{Area of bottom face of upper cuboid=}L\timesW=2.5\times1.5=3.75ft^2$

Since the upper face appears twice, it must be deducted to accurately calculate the area

The dimensions for the lower podium are

L=7.5 ft., W=1.5 ft, H=1.5 ft

$\text{Surface area=}2(7.5(1.5)+1.5(1.5)+1.5(7.5))=49.5ft^2$

Therefore, the overall surface area of the podium, including the bottom, is

19.5 + 49.5 - 3.75 = 65.25 square feet.

Thus, option C is indeed correct.

4 0

1 month ago

A freight company has shipping orders for two products. The first product has a unit volume of 10 cu ft, and it weighs 50 lbs. T

Zina [12379]

116 units of the first product and 380 units of the second product. Product 1 has a volume of 10 cu ft, while product 2 has a volume of 3 cu ft. The truck has a capacity of 2300 cu ft. Product 1 weighs 50 lbs and product 2 weighs 40 lbs, with the truck able to handle 21,000 lbs. Let X represent the units of product 1 and Y represent the units of product 2. The equations to solve are 10X+3Y=2300 and 50X+40Y=21000. Solving these equations gives us 116 for product 1 and 380 for product 2.

8 0

1 month ago

The frequency distribution of weights (in kg) of 40 persons is given below.

AnnZ [12381]

Answer: (a) 4

(b) 5

(c) 14

(d)

Class interval Class mark

30 - 35 $\dfrac{35+30}{2}=32.5$

35 - 40 $\dfrac{35+40}{2}=37.5$

40 - 45 $\dfrac{40+45}{2}=42.5$

45 - 50 $\dfrac{45+50}{2}=47.5$

50 - 55 $\dfrac{50+55}{2}=52.5$

Step-by-step explanation:

The data consists of 40 individuals:

Weights (in kg) Frequency

30 - 35 6

35 - 40 13

40 - 45 14

45 - 50 4

50 - 55 3

(a)

The lower limit refers to the smallest value in a class interval.

Accordingly, the lower limit for the fourth-class interval (45 - 50) is 4.

(b)

The class size is calculated by subtracting the lower limit from the upper limit.

For the first class interval, class size = 35 - 30 = 5.

Thus, all intervals share a consistent class size of 5.

(c)

The highest frequency in this data is 14, corresponding to the class interval of 40 - 45.

This class interval thus demonstrates the greatest frequency.

(d)

Class mark is the middle value of each class interval.

Class interval = $\dfrac{(\text{Upper limit+Lower limit})}{2}$

Class interval Class mark

30 - 35 $\dfrac{35+30}{2}=32.5$

35 - 40 $\dfrac{35+40}{2}=37.5$

40 - 45 $\dfrac{40+45}{2}=42.5$

45 - 50 $\dfrac{45+50}{2}=47.5$

50 - 55 $\dfrac{50+55}{2}=52.5$

8 0

2 months ago