The diagram show the metal piece 10mm 14mm 30mm 8mm8mm

We have shown that the null geodesics in a 3-space (x*)-(x, y, z) and line element with coordinates are given by: \(ds^2 = I(dx)^2 + 2ldxdy + 2mdxdz + 2m'dydz + ndy^2 + n'dz^2\)= 0. where I, l', m, m', n, n' are arbitrary constants satisfying \(12m^2 < 2I\)n.

Null geodesics in a 3-space (x*)-(x, y, z) and line element with coordinates are given by:

\(ds^2 = I(dx)^2 + 2ldxdy + 2mdxdz + 2m'dydz + ndy^2 + n'dz^2 = 0\)

where I, l', m, m', n, n' are arbitrary constants satisfying\(12m^2 < 2In\).

To prove this, we need to show that any solution to the above equation is a null geodesic, and any null geodesic can be expressed in this form.

Let's first assume that we have a solution to the above equation. We can write it in terms of a parameter u, such that:

x = x0 + Au
y = y0 + Bu
z = z0 + Cu

where A, B, and C are constants. Substituting these expressions into the line element equation, we get:

\(I(A^2)u^2 + 2lAuB + 2mAuC + 2m'BuC + n(B^2)u^2 + n'(C^2)u^2 = 0\)

Since this equation holds for any value of u, each term must be zero. Therefore, we get six equations:

\(I(A^2) + 2lAB + 2mAC = 0\)

\)2m'BC + n(B^2) = 0\)

\)n'(C^2) = 0\)

From the last equation, we get either C = 0 or n' = 0. If n' = 0, then the equation reduces to:

\)I(A^2) + 2lAB + 2mAC + n(B^2) = 0\)

which is the equation of a null geodesic in this 3-space. If C = 0, then we can write the line element equation as:

\)I(dx)^2 + 2ldxdy + 2m'dydz + ndy^2 = 0\)

which is also the equation of a null geodesic.

Now, let's assume we have a null geodesic in this 3-space. We can write it in the form:

x = x0 + Au
y = y0 + Bu
z = z0 + Cu

where A, B, and C are constants. Substituting these expressions into the line element equation, we get:

\)I(A^2) + 2lAB + 2mAC + n(B^2) + 2m'BC + n'(C^2) = 0\)

Since the geodesic is null, ds^2 = 0. Therefore, we get:

\)ds^2 = I(dx)^2 + 2ldxdy + 2mdxdz + 2m'dydz + ndy^2 + n'dz^2 = 0\)

which is the same as the line element equation we started with. Therefore, any null geodesic in this 3-space can be expressed in the form given by the equation above.

In conclusion, we have shown that the null geodesics in a 3-space (x*)-(x, y, z) and line element with coordinates are given by:

\)ds^2 = I(dx)^2 + 2ldxdy + 2mdxdz + 2m'dydz + ndy^2 + n'dz^2 = 0\)

where I, l', m, m', n, n' are arbitrary constants satisfying \)12m^2 < 2In.\)

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Answer:2.35

Step-by-step explanation:

The value of the expression is 2.35.

An expression is a way of writing a statement with more than two variables or numbers with operations such as addition, subtraction, multiplication, and division.

Example: 2 + 3x + 4y = 7 is an expression.

We have,

9.4 ÷ 4

= 9.4/4

= 94/(10 x 4)

= 94/40

= 2.35

Thus,

The value of the expression is 2.35.

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Answer:

  2/15

Step-by-step explanation:

You want to know the fraction that is 2/3 of 1/5.

In this context, "of" means "times."

  2/3 of 1/5 = (2/3)×(1/5) = (2·1)/(3·5) = 2/15

2/15 of the pets are alley cats.

Answer:

37° and 53°

Step-by-step explanation:

2x + 15 + 4x + 9 = 90 since the angles are complementary

6x + 24 = 90

6x = 66

x = 11

2x + 15 = 2(11) + 15 = 22 + 15 = 37

4x + 9 = 4(11) + 9 = 44 + 9 = 53

Check:  53 + 37 = 90

1. \(\boldsymbol{2x+15+4x+9=90}\)

2. \(\boldsymbol{x=11^{\circ}}\)

3. Measure of the first angle is \(\boldsymbol{37^{\circ}}\)

4. Measure of the second angle is \(\boldsymbol{53^{\circ}}\)

Two angles are said to be complementary if their sum if \(\boldsymbol{90^{\circ}}\)

Measure of the first angle\(=\boldsymbol{(2x+15)^{\circ}}\)

Measure of the second angle \(=\boldsymbol{(4x+9)^{\circ}}\)

1. \(\boldsymbol{2x+15+4x+9=90}\)

2.

\(6x+24=90\\6x=66\\\boldsymbol{x=11^{\circ}}\)

3.

Measure of the first angle \(=2(11)+15\)

                                            \(=\boldsymbol{37^{\circ}}\)

4.

Measure of the second angle \(=4(11)+9\)

                                                  \(=\boldsymbol{53^{\circ}}\)

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The variance of the length of the critical path is 5.76.

Option A is the correct answer.

We have,

To calculate the variance of the length of the critical path in the Critical Path Method (CPM) with three-time estimates (PERT), we can use the formula:

Variance = (Standard Deviation)²

Given that the standard deviation is 2.4, we can substitute it into the formula:

Variance = (2.4)² = 5.76

Therefore,

The variance of the length of the critical path is 5.76.

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The probability that xy-x-y is even is 80 when two distinct integers, x and y, are randomly chosen from the set {1,2,3,4,5,6,7,8,9,10}.

Given that,

The two distinct integers, x and y, are randomly chosen from the set {1,2,3,4,5,6,7,8,9,10}.

We have to find what is the probability that xy-x-y is even.

We know that,

x and y are two distinct integers.

So,

The odd integers are {1,3,5,7,9}

The even integers are {2,4,6,8,10}

So, we can write as the two integers are chosen from 1st set C(5,2) and the 2nd set also C(5,2)

We get,

C(5,2)C(5,2)-C(5,2)-C(5,2)

10×10-10-10

100-20

80

Therefore, The probability that xy-x-y is even is 80 when two distinct integers, x and y, are randomly chosen from the set {1,2,3,4,5,6,7,8,9,10}.

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Answer:

The balance in 2020 is $47,810,662.65576

Step-by-step explanation:

In order to find the balance in 2020, first we would have to calculate the number of year as follows:

number of years, t=2020-1626

number of years, t=394

Therefore, to calculate the balance in 2020, we would have to use the following formula:

f(t)= 24(1.0375)∧394

f(t)=24*1,992,110.94399

f(t)=$47,810,662.65576

The balance in 2020 is $47,810,662.65576

Answer: The answer is 11m + -3v.

Answer:

47.5

Step-by-step explanation:

keep 47 there

1/2 as a decimal well we know 1/2 is equivalent to 0.5 so just combine 47 & 0.5 and you'll get 47.5

Answer

I believe it is 47.5. I hope it helps!

\(\\ \Large\sf Minor\:Arc\longmapsto AOQ\)

\(\\ \Large\sf Major\:arc\longmapsto AXQ\)

Now

\(\\ \Large\sf\longmapsto m<AOQ=70°\)

\(\\ \Large\sf\longmapsto m<AXQ=2m<AOQ\)

\(\\ \Large\sf\longmapsto m<AXQ=2\times 70\)

\(\\ \Large\sf\longmapsto m<AXQ=140°\)

The values of the sides are;

41. x = 18√3. Option D

42. x = 6√3. Option A

Using the different trigonometric identities, we have;

41. Using the tangent identity, we have;

tan 60 = 9√2/y

cross multiply the values

y =9√2 ×√3

y = 9√6

Using the sine identity;

sin 45 = y/x

1/√2 = 9√6/x

cross multiply the values, we have;

x = 9√2 ×√3 ×√2

x = 18√3

42. Using the cosine identity

cos 60 = 3√3 /x

cross multiply, we have;

x = 6√3

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The value of a in the given figure is 7.5.

Given is a figure we need to simplify for the values of the angles in it,

Using the vertical angle theorem,

4a = 2b+6

2a = b+3........(i)

Also,

12b+6 + 2b+6 = 180 [linear pair angle]

14b+12 = 180

7b+6 = 90

7b = 84

b = 12

Put b = 12 in equation (i)

2a = 12+3

2a = 15

a = 7.5

Hence, the value of a in the given figure is 7.5.

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Answer:Simplify the radical by breaking the radicand up into a product of known factors, assuming positive real numbers.

6i√5

Step-by-step explanation:

I hope this helps

The given problem is,

→ √(-180)

Simplifying the given problem,

→ √(-180)

→ √(-1) × √(180)

→ i√(180)

→ i√(36) × √(5)

→ i × 6 × √(5)

→ 6i√(5)

Hence, the answer is 6i√(5).

The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

\(T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}\)

Where

\((t) = \ time\)

\(T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)\)

\(T_{s} = Surrounding \ temperature\)

\(T_{0} = Initial \ temperature \ of \ the \ body\)

\(k = constant\)

From the question,

\(T_{0} = 86 ^{o}C\)

\(T_{s} = -20 ^{o}C\)

∴ \(T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C\)

\(T_{0} - T_{s} = 106^{o} C\)

Therefore, the equation \(T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}\) becomes

\(T_{(t)}=-20+106 e^{kt}\)

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time \(t = 1 \ hour\), \(T_{(t)} = 58^{o}C\)

Then, we can write that

\(T_{(1)}=58 = -20+106 e^{k(1)}\)

Then, we get

\(58 = -20+106 e^{k(1)}\)

Now, solve for \(k\)

First collect like terms

\(58 +20 = 106 e^{k}\)

\(78 =106 e^{k}\)

Then,

\(e^{k} = \frac{78}{106}\)

\(e^{k} = 0.735849\)

Now, take the natural log of both sides

\(ln(e^{k}) =ln( 0.735849)\)

\(k = -0.30673\)

This is the value of the constant \(k\)

Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at \(t = 2 \ hours\)

From

\(T_{(t)}=-20+106 e^{kt}\)

Then

\(T_{(2)}=-20+106 e^{(-0.30673 \times 2)}\)

\(T_{(2)}=-20+106 e^{-0.61346}\)

\(T_{(2)}=-20+106\times 0.5414741237\)

\(T_{(2)}=-20+57.396257\)

\(T_{(2)}=37.396257 \ ^{o}C\)

\(T_{(2)} \approxeq 37.40 \ ^{o}C\)

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

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the person above me i correct

The rate of change of the function that has a graph that passes

through the points (-1, 3) and (-2,-4) is slope and is equal to 7.

We have given two points  (-1, 3) and (-2,-4) .

Rate of change of graph is given by slope

Using slope formula , we get

      m = -4 - 3 / -2 - (-1)

      m = -7 / -2 + 1

      m = -7 / -1

       m = 7

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To solve for the node voltages in the circuit shown, we can use the node voltage method. First, we need to label the nodes in the circuit. We will label the top node as Node 1, the middle node as Node 2, and the bottom node as Node 3.

Next, we will use Kirchhoff's current law (KCL) to write equations for each node. We can start with Node 1:
I1 - I2 = 0
where I1 is the current flowing into Node 1 from the left, and I2 is the current flowing out of Node 1 to the right.
Next, we can move on to Node 2:
I2 - I3 = 0
where I2 is the current flowing into Node 2 from the left, and I3 is the current flowing out of Node 2 to the right.
Substituting these expressions for I1 and I2 into the KCL equations, we get:
(V1 - V2)/R1 - (V2 - V3)/R2 = 0
(V2 - V3)/R2 - I3 = 0
Substituting the values of R1 and R2 given in the problem, we get:
(V1 - V2)/3 - (V2 - V3)/10 = 0
(V2 - V3)/10 - I3 = 0
Simplifying these equations, we get:
10(V1 - V2) - 3(V2 - V3) = 0
V2 - V3 = 10I3

Finally, we can use the fact that V2 = V3 to find V2:
V2 = V3 = 10V1/13
So the node voltages in the circuit are:
V1 = given
V2 = 10V1/13
V3 = 10V1/13

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The circumference of the circle is 26.4 cm and the area of the circle is 55.3896 cm².

The circumference and area of a circle of radius 4.2 cm can be calculated using the following formulas:

Circumference = 2πr, where r is the radius of the circle and π is a constant approximately equal to 3.14.

Area = πr², where r is the radius of the circle and π is a constant approximately equal to 3.14.

Circumference = 2πr = 2 × 3.14 × 4.2 cm = 26.4 cm

Area = πr² = 3.14 × (4.2 cm)² = 55.3896 cm²

Given the radius of the circle as 4.2 cm, the circumference of the circle can be found by using the formula for the circumference of a circle. The circumference of a circle is the distance around the circle and is given by the formula C = 2πr, where r is the radius of the circle and π is a constant approximately equal to 3.14. By substituting the given value of r, the circumference of the circle is calculated as follows:

Circumference = 2πr = 2 × 3.14 × 4.2 cm = 26.4 cm

Similarly, the area of the circle can be found by using the formula for the area of a circle. The area of a circle is given by the formula A = πr², where r is the radius of the circle and π is a constant approximately equal to 3.14. By substituting the given value of r, the area of the circle is calculated as follows:

Area = πr² = 3.14 × (4.2 cm)² = 55.3896 cm²

Therefore, the circumference of the circle is 26.4 cm and the area of the circle is 55.3896 cm².

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We can writey + dy = 2x1x2 / (x1+x2) + 2x1Δx2/ (x1+x2) + 2x2Δx1/(x1+x2)+ 2Δx1Δx2/ (x1+x2)On subtracting y from both sides, we getdy = 2x1Δx2/ (x1+x2) + 2x2Δx1/ (x1+x2) + 2Δx1Δx2/ (x1+x2)

Given y= x1/(x1+x2)  we need to find the total differential of y.It is given that, y= x1/(x1+x2)Let us assume, x1 = x1+Δx1, x2 = x2+Δx2. On substituting these values, we get + dy = (x1 + Δx1)/ (x1 + Δx1 + x2 + Δx2)We know that dy = y - (x1 + Δx1)/ (x1 + Δx1 + x2 + Δx2)

On further simplification, we get,dy = (Δx1(x2+Δx2))/(x1+Δx1+x2+Δx2)²-(Δx1x2)/((x1+Δx1+x2+Δx2)²)Since Δx1 and Δx2 are very small, we can neglect their squares and products, i.e., Δx1², Δx2², and Δx1.Δx2

Hence the total differential of y= x1/(x1+x2) is given by dy = (-x1x2/(x1+x2)²) dx1 + (x1²/(x1+x2)²) dx2. Note: x1 and x2 are independent variables.

Therefore, dx1 and dx2 are their differentials.Given y=2x1x2 /(x1+x2) Let us assume, x1 = x1+Δx1, x2 = x2+Δx2. On substituting these values, we gety + dy = 2(x1 + Δx1)(x2 + Δx2)/ (x1 + Δx1 + x2 + Δx2)On simplifying, we gety + dy = (2x1x2+2x1Δx2+2x2Δx1+2Δx1Δx2)/(x1+Δx1+x2+Δx2)

Since Δx1 and Δx2 are very small, we can neglect their squares and products, i.e., Δx1², Δx2², and Δx1.Δx2

Hence the total differential of y=2x1x2 /(x1+x2) is given by dy = (2x2/(x1+x2)²) dx1 + (2x1/(x1+x2)²) dx2.

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Answer:2.43

Step-by-step explanation:

Answer:

The mean increased, but the median stayed the same

Step-by-step explanation:

hope it help luvs

The mean increased, but the median remained constant If new dogs' weights are added to the data set.

The mean of a data set is defined as the ratio of the sum of observations to the total number of observations. It can be referred to as the average of a specific set of data or the arithmetic mean.

The formula for the mean of a given set of data is as follows:

Mean = Sum of Observations/Total number of observations

Given the set of observations as :

16, 34, 44, 20, 50, 19, 30, 54

Arrange the data in ascending order, we get

16, 19, 20, 30, 34, 44, 50, 54

The median of the given data would be (30+34)/2 = 32.

Here, n = 8

⇒ ∑x = Sum of Observations

⇒ ∑x = 267

Mean = ∑x/n

Mean = 267/8

Mean = 33.375

If two more dogs weighing 25 pounds and 45 pounds come to the kennel.

Arrange the given data in ascending order, we get

16, 19, 20, 25, 30, 34, 44, 45, 50, 54

The new median of the data would be (30+34)/2 = 32.

New mean = 337/10 = 33.7

Therefore, the mean increased, but the median remained constant If new dogs' weights are added to the data set.

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Kay deposited (c)$93.44  in cash and coins.

Money or coins that are legal tender can be used to pay for goods, debts, or services. Additionally, as stated by a corporation, it occasionally includes the worth of assets that can be quickly converted into cash. In its physical form, cash is another name for money. Bank accounts and marketable securities, such as government bonds and banker's acceptances, are typically included in the definition of "cash" in a business context.

We proceed from the whole backward. The amount deducted for cash received is first added back,

498.01+213.51 = 711.52

Each cheque that was deposited is then subtracted:

711.52-40.25 = 671.27

671.27-57.68 = 613.59

613.59-520.15 = 93.44

She started with (c) $93.44 in cash and coins.

Correct Question :

How much did kay deposit in cash and coins? a. $404.57 b. $213.51 c. $93.44 d. $284.50

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The given information is,

→ Length (l) = 7 ft

→ Breadth (b) = 4 ft

→ Height (h) = 4 ft

Formula we use,

→ l × b × h

The volume of cuboid will be,

→ l × b × h

→ 7 × 4 × 4

→ [ 112 ft³ ]

Hence, the volume is 112 ft³.

a) The initial amount a is 16,000.

b) The percent rate of change is 5%, the growth factor is 1.05.

c) The number of students enrolled after one year, based on the above growth factor, is 16,800.

d) The completion of the equation y = abˣ to find the number of students enrolled after x years is y = 16,000(1.05)ˣ.

e) Using the above exponential growth equation to predict the number of students enrolled after 22 years shows that 46,804 are enrolled.

An exponential growth equation shows the relationship between the dependent variable and the independent variable where there is a constant rate of change or growth.

An exponential growth equation or function is written in the form of y = abˣ, where y is the value after x years, a is the initial value, b is the growth factor, and x is the exponent or number of years involved.

a) Initial number of students enrolled at the college = 16,000

Growth rate or rate of change = 5% = 0.05 (5/100)

b) Growth factor = 1.05 (1 + 0.05)

c) The number of students enrolled after one year = 16,000(1.05)¹

= 16,800.

d) Let the number of students enrolled after x years = y

y = abˣ

y = 16,000(1.05)ˣ

e) When x = 22, the number of students enrolled in the college is:

y = 16,000(1.05)²²

y = 46,804

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The number of students enrolled at a college is 16,000 and grows 5% each year. Complete parts (a) through (e).

a) The initial amount a is ...

b) The percent rate of change is 5%, what is the growth factor?

c) Find the number of students enrolled after one year.

d) Complete the equation y = ab^x to find the number of students enrolled after x years.

e) Use your equation to predict the number of students enrolled after 22 years.

1) Considering the equation

3x -10 = -16 Add 10 to both sides

3x -10+10= -16 +10

Answer:

+.5

Step-by-step explanation:

Answer:

35$

Step-by-step explanation:

It will take 10 years and 11 months to payoff the balance. The total interest is $2,574.43