find the equation of the straight line through the point (2,3) whose intercept on x-axis is twice that on y-axis​

Answer:

The amount of frozen fruit used to make smoothies is equal to the total number of frozen fruits minus the remainder of fruits after making smoothies.

Step-by-step explanation:

Let

Total number of frozen fruits = F

Let the amount of frozen fruits used for making smoothies = S

The remaining frozen fruits = R

Then

F - S = R

To get the amount of frozen fruits used to make smoothies, we make S the subject of the formula.

-S = R - F

S = F - R

That is, the amount of frozen fruit used to make smoothies is equal to the total number of frozen fruits minus the remainder of fruits after making smoothies.

Based on the given original sample of 17, 10, 15, 21, 13, 18, none of the provided values constitute a possible bootstrap sample from the original sample.

To determine if a sample is a possible bootstrap sample, we need to check if the values in the sample are present in the original sample and in the same frequency. Let's evaluate each provided sample:
10, 12, 17, 18, 20, 21: This sample includes values (10, 17, 18, 21) that are present in the original sample, but the frequencies do not match. Thus, it is not a possible bootstrap sample.

10, 15, 17: This sample includes values (10, 17) that are present in the original sample, but it is missing the values (15, 21, 13, 18). Thus, it is not a possible bootstrap sample.

10, 13, 15, 17, 18, 21: This sample includes all the values from the original sample, and the frequencies match. Thus, it is a possible bootstrap sample.

18, 13, 21, 17, 15, 13, 10: This sample includes all the values from the original sample, but the frequencies do not match. Thus, it is not a possible bootstrap sample.

13, 10, 21, 10, 18, 17: This sample includes values (10, 17, 18, 21) that are present in the original sample, but the frequencies do not match. Thus, it is not a possible bootstrap sample.

In conclusion, only the sample 10, 13, 15, 17, 18, 21 constitutes a possible bootstrap sample from the original sample.

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Answer:

20%

Step-by-step explanation:

% increase = (final − initial) / initial × 100%

= (30 − 25) / 25 × 100%

= 20%

Answer:

The percent gain is 25%

Step-by-step explanation:

Answer:

N - 3

Step-by-step explanation:

It is a series of -3 so:

7 -3 = 4

4 - 3 = 1

1 - 3 = -2

So:

N-3

where N is the present value in the sequence.

Answer:what

Step-by-step explanation:what does this mean

The answer is 18x10^4 using simple arithmetic calculations and exponential function.

What are arithmetic calculations?
Arithmetic calculations are mathematical operations that involve the addition, subtraction, multiplication, and division of real numbers. Arithmetic is used to solve problems and execute tasks in everyday life, and is an integral part of mathematics. It is also used to analyze data and solve equations. Arithmetic involves the use of numbers, symbols, and operations in order to solve problems. Examples of arithmetic operations include addition, subtraction, multiplication, division, and exponentiation. The order of operations is an important part of arithmetic, as it dictates which operations take precedence over others. Arithmetic is used in a variety of applications, such as in engineering, finance, education, and science. It can also be used to solve puzzles and games. Arithmetic calculations can be used to help people better understand the world around them, as well as to help them make better decisions in life.

Two exponential function having same base , then powers can be added or subtracted.

In given equation, base is of 10 and powers are 8 and -4.

now the power will be 4

and 3 and 6 get multiplied i.e. 18.

So, the answer is 18x10^4.

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Answer: b = 1.5 or 3/2

Step-by-step explanation:

2 - b/3 = -5/2

Subtract 2 from both sides

-b/3 = -9/2

divide by -3

b = 1.5 or 3/2

(a)The optimal order quantity is  4609 units.

(b)The time between orders is  1.98 months.

To determine the optimal order quantity and the approximate time between orders, the Economic Order Quantity (EOQ) model. The EOQ model minimizes the total cost of inventory by balancing ordering costs and carrying costs.

Optimal Order Quantity:

The formula for the EOQ is given by:

EOQ = √[(2DS) / H]

Where:

D = Annual demand

S = Cost per order

H = Holding cost per unit per year

calculate the annual demand (D) using the 2020

sales numbers provided:

D = 651 + 808 + 514 + 174 + 435 + 426 + 687 + 582 + 662 + 1551 + 1295 + 1774

= 9559 units

To calculate the cost per order (S) and the holding cost per unit per year (H).

The cost per order (S) is given as $500.

The holding cost per unit per year (H)  calculated as follows:

H = Carrying cost percentage × Unit value

= 0.30 × $3

= $0.90

substitute these values into the EOQ formula:

EOQ = √[(2 × 9559 × $500) / $0.90]

= √[19118000 / $0.90]

≈ √21242222.22

≈ 4608.71

Approximate Time Between Orders:

To calculate the approximate time between orders, we'll divide the total number of working days in a year by the number of orders per year.

Assuming 360 days in a year and a lead time of 1 week (7 days) for shipping, we have:

Working days in a year = 360 - 7 = 353 days

Approximate time between orders = Working days in a year / Number of orders per year

= 353 / (9559 / 4609)

= 0.165 years

Converting this time to months:

Approximate time between orders (months) = 0.165 × 12

= 1.98 months

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Step-by-step explanation:

It looks as though ( from your post)  Sweden won 4 golds and 0 silver and 2 bronze medals

4:0:2     simplifies to   2 :0 : 1

The probability of having a 5-card hand that is a flush or royal flush is 0.00196

Probability: Probability is the branch of mathematics concerning numerical descriptions of how likely an event is to occur, or how likely it is that a proposition is true. The probability of an event is a number between 0 and 1, where, roughly speaking, 0 indicates impossibility of the event and 1 indicates certainty.

Flush, straight flush, and royal flush probabilities are calculated as follows: 5148/2598960≅0.00198

A flush's likelihood (while eliminating straight and royal flushes) is 5108/2598960, ≅0.00196.

Finding the fraction with the number of ways to have a flush as the numerator and the number of possible five-card hands as the denominator will allow us to determine the likelihood.

Combinations will be used to find each of these numbers (we don't care about the draw order; only about what shows up in our hand). Combinations' general formula is Cn,k=n!/(k)!(n-k)! with k=picks and n=population

Let's first determine the denominator by selecting 5 cards at random from a deck of 52 cards: C52,5=52!/(5)!(525)!

=52!/(5!)(47!)

Let's assess it!

52×51×50^10×49×48^2×47!/5×4×3×2×47!=52×51×10×49×2=2,598,960

Let's now determine the numerator.

In order to examine each hand with five cards of the same suit, we will compute all hands that feature a flush (including straight flushes, royal flushes, and flushes) (with a suit having 13 cards in total). We can say that we understand this by:

C13,5

Remember that there are 4 suites in which this might occur, but we only want 1, so multiply by C4,1. Putting it all together, we obtain:

C4,1×C13,5=4!/(1!)(4−1)!×13!/(5!)(13−5)!=4!13!/3!5!8!

Let's assess this.

4!×13×12×11×10×9^3×8!/3×2×5×4!×8!=13×12×11×3=5148

(Remember that we just calculated all hands, including straight flushes and royal flushes, that have a flush component to them!

The probability of getting a hand with a flush is:

5148/2598960≅.00198

We may exclude straight and royal flush possibilities from the 5148 flush hands by excluding those hands (which are hands with 5 consecutive value cards in the same suit, such as 3, 4, 5, 6, and 7 of hearts). Since there are four suits and 10 potential ways to get a straight (A-5, 2-6, 3-7,..., 10-A), we can subtract 4 from 5148 to get 5108 hands, which gives us the result 5108/2598960=0.00196.

The probability of having a 5-card hand that is a flush or royal flush is 0.00196

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Answer:

-5x - 32

Step-by-step explanation:

Use distributive rule :   a(b + c) = ab + ac

3x - 8(x + 4) =3x + x*(-8) + 4*(-8)

                   = 3x - 8x - 32         {combine like terms}

                    = -5x - 32

a)     |\

       |  \

  Y  |    \

      |      \

     |         \

     | ____ \

         X

b) A = 1/102

c) Marginal PDF fx(x) = (1/102) * x

Marginal PDF fy(y) = (51/102)

No, X and Y are not independent since their marginal PDFs fx(x) and fy(y) are not separable (i.e., they cannot be expressed as the product of individual PDFs)

d) P(0 < X < 2, 1) = 1.

a) To sketch the sample space, we need to consider the ranges of X and Y as defined in the problem statement: 0 < x < 1 and 0 < y < 2x. This means that X ranges from 0 to 1 and Y ranges from 0 to 2X. The sample space can be represented by a triangular region bounded by the lines Y = 0, X = 1, and Y = 2X.

       |\

       |  \

  Y  |    \

      |      \

     |         \

     | ____ \

         X

b) To find the value of A so that fx,y(x, y) is a valid joint PDF, we need to ensure that the joint PDF integrates to 1 over the entire sample space.

The joint PDF is given by fx,y(x, y) = A(48 + 3), where 0 < x < 1 and 0 < y < 2x.

To find A, we integrate the joint PDF over the sample space:

∫∫fx,y(x, y) dy dx = 1

∫∫A(48 + 3) dy dx = 1

A∫∫(48 + 3) dy dx = 1

A(48y + 3y)∣∣∣0∣∣2xdx = 1

A(96x + 6x)∣∣∣0∣∣1 = 1

A(96 + 6) = 1

102A = 1

A = 1/102

Therefore, A = 1/102.

c) To find the marginal PDFs fx(x) and fy(y), we integrate the joint PDF over the respective variables.

Marginal PDF fx(x):

fx(x) = ∫fy(x, y) dy

Since 0 < y < 2x, the integral limits for y are 0 to 2x.

fx(x) = ∫A(48 + 3) dy from 0 to 2x

fx(x) = A(48y + 3y)∣∣∣0∣∣2x

fx(x) = A(96x + 6x)

fx(x) = 102A * x

fx(x) = (1/102) * x

Marginal PDF fy(y):

fy(y) = ∫fx(x, y) dx

Since 0 < x < 1, the integral limits for x are 0 to 1.

fy(y) = ∫A(48 + 3) dx from 0 to 1

fy(y) = A(48x + 3x)∣∣∣0∣∣1

fy(y) = A(48 + 3)

fy(y) = A(51)

fy(y) = (51/102)

No, X and Y are not independent since their marginal PDFs fx(x) and fy(y) are not separable (i.e., they cannot be expressed as the product of individual PDFs).

d) To find P(0 < X < 2, 1), we need to integrate the joint PDF over the given range.

P(0 < X < 2, 1) = ∫∫fx,y(x, y) dy dx over the region 0 < x < 2 and 0 < y < 1

P(0 < X < 2, 1) = ∫∫A(48 + 3) dy dx over the region 0 < x < 2 and 0 < y < 1

P(0 < X < 2, 1) = A(48y + 3y)∣∣∣0∣∣1 dx over the region 0 < x < 2

P(0 < X < 2, 1) = A(48 + 3) dx over the region 0 < x < 2

P(0 < X < 2, 1) = A(51x)∣∣∣0∣∣2

P(0 < X < 2, 1) = A(102)

P(0 < X < 2, 1) = (1/102)(102)

P(0 < X < 2, 1) = 1

Therefore, P(0 < X < 2, 1) = 1.

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A plane traveled 630 kilometers each way to Toronto and back. The trip there was with the wind. It took 7 hours. The trip back was into the wind. The trip back took 15 hours. The speed of the plane in still air is 66 km/h, and the speed of the wind is 24 km/h.

Let's use the following variables:

When the plane is flying with the wind, its effective speed is p + w. When it is flying against the wind, its effective speed is p - w.

We know that the distance to Toronto and back is 630 km each way, for a total of 1260 km. We also know that the time it took to fly to Toronto with the wind was 7 hours, and the time it took to fly back against the wind was 15 hours.

Using the formula distance = speed x time, we can set up the following equations:

630 = (p + w) x 7 (equation 1)

630 = (p - w) x 15 (equation 2)

We now have two equations with two unknowns. We can solve for either p or w in terms of the other variable. Let's solve for p in terms of w.

From equation 1, we get:

p + w = 90

From equation 2, we get:

p - w = 42

Adding the two equations, we get:

2p = 132

So,

p = 66 km/h

Now we can use either equation 1 or equation 2 to solve for w. Let's use equation 1:

630 = (66 + w) x 7

Simplifying this equation, we get:

90 = 66 + w

w = 24 km/h

Therefore, the speed of the plane in still air is 66 km/h, and the speed of the wind is 24 km/h.

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It is proved that Point A is situated equal distance from Triangle PQA and RQA.

It is a line that divide an angle or a line in two equal parts.  In the given figure, QA is a bisector as it divides the angle PQR.

Triangles are called congruent when they have equal corresponding sides and equal corresponding angles. As shown in figure, the bisector QA creates two congruent triangle PQA and RQA.

it is given, QA is bisector of angle Q, so point A lies equal distance from P and R. According to the figure, APQ and ARQ are right angle.

As per the rules of congruence theorem, triangle PQA and Triangle RQA are two congruent triangles.

hence, A is the same distance from P and R

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Statement                                                                  Reason

1. QA is the bisector of ∠PQR                                   Given

2. ∠PQA   ≅ ∠RQA                                          Definition of bisector

4. ∠QXA ≅ ∠QYA                                   All right angles are congruent

5. QA ≅ QA                                                                Given

6.△PQA ≅ △RQA                                            AAS Postulate

8. Point A is equidistance                              Definition equidistance

from the side of ∠PQR        

What is right triangle?

A right triangle, also known as a right-angled triangle, right-perpendicular triangle, orthogonal triangle, or formerly rectangle triangle, is a triangle with one right angle, or two perpendicular sides. The foundation of trigonometry is the relationship between the sides and various angles of the right triangle.

Given that point A is lie on the bisector of ∠PQR.

When anything is divided into two equal or congruent portions, usually by a line, it is said to have been bisected in geometry. The line is then referred to as the bisector. Segment bisectors and angle bisectors are the sorts of bisectors that are most frequently taken into consideration.

Therefore QA is the bisector of ∠PQR

Then ∠PQA   ≅ ∠RQA  

Since ∠QXA and ∠QYA are right angle, therefore they are congruent.

∠QXA ≅ ∠QYA

QA ≅ QA

AAS postulate: The triangles are congruent if two angles and the excluded side of one triangle match two angles and the excluded side of another triangle.

According to AAS postulate,

△PQA ≅ △RQA  

Point A is equidistance from ∠PQR.

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The equivalent logical expressions are (I) and (III) only. The correct answer is D.

Let's analyze each expression:

1. (x == y && x != z) || (x != y && x == z):

This expression evaluates to true if either (x == y) and (x != z) are both true, or (x != y) and (x == z) are both true. It implies that either x is equal to y and not equal to z, or x is not equal to y and equal to z. This can also be written as (x == y) XOR (x == z), where XOR represents the exclusive OR operation. However, this expression is not equivalent to the other options.

2. (x == y || x == z) && (x != y || x != z):

This expression evaluates to true if both (x == y || x == z) and (x != y || x != z) are true. It implies that either x is equal to y or x is equal to z, and at the same time, either x is not equal to y or x is not equal to z. However, this expression is not equivalent to the others.

3. (x == y) != (x == z):

This expression evaluates to true if (x == y) and (x == z) have different truth values. It implies that x is equal to y and not equal to z, or x is not equal to y and equal to z. This expression is equivalent to expression (I).

Therefore, the correct answer is D. I and III only.

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The number of people are 72.

Highest Common Factor is the full name for HCF in mathematics.

According to the laws of mathematics, the highest positive integer that divides two or more positive integers without leaving a residual is known as the greatest common divisor, or gcd.

Least Common Multiple is the full name for LCM in mathematics.

LCM (a,b) in mathematics stands for the least common multiple, or LCM, of two numbers, such as a and b. The smallest or least positive integer that is divisible by both a and b is known as the LCM.

We need to know the size of the pieces in order to calculate how many people can be fed equal-sized pieces of cake from a 16 by 18-inch cake.

Assume that each piece is a square with sides measuring "x" in length. Each piece's area would be x², and the cake's overall area would be 16 x 18 inches, or 288 square inches.

We must divide the entire area of the cake by the area of each piece to determine the maximum number of pieces that can be cut from the cake:

288 / x² = (16*18) / x² = 288 / x²

Choosing a "x" value that evenly splits into 16 and 18 is necessary if we want every piece to be the exact same size. Since 2 is the greatest possible value, we can divide the cake into 8 rows of 9 squares, yielding a total of 72 pieces.

As a result, a cake measuring 16 inches by 18 inches can be divided into 72 equal-sized pieces, each measuring 2 inches by 2 inches.

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We can serve 18 people equal-sized pieces of cake from a cake that is 16 inches by 18 inches, assuming each piece is a square and we want the pieces to be exactly the same size.

A two-dimensional figure, form, or planar lamina's area is a measurement of how much space it takes up in the plane.

To find out how many people can be served equal-sized pieces of cake from a cake that is 16 inches by 18 inches, we need to first determine the size of each piece of cake.

The total area of the cake is:

16 inches x 18 inches = 288 square inches

To divide the cake into equal-sized pieces, we need to determine the size of each piece. Let's assume we want each piece to be a square. To find the size of each square, we need to find the square root of the total area of the cake:

√(288 square inches) ≈ 16.97 inches

Since we want the pieces to be exactly the same size, we'll round down to the nearest inch, which gives us:

Each piece of cake will be approximately 16 inches by 16 inches.

To find out how many people can be served equal-sized pieces of cake, we need to divide the total area of the cake by the area of each piece:

288 square inches ÷ (16 inches x 16 inches) = 18

Therefore, we can serve 18 people equal-sized pieces of cake from a cake that is 16 inches by 18 inches, assuming each piece is a square and we want the pieces to be exactly the same size.

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The values of the sets are:

I. U − B = {0, 2, 4, 6, 8, 10}

II. B ∩ (B c − A) = {}

III. (A ∪ B) − (B − A) = {0, 2, 4, 6, 8, 10}

IV. (A ∪ Ac) = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

V. ((A – B)c = {1, 3, 5, 7, 9}

VI. (A ∪ B c) ∩ B = {}

VII. (A ∩ B) ∪ B c = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

VIII. Ac ∩ B c = {}

IX. B − Ac = {}

X. (Ac − Bc)c = {0, 2, 4, 6, 8, 10}

I. U − B:

The set U − B represents the elements in the universal set U that are not in the set B.

In this case, B consists of odd numbers in the range of U. Therefore, U − B would include all the even numbers in the universal set U.

U − B = {0, 2, 4, 6, 8, 10}

II. B ∩ (B c − A):

B c = {0, 2, 4, 6, 8, 10}

A = {0, 2, 4, 6, 8, 10}

(B c − A) = {}

B ∩ (B c − A) = {}

III. (A ∪ B) − (B − A):

(A ∪ B) represents the union of sets A and B, and (B − A) represents the elements in set B that are not in A.

So, (A ∪ B) = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(B − A) = {1, 3, 5, 7, 9}

(A ∪ B) − (B − A) = {0, 2, 4, 6, 8, 10}

IV. (A ∪ Ac):

A = {0, 2, 4, 6, 8, 10}

Ac = {1, 3, 5, 7, 9}

So, (A ∪ Ac) = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

V. (A – B)c:

(A – B) = {0, 2, 4, 6, 8, 10}

So, (A – B)c = {1, 3, 5, 7, 9}

VI. (A ∪ B c) ∩ B:

B c = {0, 2, 4, 6, 8, 10}

(A ∪ B c) = {0, 2, 4, 6, 8, 10}

So,  (A ∪ B c) ∩ B = {}

VII. (A ∩ B) ∪ B c

(A ∩ B) = {}

So, (A ∩ B) ∪ B c = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

VIII. Ac ∩ B c:

Ac = {1, 3, 5, 7, 9}

B c = {0, 2, 4, 6, 8, 10}

So, Ac ∩ B c = {}

IX. B − Ac:

B − Ac represents the elements in set B that are not in set Ac.

B = {1, 3, 5, 7, 9}

Ac = {1, 3, 5, 7, 9}

So, B − Ac = {}

X. (Ac − Bc)c:

Ac = {1, 3, 5, 7, 9}

Bc = {0, 2, 4, 6, 8, 10}

(Ac − Bc) = {1, 3, 5, 7, 9}

So, (Ac − Bc)c = {0, 2, 4, 6, 8, 10}

(b) Proving or disproving the statements:

I. A = B:

The statement is not true.

Set A consists of even numbers obtained by the equation x = 5a − 12, while set B consists of odd numbers obtained by the equation y = 5b + 8.

II. B ⊆ C:

The statement is not true.

Set B consists of odd numbers obtained by the equation y = 5b + 8, while set C consists of numbers obtained by the equation z = 10c + 2.

Since there are no values that satisfy the equation y = 5b + 8 and z = 10c + 2 simultaneously, B is not a subset of C.

III. C ⊆ A:

The statement is not true. Set C consists of numbers obtained by the equation z = 10c + 2, while set A consists of even numbers obtained by the equation x = 5a − 12.

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The partitioning of the array A = [3, 6, 2, 8, 7, 9, 5, 1, 4] with Partition(A, 1, 9) manually results in [3, 2, 1, 4, 7, 9, 5, 8, 6].

We are given an array A containing 9 elements. We need to perform the following tasks:

5a. Compute the Partition function on A, where the function takes in the array A and two indices (1 and 9 in this case) as arguments. Partition function is a part of the Quick Sort algorithm that partitions the array into two parts based on a pivot element.

5b. We need to consider two cases where the last element of the array A, A[9], is 14 and 0 respectively, and see how it affects our computation in 5a.

5c. Finally, we need to sort array A using the Bucket Sort algorithm with min-max scaling. The bucket Sort algorithm works by dividing the range of values into a series of buckets and then distributing the elements into those buckets. Min-max scaling is a technique used to scale the values of an array between 0 and 1.

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The area of the quadrilateral formed by points P, Q, R, and S is approximately 10.55 units.

Given four points P(2, 4, 4), Q(3, 1, 6), R(2, 8, 0), and S(8, −1, 3)We have to check if these four points lie in the same plane or not. If they lie in the same plane, then it's a quadrilateral shape. If not, then it's not a quadrilateral shape.

Let's first form vectors using three of these points to determine whether they are collinear or not.

We have used P, Q, and R points to find the vector

\(n→.PQ = Q - P = (3-2) i + (1-4) j + (6-4) k = i - 3j + 2kPR = R - P = (2-2) i + (8-4) j + (0-4) k = 4j - 4k\)

Let's find the cross product of these two vectors,

\(n→= PQ × PR = i j k \(\begin{vmatrix} i & j & k \\ 1 & -3 & 2 \\ 0 & 4 & -4 \end{vmatrix}\) = i(-8) - j(2) + k(4) = -8i - 2j + 4k\)

Now, Let's plug in the coordinates of point S to the equation of the plane,\(-8i - 2j + 4k . (8, -1, 3) + D = 0= > (-8)(8) - (2)(-1) + (4)(3) + D = 0= > D = 150\)

Now, the equation of the plane is -8x - 2y + 4z + 150 = 0.

Now that we know that all points lie on the same plane, we can use vectors to find the area of the quadrilateral defined by them.We can find the area of the parallelogram formed by the vectors PQ and PR using cross product.

Let's call it \(/n1→.n1→= PQ × PR= \(\begin{vmatrix} i & j & k \\ 1 & -3 & 2 \\ 0 & 4 & -4 \end{vmatrix}\)= -8i - 2j + 4k\)

Now, we can find the magnitude of n1→ by using formula: \(|n1→| = √((-8)² + (-2)² + 4²) = √(84)\)

Now, let's calculate another cross-product for vectors PQ and PS to find the area of the parallelogram formed by them.

We can call this

\(n2→.n2→ = PQ × PS= \(\begin{vmatrix} i & j & k \\ 1 & -3 & 2 \\ 5 & -5 & -1 \end{vmatrix}\)= 17i + 7j + 2k\)

Now, we can find the magnitude of n2→ by using formula: \(|n2→| = √(17² + 7² + 2²) = √(378)\)

Finally, let's use the area of the parallelograms formed by these two vectors to calculate the area of the quadrilateral defined by the four points, \(Area = 1/2 × |n1→| + |n2→|Area = 1/2 × √84 + √378Area = 0.5 × 2√21 + 3√14Area = √21 + (3/2)√14Area = 10.55 units\) (approx)

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Step-by-step explanation:

f(x) = x³ = 2³ = is 8

hope it's helpful

The volume of the box with surface area 10 is given by the formula V = 2.5x^2 - 0.25x^4, where x is the length of a side of the square base.

To find a formula for the volume of the box with surface area A and square base with side x, we first need to find the height of the box. Since the box has a square base, the area of the base is x^2. The remaining surface area is the sum of the areas of the four sides, each of which is a rectangle with base x and height h. Therefore, the surface area A is given by:

A = x^2 + 4xh

Solving for h, we get:

h = (A - x^2) / 4x

The volume V of the box is given by:

V = x^2 * h

Substituting the expression for h, we get:

V = x^2 * (A - x^2) / 4x

Simplifying, we get:

V = (Ax^2 - x^4) / 4

The domain of V is all non-negative real numbers, since both x^2 and A are non-negative.

To graph V as a function of x, we can use a graphing calculator or plot points using a table of values. The graph will be a parabola opening downwards, with x-intercepts at 0 and sqrt(A) and a maximum at x = sqrt(A) / sqrt(2).

To find the maximum value of V, we can take the derivative of V with respect to x and set it equal to 0:

dV/dx = (2Ax - 4x^3) / 4

Setting this equal to 0 and solving for x, we get:

x = sqrt(A) / sqrt(2)

Plugging this value of x into the formula for V, we get:

V = A^1.5 / (4sqrt(2))

Therefore, the maximum value of V is A^1.5 / (4sqrt(2)).

To find the formula for the volume of a box having surface area 10, we simply replace A with 10 in the formula we derived earlier:

V = (10x^2 - x^4) / 4

Simplifying, we get:

V = 2.5x^2 - 0.25x^4

Therefore, the volume of the box with surface area 10 is given by the formula V = 2.5x^2 - 0.25x^4, where x is the length of a side of the square base. The domain of V is all non-negative real numbers.

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The probability of the arrow landing on an odd number is the number of odd numbers divided by the total number of possible outcomes. Therefore, the probability of the arrow landing on an odd number is  0.5 or 50%.


To find the probability that the arrow will point at an odd number on a circular board with 8 equal parts, we'll first determine the total number of odd numbers present and then divide that by the total number of possible outcomes.

Step 1: Identify the odd numbers on the board. They are 1, 3, 5, and 7. The game consists of spinning the arrow on a circular board with 8 equal parts, which means there are 8 possible outcomes or numbers. Since we want to know the probability of landing on an odd number, we need to count how many odd numbers are on the board. In this case, there are four odd numbers: 1, 3, 5, and 7.

Step 2: Count the total number of odd numbers. There are 4 odd numbers.

Step 3: Count the total number of possible outcomes. Since the board is divided into 8 equal parts, there are 8 possible outcomes.

Step 4: Calculate the probability. The probability of the arrow pointing at an odd number is the number of odd numbers divided by the total number of possible outcomes.

Probability = (Number of odd numbers) / (Total number of possible outcomes)
Probability of landing on an odd number = Number of odd numbers / Total number of possible outcomes
Probability of landing on an odd number = 4 / 8

Step 5: Simplify the fraction. The probability of the arrow pointing at an odd number is 1/2 or 50%.

So, the probability that the arrow will point at an odd number is 1/2 or 50%.

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Given:

The equation is:

\((2x+3y)(2x-3y)=4x^2+6y^2\)

To find:

The error in the given equation and correct it.

Solution:

We have,

\((2x+3y)(2x-3y)=4x^2+6y^2\)

Taking left-hand side, we get

\(L.H.S.=(2x+3y)(2x-3y)\)

\(L.H.S.=(2x)^2-(3y)^2\)                 \([\because a^2-b^2=(a-b)(a+b)]\)

\(L.H.S.=(2)^2(x)^2-(3)^2(y)^2\)       \([\because (ab)^x=a^xb^x]\)

\(L.H.S.=4x^2-9y^2\)

It is not equal to right-hand side \(4x^2+6y^2\). In the right hand side, there must be a negative sign instead of positive sign.

Therefore, \((2x+3y)(2x-3y)=4x^2-6y^2\).

Answer:

The answer is 5.

Step-by-step explanation:

That dot is basically a multiplication sign.

Answer:

the dot is a sign for multiplication. Sometimes people use "•" because "x" can sometimes confuses people because "x" is a variable in algebra as well

Step-by-step explanation:

Answer:

3/4 ÷ 1/5 = 3.75

Step-by-step explanation:

3/4=_/20, to get that multiply the bottom number by each other, in doing so,you will get 15/20, now lets make 1/5 equal to 20, so we get 4/20 therefore saying 4/20+4/20+4/20=12/20, so you can make 3 1/5 cups and you will have 3/20.

I hope this helps and have a good day!

A small page can fit 4 photographs, and a large one can fit 2 photographs.

From the question, we have the following information available is:

Harper plans to assemble 4 small pages and 1 large page and brought a total of 12 photographs to use.

Savannah brought 28 photographs, which is enough to assemble 4 small pages and 5 large pages.

Let the number of photographs on small pages be x and number of large pages be y

=> 4x + y = 12 __(eq.1)

=> 4x + 5y = 28__(eq.2)

Subtracting equation 2 from equation 1 we get:

4x + y - (4x + 5y) = 12 - 28

4x + y - 4x - 5y = -16

-4y = -16

y = 4

Put the value of y in eq.1

4x + y = 12

4x + 4 = 12

4x = 8

x = 2

Therefore, a small page can fit 4 photographs, and a large one can fit 2 photographs.

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Answer:

73°

Step-by-step explanation:

To find ∠QPS, just add ∠1 and∠2

∠QPS = ∠1 + ∠2

          = 44° + 29°

          = 73°

Answer:

∠QPS = 73°

Step-by-step explanation:

∠1 = 44°

∠2 = 29°

∠QPS = ∠1 + ∠2 = 44° + 29° = 73°

The volume of the sample of the gas  at the final temperature ( assuming constant pressure)  is approximately 54.6 mL.

To find the volume at the final temperature, we can use the formula for Charles's Law, which states that:

V1/T1 = V2/T2,

( where V1 is the initial volume, T1 is the initial temperature, V2 is the final volume, and T2 is the final temperature.) Assume constant pressure throughout the process. Given a 45.1-mL sample of gas in a cylinder warmed from 27°C to 91°C, follow these steps:

1. Convert the temperatures to Kelvin: T1 = 27°C + 273.15 = 300.15 K, T2 = 91°C + 273.15 = 364.15 K.
2. Substitute the known values into Charles's Law formula: (45.1 mL / 300.15 K) = (V2 / 364.15 K).
3. Solve for V2: V2 = (45.1 mL / 300.15 K) * 364.15 K.
4. Calculate V2: V2 ≈ 54.6 mL.

The volume at the final temperature is approximately 54.6 mL.

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