the volume of the indian ocean is about 2.1 X 10^17 cubic meters. How many years would it take for the Narmada River to fill the Indian, assuming it has a water flow of 6.3 X 10^11 cubic meters per year? Answer in scientific notation

Answer:

2x^2 = (2 x 2) x (2 x 2)

= 4 x 4

= 16

I hope this helps you

Answer:

Triangle A = scalene

Triangle B = scalene

Triangle C = equilateral

Step-by-step explanation:

scalene triangle = all lengths are different

isosceles triangle = 2 side lengths are the same

equilateral triangle = all 3 side lengths are the same

In the given diagram, the volume of the composite shape is 12,672 ft³

From the question, we are to calculate the volume of the given composite shape

The composite shape is made up of a pyramid and a cuboid

Thus,

Volume of the shape = Volume of pyramid + Volume of cuboid

Volume of pyramid = 1/3 × base area × height

Volume of cuboid = Length × Width × Height

Calculating the volume of the pyramid

Volume of pyramid = 1/3 × 24² ×15

Volume of pyramid = 1/3 × 576 ×15

Volume of pyramid = 576 × 5

Volume of pyramid = 2880 ft³

Calculating the volume of the cuboid

Volume of the cuboid = 24² × 17

Volume of the cuboid = 9792 ft³

Thus,

Volume of the shape = 2880 ft³ + 9792 ft³

Volume of the shape = 12,672 ft³

Hence, the volume of the shape is 12,672 ft³

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a. Using Taylor series d(x³eˣ²)/dx about x = 0 is  x⁴.

b. Using Taylor series d⁷(x³eˣ²)/dx⁷ about x = 0 is  x¹⁰.

A Taylor series is a polynomial expansion of a function about a given point. It is given by f(x - a) = ∑(x - a)ⁿfⁿ(x - a)/n! where

Given that the Taylor series of the function f(x) = x³eˣ² about x = 0 is

f(x) = x³ + x⁵ + x⁷/2! + x⁹/3! + x¹¹/4!, (1) we proceed to find the given variables

a. To find d( x³eˣ²)/dx about x = 0, the Taylor series expansion about x = 0 is given by

f(x - a) = ∑(x - a)ⁿfⁿ(a)/n!

f(x - 0) = ∑(x - 0)ⁿf(0)/n!

f(x) = ∑xⁿf(0)/n!

f(x) = x⁰f(x)/0! + xf(x)/1! + x²f(x)/2! + x³f(x)/3! + ....

f(x) = f(x) + xf¹(x) + x²f²(x)/2! + x³f³(x)/3! + ....(2)

Since fⁿ(x) is the nth derivative of f(x), and we desire f¹(x) which is the first derivative of f(x). Comparing equations (1) and (2), we have that

x⁵ =  xf¹(x)

f¹(x) =  x⁵/x

= x⁴

So, d( x³eˣ²)/dx about x = 0 is  x⁴.

b. To find d⁷( x³eˣ²)/dx⁷ about x = 0, the Taylor series expansion about x = 0 is given by

f(x - a) = ∑(x - a)ⁿfⁿ(a)/n!

f(x - 0) = ∑(x - 0)ⁿf(0)/n!

f(x) = ∑xⁿf(0)/n!

f(x) = x⁰f(x)/0! + xf(x)/1! + x²f(x)/2! + x³f(x)/3! + ....

Expanding it up to the 8 th term, we have that

f(x) = f(x) + xf¹(x) + x²f²(x)/2! + x³f³(x)/3! + x⁴f⁴(x)/4! + x⁵f⁵(x)/5! + x⁶f⁶(x)/6!  + x⁷f⁷(x)/7!.....(3)

Now expanding equation (1) above to the 8th term by following the pattern, we have that

f(x) =   x³ + x⁵ + x⁷/2! + x⁹/3! + x¹¹/4! + x¹³/5! + x¹⁵/6!  + x¹⁷/7!.....(4)

Since fⁿ(x) is the nth derivative of f(x), and we desire f⁷(x) which is the seventh derivative of f(x). Comparing equations (3) and (4), we have that

x⁷f⁷(x)/7! = x¹⁷/7!  

f⁷(x) =  x¹⁷/x⁷

= x¹⁰

So, d⁷( x³eˣ²)/dx⁷ about x = 0 is  x¹⁰.

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The shortest path from vertex A to vertex L is ACFIL. The total distance of the path is 27 units.

To find the shortest path from vertex A to vertex L, we can use Dijkstra's algorithm. We start by marking the distance of each vertex from A as infinite except for A, which is 0. Then, we choose the vertex with the smallest marked distance and update the distances of its neighbors. We repeat this process until we reach L.

In this case, we would start at A and update the distances of B and D to 2 and 19, respectively. We would then choose B and update the distances of C, F, and E to 11, 10, and 18, respectively.

Next, we would choose F and update the distances of I and L to 27 and 30, respectively. Finally, we would choose L and have found the shortest path from A to L: ACFIL with a total distance of 27.

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Answer:

$2,631.55

Step-by-step explanation:

To find the total amount in the account after 6 years, we can use the compound interest formula.

[tex]\boxed{\sf A=P\left(1+\dfrac{r}{n}\right)^{nt}}[/tex]

where:

Given values:

Substitute the given values into the formula and solve for A:

[tex]\implies \sf A=2000\left(1+\dfrac{0.046}{4}\right)^{4 \cdot 6}[/tex]

[tex]\implies \sf A=2000\left(1+0.0115\right)^{24}[/tex]

[tex]\implies \sf A=2000\left(1.0115\right)^{24}[/tex]

[tex]\implies \sf A=2000\left(1.3157739...\right)[/tex]

[tex]\implies \sf A=2631.54794...[/tex]

Therefore, the total amount after 6 years is $2,631.55 rounded to the nearest cent.

Note that in order to map A onto B, Kyle would have to dilate the given figure by a scale factor or 3.

A dilation is a function f from a metric space M into itself that fulfills the identity d=rd for all locations x, y in M, where d is the distance between x and y and r is some positive real integer. Such a dilatation is a resemblance of space in Euclidean space.

The original point of figure A which has 4 points are


(0,02)
(-1, 2)
(0, 1)
(1, 2)

Multiply all th e points by 3, and you get,

(0,02) x 3 = (0, -6) =
(-1, 2) x 3 = (-3, 6)
(0, 1) x 3 = (0, 3)
(1, 2) x3 = (3, 6)

Plotting the new values will give us the transformation (dilation) required. See the attached image.

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Answer: We can solve this problem by finding the total weight of the meat and then dividing by the weight of each patty to see if we can make the desired number of patties.

Let's add the weights of the two packages of meat:

10 + 11.5 = 21.5

For option A, we need 87 1/4-pound patties.

Each pound of meat will yield 4 patties of this size, so 21.5 pounds of meat will yield:

21.5 x 4 = 86 patties

Since we need 87 patties, Claude does not have enough meat for option A.

For option B, we need 45 1/2-pound patties.

Each pound of meat will yield 2 patties of this size, so 21.5 pounds of meat will yield:

21.5 x 2 = 43 patties

Since we need 45 patties, Claude does not have enough meat for option B.

For option C, we need 30 1/2-pound patties and 2 74-pound patties.

Each pound of meat will yield 2 patties of the smaller size and 1 patty of the larger size, so 21.5 pounds of meat will yield:

(21.5 x 2) + (2 x 1) = 43 + 2 = 45 patties

Since Claude only has 21.5 pounds of meat, he does not have enough for option C.

For option D, we need 29 3/4-pound patties.

Each pound of meat will yield 1.5 patties of this size, so 21.5 pounds of meat will yield:

21.5 x 1.5 = 32.25 patties

Since we only need 29.75 patties, Claude has enough meat for option D.

Therefore, the answer is Yes, Claude can make 29 3/4-pound hamburger patties.

Using the properties of a parallelogram, the values of x and y are:

x = 9; y = 22.

Since the opposite sides are parallel and equal to each other in the parallelogram above, therefore, Opposite angles are equal (or congruent) while the consecutive angles are supplementary to each other.

Therefore, we have:

6y = 180 - 48

6y = 132

Divide both sides by 6:

6y/6 = 132/6

y = 22

(5x + 3) + 6y = 180

5x + 3 + 6(22) = 180

5x + 135 = 180

5x = 180 - 135

5x = 45

x = 9

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Answer: 3.6 is 5% of 72

Step-by-step explanation: If 3.6 is 5% then just multiply 3.6 by 95 to get 68.4 and add to get 72, Yw.

To explain why A + B is irrational, we need to show that it cannot be expressed as a ratio of two integers.

Suppose that A + B is rational, which means we can write it as the ratio of two integers p and q (where q is not zero):

A + B = p/q

Now, we can substitute the values of A and B and simplify:

√363 + √27 = p/q

We can then rearrange the terms to isolate one of the square roots:

√363 = p/q - √27

We can square both sides of this equation to eliminate the square roots:

363 = p^2/q^2 + 27 - 2(p/q)√27

Notice that the right-hand side of this equation has a term with a square root. This means that if we assume that A + B is rational, we arrive at a contradiction: we have shown that √363 (which is equal to A) is irrational, which means that p^2/q^2 + 27 must also be irrational. However, the left-hand side of the equation is rational. Therefore, our assumption that A + B is rational must be false, and we conclude that A + B is irrational.

To explain why AB is rational, we can use the fact that the product of two rational numbers is rational.

We can rewrite A and B as follows:

A = √(363) = √(121 x 3) = √(11^2 x 3) = 11√3

B = √(27) = √(9 x 3) = √(3^2 x 3) = 3√3

Therefore, AB = 11√3 x 3√3 = 33 x 3 = 99.

Since 99 is a rational number (which can be expressed as the ratio of the integers 99 and 1), we conclude that AB is rational.

Answer:

Set your calculator to degree mode.

cos(37°) = 52/x

x cos(37°) = 52

x = 52/cos(37°) = 65.1

Answer:

$162.01

Step-by-step explanation:

amount of money she earns: 12($18.50) = $222

spends $59.99

amount of money after her purchase: $222-$59.99=$162.01

Where Rectangle ABCD and A’B’C’D’ are similar.

a. the scale factor from ABCD to A’B’C’D’ is 1/2

b. What is the scale factor from A’B’C’D’ to  ABCD is 2

In mathematics, a scale factor is the ratio of matching measurements of an item to a representation of that thing. The copy will be bigger if the scaling factor is a full number. The duplicate will be smaller if the scaling factor is a fraction.

When the scale factor is less than one, you are going from big to small. You are dilating negatively.

When you are going from small to big, you scale factor is reversed. In this case, we had 1/2 in a, in b it became 2/1 which = 2.

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Full Question:

Rectangle ABCD and A’B’C’D’ are similar.

a. What is the scale factor from ABCD to A’B’C’D’?

b. What is the scale factor from A’B’C’D’ to ABCD?

See attached image.

The transformation of f(x) to g(x) is (a) f(x) is shifted up 1 unit to g(x).

From the question, we have the following parameters that can be used in our computation:

The functions f(x) and g(x)

In the graph, we can see that

So, we have

Difference = 1 - 0

Evaluate

Difference = 1

This means that the transformation of f(x) to g(x) is (a) f(x) is shifted up 1 unit to g(x).

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1) The amount is php 340,060

2) The rate is 13%

3) The amount is  php 137,622

Compound interest is commonly used in many types of financial products, such as savings accounts, CDs, and loans.

1)

[tex]A = P(1 + r/n)^nt\\A = 80000( 1 + 0.16)^9.75\\A = php 340,060[/tex]

2)

[tex]12300 = 9750(1 + r)^2\\12300/9750 = (1 + r)^21.26 = (1 + r)^2\\ln 1.26 = 2 ln1 + r\\ln 1.26/2 = ln1 + r\\0.12 = ln1 + r\\e^{0.12} = 1 + r\\r = e^{0.12} - 1\\r = 0.13\\r = 13%[/tex]

3)

[tex]A = 50000(1 + 0.15/2)^2 * 7\\A = php 137,622[/tex]

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The missing part of the attached proof:

a) definition of angle congruence

b) ∠CPA ≅ ∠CPB

c) CP ≅ CP

The complete paragraph proof would be,

Because CP is perpendicular bisector of AB, CP is perpendicular to AB and point P is the midpoint of AB.

By definition of midpoint,

AP = BP

and by the definition of perpendicuar lines,

m∠CPA = m∠CPB = 90°

Then by definition of segment congruence,

AP ≅ BP

and by definition of angle congruence,

∠CPA ≅ ∠CPB

By the reflexive property of segment congruene,

CP ≅ CP

So, ΔCPA ≅ ΔCPB         ........by SAS congruence theorem

and CA ≅ CB because corresponding parts of congruent triangles are congruent.

So, CA = CB by the definition of segment congruence.

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The tangent function for the given terminal point is undefined.

The given terminal point of Θ is (0,-1). The tangent of an angle is defined as the quotient between the y-component and the x- component of the terminal point.

So, according to the question, tan Θ = -1/0. Hence, tan Θ is undefined as -1/0 is not defined.

Hence, tan Θ for the given function is not defined.

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Answer:

A) Sequence 2: 15,13,11,9,7

A) Ordered pairs (12,15), (16,13), (19,11), (22,9), (25,7)

B)  Sequence 1: 1,5,17,53,161

B) Sequence 2: 6,15,33,69,141

B) Ordered pairs (1,6), (5,15), (17,33), (53, 69) (161,141)

Step-by-step explanation:

Helping in the name of Jesus.

Answer:

A) Sequence 2: 15,13,11,9,7

A) Ordered pairs (12,15), (16,13), (19,11), (22,9), (25,7)

B)  Sequence 1: 1,5,17,53,161

B) Sequence 2: 6,15,33,69,141

B) Ordered pairs (1,6), (5,15), (17,33), (53, 69) (161,141)

Step-by-step explanation:

Helping in the name of typing.

It will take 15.03 years for Bentley's money to triple.

For Bentley:

We have

Principal (P) = $670

Interest rate (r) = 6 7/8% = 0.06875 (compounded quarterly)

Time (t) = unknown

Amount after t years (A) = [tex]P(1 + r/4)^{4t}[/tex]

For Camden:

We have

Principal (P) = $670

Interest rate (r) = 7 3/8% = 0.07375 (compounded annually)

Time (t) = unknown

Amount after t years (A) = [tex]P(1 + r)^t[/tex]

We have to find t such that 3P_Bentley = Camden.

[tex]2010 = 2010(1 + r/4)^{4t} / (1 + r)^t[/tex]

Simplifying, we get:

[tex](1 + r/4)^{4t} / (1 + r)^t = 1/3[/tex]

Taking natural logarithm:

4t ln(1 + r/4) - t ln(1 + r) = ln(1/3)

4 ln(1 + r/4) - ln(1 + r) = ln(1/3) / t

t = (4 ln(3.88356) - 4 ln(16)) / ln(3)

Solving for t, we get:

t = 15.034752564

t = 15.03 years.

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The inequality to represent the situation is 11x + 18 ≤  40.

The rental car company charges $22 per day to rent a car at 0.08 for every mile driven. Samuel wants to rent a car knowing that he plans to drive 450 miles he has at most to spend $80.

Therefore, the inequality to represent the situation of Samuel when x is the number of days can be represented as follows:

Hence,

22x + 0.08 × 450 ≤ 80

22x + 36 ≤  80

Divided both sides of the inequality by 2

11x + 18 ≤  40

Hence, the required inequality expression is 11x + 18 ≤  40

where

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Answer:

See explanation.

Step-by-step explanation:

Let's analyze each solution to determine if it's viable or non-viable based on the given constraints: Johnnie and Amy have $38.50, and they cannot buy more than 7 lunches.

1. 8 slices of pizza and 0 hamburgers

This solution is non-viable because they are not allowed to buy more than 7 lunches. In this case, they would be buying 8 lunches.

2. 5 slices of pizza and 2 hamburgers

Cost: 5×4.75+2×5.55=23.75+11.10=34.85

This solution is viable because the total cost is within their budget, and they are buying 7 lunches.

3. 4 slices of pizza and 3 hamburgers

Cost: 4×4.75+3×5.55=19.00+16.65=35.65

This solution is viable because the total cost is within their budget, and they are buying 7 lunches.

4. 3 slices of pizza and 2 hamburgers

Cost: 3×4.75+2×5.55=14.25+11.10=25.35

This solution is non-viable because they are only buying 5 lunches, which is less than the maximum allowed. But if it is acceptable to only buy below 7 lunches, then this solution can be viable.

5. 2 slices of pizza and 6 hamburgers

Cost: 2×4.75+6×5.55=9.50+33.30=42.80

This solution is non-viable because the total cost exceeds their budget.

6. 0 slices of pizza and 7 hamburgers

Cost: 7×5.55=38.85

This solution is non-viable because the total cost exceeds their budget.

In conclusion, the viable solutions are:

5 slices of pizza and 2 hamburgers

4 slices of pizza and 3 hamburgers

Possible in being viable:

3 slices of pizza and 2 hamburgers

The number of sun flowers with a height  of 27¹/₂ inches or more are 4 more than those with a height of less than 27¹/₂ inches

A dot plot, which is also known as a strip plot or dot chart, is a simple form of data visualization that comprises of data points plotted as dots on a graph with an x- and y-axis. These types of charts are used to graphically depict certain data trends or groupings.

The number of heights below 27¹/₂ inches that exists in the given dot plot is seen to be 8 in number.

Similarly, the number of dot plots that exists above or equel to 27¹/₂ inches from the given dot plot are 12 in number.

Therefore, we can say that:

Difference in total number for both parameters = 12 - 8= 4

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The dimensions of each rectangle are x by (x- y) and y by (x - y); option A and D

The area of each rectangle is x² - 2xy + y²  and x² - xy

The total area of the remaining figure is 2x² - 4xy + 2y²

This figure represents the difference of two squares in that x² - y² = (x + y)(x -y)

The area of each rectangle is:

x * (x- y) =  x² -xy

and

y * (x - y) = xy - y²

The total area of the remaining figure is:

x² -xy + xy - y² = x² - y²

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The sine of the angle ANM is equal to 3/5 as the triangles are similar.

Given that the sine of angle ACB is equal to 3/5, line MN is parallel to BC and MN is midsegment of triangle ABC which implies that the triangles ABC and ANM are similar so;

sine ANM = 3/2 ÷ 5/2

sine ANM = 3/2 × 2/5

sine ANM = 3/5.

Therefore, the sine of the angle ANM is equal to 3/5 as the triangles are similar.

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Answer: C: 5/8

Step-by-step explanation:

Simply 5/8 of the bar is filled in

Using probability, we can find probability of the random variable, x falling in the shaded region as to be 5/8.

Here, we have,

Probability is the ratio of favorable outcomes to all other potential outcomes of an event. The symbol x can be used to express the quantity of successful outcomes for an experiment with 'n' outcomes. The following formula can be used to determine an event's probability.

Positive Outcomes/Total Results = x/n = Probability(Event)

Let's look at a simple example to better understand probability. Imagine that we need to predict whether it will rain or not. The right response to this question is "Yes" or "No." Whether it rains or not is uncertain. Probability is used to predict the outcomes when tossing coins, rolling dice, or drawing cards from a deck of cards.

Here in the question,

Total region = 8.

Shaded region = 5

So, probability of falling in the shaded region = 5/8

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In order to do a comparison for percents, you must convert them to decimal form and then juxtapose the decimal values side by side to see which is greater or less.

A percentage is a figure or ratio stated as a fraction of 100 in mathematics. The percent sign is commonly used, however the acronyms pct., pct, and sometimes pc are also used. A % is a number with no dimensions; it has no unit of measurement.

To calculate the percentage, divide the amount by the total value and multiply the result by 100.

Divide a percentage by 100 to get a decimal. So 25% is equal to 25/100, or 0.25. Multiply a decimal by 100 (simply shift the decimal point two places to the right) to convert it to a percentage. For example, 0.065 equals 6.5% and 3.75 equals 375%.

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1. The probability that P(X > 979) = 0.567

2. The probability that P(M > 979) = 0.956

To calculate the probability that  a single randomly selected value is more than 979 dollars.

 979-1016

= -37/209

= -0.177

P(X > -0.177) = 0.5675

Therefore P(X > 979) = 0.568

To calculate that a sample of size is randomly selected with a mean that is more than 979 dollars.

209 / √94 = 21.557

-37/ 21.557 = -1.7164

P(Z > -1.714) = 0.9564

therefore P(M > 979) = 0.956

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The value of proᵥᵥv is (-30/37)i + (42/37)j and the decomposed vector v into v₁ and v₂ are  6 + 6j and -5i + j - 6 respectively.

a. To find the projection of w onto v (proᵥᵥw), we can use the formula,

proᵥᵥw = (w · ȳ)ȳ where ȳ is the unit vector in the direction of v.

First, let's find the unit vector in the direction of v,

|v| = √((-5)² + 7²) = √(74)

ȳ = v/|v| = (-5/√(74))i + (7/√(74))j

Next, let's find the dot product of w and ȳ,

w · ȳ = (-1)(-5/√(74)) + (-1)(7/√(74))

w · ȳ  = 12/√(74)

Finally, we can find the projection of w onto v,

proᵥᵥw = (w · ȳ)ȳ = (12/√(74))((-5/√(74))i + (7/√(74))j)

(w · ȳ)ȳ = (-60/74)i + (84/74)j

(w · ȳ)ȳ = (-30/37)i + (42/37)j

Therefore, the projection of w onto v is (-30/37)i + (42/37)j.

b. v₁ = ((v · w)/|w|²)w

v₂ = v - v₁ where |w| is the magnitude of w. First, let's find |w|,

|w| = √((-1)² + (-1)²)

|w| = √(2)

Next, let's find v · w,

v · w = (-5)(-1) + (7)(-1)

v · w = -12

Using the formula, we can find v₁,

v₁ = ((v · w)/|w|²)w = (-12/2)(-1 - j)

v₁ = 6 + 6j

Now, finding v₂,

v₂ = v - v₁

v₂ = (-5i + 7j) - (6 + 6j)

v₂ = -5i + (7 - 6)j - 6

v₂ = -5i + j - 6

As a result, the vector v can be divided into two parts: v₁ = 6 + 6j and v₂ = -5i + j - 6, where v₁ is parallel to w and v₂ is orthogonal to w.

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Complete question - Given v = - 5i + 7j and w = - 1 - j.

a. Find proᵥᵥv

b. Decompose v into two vectors v₁ and v₂ where v₁ is parallel to w and v₂ is orthogonal to w.

The class mark of the modal class is: 25

From the given histogram, the following Frequency Distribution is obtained:

Class Interval Mid point Frequency

625-675 650 3

676-726 701 5

727 - 777 752 7

778 - 828 803 8

829 - 879 854 6

880 - 930 905 2

931 - 981 956 0

982 - 1032 1007 1

b. To find the lower class limit of the first class:

First class: 625

c. The upper limit of the first class is:

First class: 676

The class mark of the modal class is:

The modal class is: 803

The class mark is upper limit + lower limit/2

Thus, 828-778/2

=> 50/2

The class mark of the modal class is: 25

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