Find the distance between the two points rounding to the nearest tenth (if necessary).
(5,7) and (-4,3)

A base must be raised to a certain exponent or power, or logarithm, in order to produce a given number. If bx = n, then x is the logarithm of n to the base b, which is expressed mathematically as x = logb n. For instance, 23 = 8; consequently, 3 is the base-2 logarithm of 8 or 3 = log2 8.

When measuring the relative change of a value rather than its absolute difference, logarithmic scales are helpful. Moreover, logarithmic scales are used to compress large-scale scientific data because the logarithmic function log(x) grows very slowly for large x.

\(log_{2} 2=1\\log_{4}4=1\\ log_{100}100=1\\log10=1\)

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The area of this rectangular figure is equal to 71.3 square centimeters.

In Mathematics and Geometry, the area of a rectangle can be calculated by using the following mathematical equation (formula):

A = LB

Where:

By substituting the given dimensions into the formula for the area of a rectangle, we have the following;

Area of rectangle = 6.2 cm × 11.5 cm

Area of rectangle = 71.3 square centimeters.

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Complete Question:

Find the area of the rectangular figure shown above.

The linear program has an optimal solution of 28.5 at (A, B) = (2, 5) when the right-hand side for constraint 1 is 10. If the right-hand side is increased to 11, the optimal solution remains 28.5 at the same point.

The dual value for constraint 1 stays the same as long as the right-hand side stays within the range 8.4 to 11. The value of the optimal solution will increase by 0.5 for every unit increase in the right-hand side of constraint 2 as long as the right-hand side is within the range 23 to 27.

a) To find the new optimal solution, we need to graph the feasible region and find the new corner points. After the change, the first constraint becomes 1A + 1B ≤ 11. The other two constraints remain the same. The corner points can be found by substituting the bounds of each variable into the constraints and solving for the other variable. The feasible region can then be graphed, and the maximum of the objective function can be found at one of the corner points.

The new corner points are (0,0), (0,9), (6,0), and (2,5). The feasible region can be graphed as a triangle with these corner points. The maximum of the objective function 3A + 2B occurs at the point (2,5), where the value of the objective function is 28.5.

b) The value of the objective function at the optimal solution is 28.5, at (A, B) = (2, 5).

c) The right-hand side range information for constraint 1 tells us that if the right-hand side is increased from 10 to 11, the dual value will remain the same (since the allowable increase is 2.6 and the right-hand side only increases by 1). The range for constraint 1 is 8.4 to 11. As long as the right-hand side stays within this range, the dual value of 2.6 is applicable.

d) The dual value for constraint 2 is 0.5. Using this dual value and the right-hand side range information, we can conclude that the value of the optimal solution will increase by 0.5 for every unit increase in the right-hand side of constraint 2 as long as the right-hand side is within the range 23 to 27.

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Complete question:

Consider the following linear program.

Max 3A + 2B s.t.

1A + 1B ≤ 10

3A + 1B ≤ 27

1A + 2B ≤ 18

A, B ≥ 0

The value of the optimal solution is 28.5. Suppose that the right-hand side for constraint 1 is increased from 10 to 11.

(a) Use the graphical solution procedure to find the new optimal solution.

What is the value of the objective function at the optimal solution?

(blank) at (A, B) =

Use the solution to part (a) to determine the dual value for constraint 1.

(c) The computer solution for the linear program provides the following right-hand side range information

. Constraint RHS Value Allowable Increase Allowable Decrease

1 10.00000 2.60000 1.00000

2 27.00000 3.00000 13.00000

3 18.00000 Infinite 6.50000

What does the right-hand side range information for constraint 1 tell you about the dual value for constraint 1?

The range for constraint 1 is (blank) to (blank) . As long as the right-hand side stays (within/outside) this range, the dual value of (blank) is applicable.

(d) The dual value for constraint 2 is 0.5. Using this dual value and the right-hand side range information in part (c), what conclusion can be drawn about the effect of changes to the right-hand side of constraint 2?

The value of the optimal solution will increase by (blank) for every unit increase in the right-hand side of constraint 2 as long as the right-hand side is (within/outside) the range (blank) to (blank).

In problem 4.5, we need to determine if vector b is a linear combination of the columns of the given matrix A. In problem 4.6, we need to find the value(s) of h for which vector y is in the span of v₁ and v₂.

Problem 4.5:
To determine if vector b is a linear combination of the columns of matrix A, we need to check if there exist coefficients such that b can be expressed as a linear combination of the columns of A. We can set up the equation Ax = b, where A is the matrix with columns as the given vectors, and x is the vector of coefficients. Then we can solve this equation to check if a solution exists.
Problem 4.6:
To find the value(s) of h for which vector y is in the span of v₁ and v₂, we need to check if there exist coefficients such that y can be expressed as a linear combination of v₁ and v₂. We can set up the equation y = av₁ + bv₂, where a and b are the coefficients. Then we can solve this equation to find the values of h for which the equation holds true.
In both problems, the steps involve setting up appropriate equations and solving them to determine if the given conditions are satisfied. By following the process, we can find the answers to these problems and show all the intermediate steps.

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Answer:

Region B & C

Step-by-step explanation:

Usually, in inequality graphs, the solutions are usually located in the shaded region of the graph.

Also, we make the boundary line thick if the points on that line also contain solutions to the inequality.

In this question, the boundary line of function g(x) is thick and the shaded region is B and C.

Using the place value technique, the product of 4457×2 is 8914.

We need to use the place value technique to find the product of 4457*2.

When multiplying two integers, the place value of each digit is examined, and each multiplication is performed. Individual multiplication results are later put together to provide the final result.

The number 4457 can be written as below :

4457 = 4*1000 + 4*100 + 5*10 + 7*1

We will multiply the digit and the place value one by one with the number.

7*2 = 14

50*2 = 100

400*2 = 800

4000*2 = 8000

Now add all the results together.

M = 8000 + 800 + 100 + 14

M = 8914

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The answer to this question is 5 days

Answer:

5

Step-by-step explanation:

7+4=11

11+4=15

15+4=19

19+4=23

23+4=27

Based on the information provided, we have a random sample of 60 measurements taken from a population with a mean (µ) of 25 and a variance (σ²) of 200.

The mean of the sample (X) will be equal to the population mean, so X = 25.

To calculate the standard error (SE) of the sample, we use the formula SE = σ / √n, where σ is the population standard deviation and n is the sample size. Since the variance is given as 200, the standard deviation (σ) is the square root of 200, which is approximately 14.14.

Now, we can calculate the standard error: SE = 14.14 / √60 ≈ 1.82.

So, the pair for the mean and standard error of x is (25, 1.82).

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Taking into acount the square of a sum and common factor, you obtain that

(x²+y²)² + 2x²y² - y⁴= x²(x² + 4y²)

The square of a sum is the sum of the squares plus twice the product. That is, the result of the square of a sum of two terms is equal to the square of the first plus twice the product of the first and the second, plus the square of the second.

Calling the terms of the sum "a" and "b", mathematically that is expressed by:

(a+b)²= (a+b)×(a+b)= a×a + a×b + b×a + b×b= a² + 2a×b + b²

In this case, you can express (x²+y²)², being a=x² and b=y², as:

(x²+y²)²= (x²)² + 2x²y² + (y²)²

(x²+y²)²=  x⁴ + 2x²y² + y⁴

So, the expression (x²+y²)²  2x²y² - y⁴ can be written as:

(x²+y²)² + 2x²y² - y⁴=  x⁴ + 2x²y² + y⁴ + 2x²y² - y⁴

(x²+y²)² + 2x²y² - y⁴= x⁴ + 4x²y²

On the other side, the common factor is the number or variable found in all the terms of an expression.

If a common factor appears in all the terms of an expression, this expression can be written as the product of that factor by the expression that results from dividing each term by that factor.

It should be taken into account that when a factor with different exponents is repeated, then the one with the lowest exponent must be considered as a common factor.

In this case, the common factor is x². So, finally the expression (x²+y²)²  2x²y² - y⁴ can be written as:

(x²+y²)² + 2x²y² - y⁴=  x⁴ + 4x²y²= x²(x² + 4y²)

(x²+y²)² + 2x²y² - y⁴= x²(x² + 4y²)

Learn more about common factor and square of a sum: 

Answer:

6(4+x)

Step-by-step explanation:

Answer:

6(x+4)

Step-by-step explanation:

We want to write an algebraic expression for "six times the sum of four and a number"

Let's break this down.

The sum of four and a number:

The words "sum" indicates addition. We can call " a number" x, since it is unknown.

So, the sum of four and a number is: (x+4).

Six times:

The word "times indicates multiplication. We are multiplying six and the sum of four and a number.

So, we have: 6(x+4)

The algebra expression for "six times the sum of four and a number" is 6(x+4)

From the given data, the pH change upon adding 0.085 mol of HNO₃ to 1.00 L of the buffer solution is 0.14.

To determine the pH change upon adding HNO₃ to the buffer solution, we need to first calculate the initial pH of the buffer solution using the Henderson-Hasselbalch equation:

pH = pKa + log([base]/[acid])

where pKa is the acid dissociation constant of the weak acid in the buffer solution (in this case, KHSO₃), [base] is the concentration of the conjugate base (in this case, SO₃²⁻), and [acid] is the concentration of the weak acid (in this case, KHSO₃).

The pKa of KHSO₃ can be found in a table or calculated from its Ka value (which can be found in a similar manner). For simplicity, we will assume that the pKa of KHSO₃ is 1.92 (which is approximately the value reported in some tables).

Using the given concentrations of KHSO₃ and K₂SO₃, we can calculate the concentrations of HSO₃⁻ and SO₃²⁻ ions in the buffer solution:

[HSO₃⁻] = 0.314 M

[SO₃²⁻] = 0.283 M

Now we can use the Henderson-Hasselbalch equation to calculate the initial pH of the buffer solution:

pH = 1.92 + log(0.283/0.314)

pH = 1.92 - 0.048

pH = 1.87

Therefore, the initial pH of the buffer solution is 1.87.

When 0.085 mol of HNO₃ is added to the buffer solution, it reacts with the HSO₃⁻ ions to form H₂SO₃:

HNO₃ + HSO₃⁻ → H₂SO₃ + NO₃⁻

The amount of HSO₃⁻ ions consumed by the reaction can be calculated using the stoichiometry of the reaction:

0.085 mol HNO₃ × 1 mol HSO₃⁻/1 mol HNO₃ = 0.085 mol HSO₃⁻

The new concentration of HSO₃⁻ ions in the buffer solution is:

[HSO₃⁻] = 0.314 M - (0.085 mol/1.00 L) = 0.229 M

The new concentration of H₂SO₃ (formed from the reaction of HNO₃ with HSO₃⁻) can be calculated as:

[H₂SO₃] = (0.085 mol/1.00 L) = 0.085 M

The new concentration of SO₃²⁻ ions is still 0.283 M, as it is not consumed in the reaction.

Now we can use the Henderson-Hasselbalch equation again to calculate the new pH of the buffer solution:

pH = 1.92 + log(0.283/0.229)

pH = 1.92 + 0.090

pH = 2.01

Therefore, the pH of the buffer solution after adding 0.085 mol of HNO₃ is 2.01.

The pH change can be calculated as the difference between the new pH and the initial pH:

pH change = 2.01 - 1.87 = 0.14

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The number of cubes that would be needed to be added to turn the shape into a solid 3x4x4 cuboid would be = 24 cubes.

A cuboid is defined as the type of quadrilateral that has 6 faces, 8 vertices, and 12 edges.

The volume of the cuboid;

volume of the cuboid;= l×w×h

= 3x4x4

= 48cm³

The volume of each cube = 1cm×1cm×1cm

The number of cubes available= 24 cubes

Therefore the remaining cubes = 48-24 = 24 cubes.

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For lower cuboid

l=16m

b=9n

h=6m

For upper cuboid

l=9m

b=2m

h=6m

Note: volume of cuboid =l×b×h

total volume =lower cuboid +upper cuboid

=16*9*6+2*6*9=972m³

Answer:

Step-by-step explanation:

The volume is the sum of volumes of two rectangular prisms:

Answer:

2nd Base (I think)

Step-by-step explanation:

The right fielder is closer to the 2nd base.

As the probability of flooding any given year is 0.05, the probability of not flooding in the next year is 0.95 or 95%.

The probability of flooding in a region once every 20 years means that the probability of flooding any given year is 1/20 or 0.05.

Given that the probability of flooding in any one year is greater than zero, we know that the probability of not flooding is less than 1.

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Answer:

the answer to this question is 1/4

Answer:

jgcoyc9t9yf9ycrcuflc

Answer:

Part 1) The fractions in the list are not equal to the decimal number indicated

Part 2) The fractions 401/99 and 802/198 are equal to the decimal number indicated.

Part 1)

we have the decimal number

4.05 -----> is a terminating decimal ( t's a decimal with a finite number of digits)

Convert to fraction number

Multiply and divide by 100

Simplify

Divide by 5 both numerator and denominator

therefore

The fractions in the list are not equal to the decimal number indicated.

Part 2) we have

4.050505... ------> is a repeating decimal (is a decimal that has a digit, or a block of digits, that repeat without ever ending)

Let

x=4.050505...

we have that

Multiply by 2 both numerator and denominator

therefore

The fractions 401/99 and 802/198 are equal to the decimal number indicated.

Step-by-step explanation:

Answer:

if the cost on gas is 1.64 you can divide 23.78 by 1.64 which is 14.5

Step-by-step explanation:

23.78 divided by 1.64 =14.5

Answer:

14.45

Step-by-step explanation:

The forecast for May using a five-month moving average is 39 (Option E).

Moving average is used for smoothing out time series data to find any trends or cycles within the data. A five-month moving average is the average of the past five months. To calculate the moving average, add up the sales for the previous five months and divide it by five.

According to the question, the sales for the previous five months are: Nov. 39 Dec. 27 Jan. 40 Feb. 42 Mar. 41 April 47

We have to add the sales of these five months, which gives:

27 + 40 + 42 + 41 + 47 = 197

To find the moving average for May, we divide this sum by 5:

197 / 5 = 39.4

Since we have to round the answer to the nearest whole number, we round 39.4 to 39, which is option E.

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Answer:

5(3x+4)=23

5*3=15 5*4=20

15x+20=23

      -20=-20

15x=3

divide by 15 from both sides

x=0.2

Step-by-step explanation:

Answer:

5(3x+4)=23

5*3=15 5*4=20

15x+20=23

     -20=-20

15x=3

divide by 15 from both sides

x=0.2

Explanation: In a fraction, the number of equal parts being described is the numerator (from Latin: numerātor, "counter" or "numberer"), and the type or variety of the parts is the denominator (from Latin: dēnōminātor, "thing that names or designates"). As an example, the fraction 8/5

amounts to eight parts, each of which is of the type named "fifth". In terms of division, the numerator corresponds to the dividend, and the denominator corresponds to the divisor.

Informally, the numerator and denominator may be distinguished by placement alone, but in formal contexts they are usually separated by a fraction bar. The fraction bar may be horizontal (as in 1/3), oblique (as in 2/5), or diagonal (as in 4⁄9). These marks are respectively known as the horizontal bar; the virgule, slash (US), or stroke (UK); and the fraction bar, solidus, or fraction slash. In typography, fractions stacked vertically are also known as "en" or "nut fractions", and diagonal ones as "em" or "mutton fractions", based on whether a fraction with a single-digit numerator and denominator occupies the proportion of a narrow en square, or a wider em square. In traditional typefounding, a piece of type bearing a complete fraction (e.g. 1/2) was known as a "case fraction", while those representing only part of fraction were called "piece fractions".

On solving the given problem by help of mathematical operations, we got - After Ray poured 0.47L, he was left with 2.201L.

The term "operation" in mathematics refers to the process of calculating a value using operands and a math operator. For the given operands or numbers, the math operator's symbol has predetermined rules that must be followed.

In mathematics, there are five basic operations: addition, subtraction, multiplication, division, and modular forms.

Total amount of tea Ray has - 2.671L

Amount of tea Ray poured - 0.47

Therefore,

amount he was left with was  = 2.671- 0.47 = 2.201L

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Answer:

B

Step-by-step explanation:

Option C: 0.4 for p and 32 for n Using a normal curve, one may approximate the sampling distribution using the sample size (n) and sample proportion (n is 6).

If np > 10 or n(1 - p) > 10 is true, the normal curve can be applied in this situation.

For example, if n = 28 and p = 0.3;

np = 28 × 0.3 = 8.4 < 10

It is thus useless.

B) In case n = 28 and p = 0.9;

np = 28 × 0.9 = 25.2 > 10 Ok

Not acceptable: n(1 - p) = 28(1 - 0.9) = 2.8

Thus, it cannot be applied.

C) With n = 32 and p = 0.4

np = 32 × 0.4 = 12.8 > 10 Ok

n(1 - p) = 32(1 - 0.4) = 19.2 > 10 Ok

Thus, it can be utilized.

D) When n is 32 and p is 0.2

np = 32 × 0.2 = 6.4 < 10 Not Ok

So it is useless.

Option C: 0.4 for p and 32 for n Using a normal curve, one may approximate the sampling distribution using the sample size (n) and sample proportion (n is 6).

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Answer:

t=taco

b=burrito

4.5t+7.75b=519.25

b2=t

Step-by-step explanation:

a) U is subspace of R^3.

b) The set {(3, -2, 0), (0, 1/2, 1)} is a basis for U.

c) 2.

a) To show that U is a subspace of R^3, we need to verify the following three conditions:

i) The zero vector (0, 0, 0) is in U.

ii) U is closed under addition.

iii) U is closed under scalar multiplication.

i) The zero vector is in U since 0 + 2(0) - 3(0) = 0.

ii) Let (x1, y1, z1) and (x2, y2, z2) be two vectors in U. Then we have:

x1 + 2y1 - 3z1 = 0 (by definition of U)

x2 + 2y2 - 3z2 = 0 (by definition of U)

Adding these two equations, we get:

(x1 + x2) + 2(y1 + y2) - 3(z1 + z2) = 0

which shows that the sum (x1 + x2, y1 + y2, z1 + z2) is also in U. Therefore, U is closed under addition.

iii) Let (x, y, z) be a vector in U, and let c be a scalar. Then we have:

x + 2y - 3z = 0 (by definition of U)

Multiplying both sides of this equation by c, we get:

cx + 2cy - 3cz = 0

which shows that the vector (cx, cy, cz) is also in U. Therefore, U is closed under scalar multiplication.

Since U satisfies all three conditions, it is a subspace of R^3.

b) To find a basis for U, we can start by setting z = t (where t is an arbitrary parameter), and then solving for x and y in terms of t. From the equation x + 2y - 3z = 0, we have:

x = 3z - 2y

y = (x - 3z)/2

Substituting z = t into these equations, we get:

x = 3t - 2y

y = (x - 3t)/2

Now, we can express any vector in U as a linear combination of two vectors of the form (3, -2, 0) and (0, 1/2, 1), since:

(x, y, z) = x(3, -2, 0) + y(0, 1/2, 1) = (3x, -2x + (1/2)y, y + z)

Therefore, the set {(3, -2, 0), (0, 1/2, 1)} is a basis for U.

c) Since the basis for U has two elements, the dimension of U is 2.

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\(500/60=8\frac{1}{3\\}\) so either \(8\) or \(8\frac{1}{3}\)

The stock bottle of 500 tablets  fill a maximum of 8 prescription orders for 60 tablets each.

To determine how many prescription orders for 60 tablets filled with the stock bottle of 500 tablets,  to perform a division.

Number of prescription orders that  filled = Total number of tablets in stock bottle / Number of tablets per prescription order

Number of prescription orders = 500 tablets / 60 tablets per prescription order

Number of prescription orders = 8.333...

Since  have a fraction of a prescription order,  round down to the nearest whole number because out partial orders.

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Answer:

r = su/tv

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Equality Properties

Step-by-step explanation:

Step 1: Define

su = rtv

Step 2: Solve for r

The probability that the person selected will be one who plays baseball but NOT soccer is 12%

First, we have to find the percentage of people who play baseball but not soccer.We know that:
Total number of people = 300
Number of people who play soccer = 15% of 300 = 45
Number of people who play baseball = 21% of 300 = 63
Number of people who play both soccer and baseball = 9% of 300 = 27
Therefore,Number of people who play only baseball = Number of people who play baseball - Number of people who play both soccer and baseball= 63 - 27= 36
So, the percentage of people who play baseball but not soccer is:
Percent of people who play baseball but not soccer = Number of people who play only baseball/ Total number of people= 36/300= 0.12 = 12%

Therefore, the probability that the person selected will be one who plays baseball but NOT soccer is 12%

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