. Luong collects $1,300 from a customer in 2020 for services to be performed in 2021. select a type of adjusting entry 2. Luong incurs utility expense which is not yet paid in cash or recorded. select a type of adjusting entry 3. Luong’s employees worked 3 days in 2020 but will not be paid until 2021. select a type of adjusting entry 4. Luong performs services for customers but has not yet received cash or recorded the transaction. select a type of adjusting entry 5. Luong paid $2,800 rent on December 1 for the 4 months starting December 1.

There will be a significant difference between the two proportions. A further explanation is given below.

Thus the above is the appropriate solution.

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Answer:

true and I know its right

The values of x and y after solving this system of equations will be 32/35 and -88/105 respectively.

System of equations are a set of two or more equations that contain variables. The values of the variables can be calculated using either method of elimination or substitution. The values of the variables must then satisfy all the equations.

For a set of two equations, in elimination we make the coefficient of any one variable equal to the coefficient of the same variable in the other equation. This can be done by multiplying or dividing the whole equation accordingly. Then we subtract the equations from each other and solve for one variable.

6x - 3y = 8 (multiplying this equation by -2 to make the coefficient of y equal to that of in the equation 23x + 6y = 16)

Now are equations are

-12x + 6y = -16 and 23x + 6y = 16

subtracting eq 1 from eq 2 gives,

35x = 32 (Note that y has now been eliminated)

x = 32/35

Substituting this value of x in any one of the equations will then give the value of y to be -88/105

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Answer: See below

Step-by-step explanation:

A.

Let's split the integral into two parts, by the Sum Rule.

\(\int\limits {x-4x^3} \, dx\)                              [split into 2 integrals]

\(\int\limits {x} \, dx -\int\limits {4x^3} \, dx\)                      [solve integral for each part]

\(\frac{1}{2} x^2-x^4+C\)                            [Remember, we need to add C for constant]

-------------------------------------------------------------------------------------------------

B.

\(\int\limits {\frac{1+x}{\sqrt{x} } } \, dx\)                                    [expand into 2 integrals]

\(\int\limits {\frac{1}{\sqrt{x} } } \, dx +\int\limits {\frac{x}{\sqrt{x} } } \, dx\)                    [simplify second integral]

\(\int\limits {\frac{1}{\sqrt{x} } } \, dx +\int\limits {\sqrt{x} } \, dx\)                    [solve integral for each part]

\(2\sqrt{x} +\frac{2}{3}x^3^/^2+C\)

-------------------------------------------------------------------------------------------------

C.

\(\int\limits^4_0 {z(z^1^/^2-z^-^1^/^2)} \, dz\)                [distribute]

\(\int\limits^4_0 {z^3^/^2-z^1^/^2} \, dz\)                       [split into 2 integrals]

\(\int\limits^4_0 {z^3^/^2} \, dz -\int\limits^4_0 {z^1^/^2} \, dxz\)              [solve integral for each part]

\(\frac{64}{5} -\frac{16}{3}\)                                     [solve]

\(\frac{112}{15}\)

-------------------------------------------------------------------------------------------------

D. *Note: I can't put -1 for the interval, but know that the 1 on the bottom is supposed to be -1.

\(\int\limits^1_1 {(1+u)(1-u)} \, du\)                  [expand]

\(\int\limits^1_1 {1-u^2} \, du\)                              [split into 2 integrals]

\(\int\limits^1_1 {1} \, du-\int\limits^1_1 {u^2} \, du\)                      [solve integral for each part]

\(2-\frac{2}{3}\)                                      [solve]

\(\frac{4}{3}\)

Answer:

As x gets smaller, pointing to negative infinity, the value of p increases, pointing to positive infinity.

As x increases, pointing to positive infinity, the value of p increases, pointing to positive infinity.

Step-by-step explanation:

To find the end behaviour of a function f(x), we calculate these following limits:

\(\lim_{x \to +\infty} f(x)\)

And

\(\lim_{x \to -\infty} f(x)\)

At negative infinity:

\(\lim_{x \to -\infty} (4x^{8} - 6x^{7} + 3x^{3} - 10)\)

When the variable points to infinity, we only consider the term with the highest exponent. So

\(\lim_{x \to -\infty} (4x^{8} - 6x^{7} + 3x^{3} - 10) = \lim_{x \to -\infty} 4x^{8} = 4*(-\infty)^{8} = \infty\)

Plus infinity, because the exponent is even.

So as x gets smaller, pointing to negative infinity, the value of p increases, pointing to positive infinity.

At positive infinity:

\(\lim_{x \to \infty} (4x^{8} - 6x^{7} + 3x^{3} - 10) = \lim_{x \to \infty} 4x^{8} = 4*(\infty)^{8} = \infty\)

As x increases, pointing to positive infinity, the value of p increases, pointing to positive infinity.

Answer:

A - As x -> infinity, p(x) -> infinity, and as x -> -infinity, p(x) -> infinity.

Step-by-step explanation:

Let's begin by listing out the information given to us:

height (opposite) = 21, breadth (adjacent) = 16, x = ?

We have the length of the side opposite angle x and we have the length of the side adjacent to angle x. This informs us of the trigonometric ratio to us

We will use the trigonometric ratio to calculate for x as shown below:

a. The answer is: The fewest number of multibase blocks required to represent 20 longs in base seven is 2 flats.

b. The answer is: The fewest number of multibase blocks required to represent 10 longs in base three is 3 flats and 1 unit.

a. To represent 20 longs in base seven, we need to find the fewest number of multibase blocks required.

In base seven, we have the following conversions:

1 long = 1 unit

1 flat = 10 units

1 block = 10 flats

To represent 20 longs, we can use 2 flats (each flat representing 10 units) and 0 units since there are no remaining units.

So, the fewest number of multibase blocks required would be 2 flats.

Therefore, the answer is: The fewest number of multibase blocks required to represent 20 longs in base seven is 2 flats.

b. To represent 10 longs in base three, we need to find the fewest number of multibase blocks required.

In base three, we have the following conversions:

1 long = 1 unit

1 flat = 3 units

1 block = 3 flats

To represent 10 longs, we can use 3 flats (each flat representing 3 units) and 1 unit since there is one remaining unit.

So, the fewest number of multibase blocks required would be 3 flats and 1 unit.

Therefore, the answer is: The fewest number of multibase blocks required to represent 10 longs in base three is 3 flats and 1 unit.

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To make the dough, he'll need 5 3/4 cups of floor.

A fraction is a portion of a whole or, more broadly, any number of equal parts. In everyday English, a fraction describes the number of parts of a specific size, such as one-half, eight-fifths, or three-quarters. In mathematics, a fraction is used to represent a portion or part of a whole. It symbolizes the equal parts of the whole. A fraction is made up of two parts: the numerator and the denominator. The top number is known as the numerator, and the bottom number is known as the denominator.

Here,

10 1/4 cups of flour needed to make a batch=41/4 cups

4 1/2 cups of floor is there=9/2 cups

flour does he need to make the dough,

=41/4-9/2

=(41-18)/4

=23/4

=5 3/4 cups

He needs 5 3/4 cups of floor to make the dough.

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The equation of the line in point slope form is  y = M(x - x₀) + y₀ .

The Point slope form of the line passing through the point (x₁,y₁) with slope m is given by the formula

(y - y₁) = m(x - x₁)

In the question ,

it is given that

the required line passes through the points (x₀ , y₀)

and the slope is M .

So, the point slope form equation of the line will be

( y - y₀) = M(x - x₀)

y - y₀ = M(x - x₀)

y = M(x - x₀) + y₀

Therefore , the equation of the line in point slope form is  y = M(x - x₀) + y₀ .

The given question is incomplete , the complete question is

Write an equation for the line through the point (x₀,y₀) with a slope of M in point slope form. Your equation should be in the form of y = ...    ?

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The amount in the account after 4 years, rounded to the nearest cent, will be approximately $17,332.8.

Recall the compounding formula:

A = P * \(e^{rt}\)

Where:

A = Final amount in the account

P = Initial principal (deposit)

e = Euler's number (approximately 2.71828)

r = Annual interest rate (as a decimal)

t = Time in years

Given:

Initial principal (P) = $16,000

Annual interest rate (r) = 2%  = 0.02

time (t) = 4 years.

Substitute these values into the formula, we get:

A = $16,000 * \(e^{0.02 * 4}\)

Using a calculator, we can calculate:

A = $16,000 * \(e^{0.08}\)

A = $16,000 * 1.0833

A = $17,332.8

Therefore, the amount in the account after 4 years, rounded to the nearest cent, will be approximately $17,332.8.

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To determine which ordered pair is a local minimum of the function, f(x), we need to look for a point where the function value is smaller than the values of the function at the surrounding points.

From the table, we can see that the function value at x = 2 is -15, which is smaller than the function values at x = 1 and x = 3, which are both 0. Therefore, the ordered pair (2, -15) is a local minimum of the function, f(x).

Note that the ordered pairs (0, 9), (4, 105), and (-1, 0) are not local minima, as the function values at these points are all larger than the values at their surrounding points.

Answer:

D

Step-by-step explanation:

Answer:

3/4

Step-by-step explanation:

You have to Subtract 6 ¼ by 5 ½

Answer:

y is 31

Step-by-step explanation:

we need to substitute first

x + y = 10

x= -21 so...

-21 + y = 10

now we need to add 21 on both sides of the equation, we are adding because  21 because we want that 21 gone, and it's the oppisite of subtracting

-21 + 21 + y = 10 + 21

0 + y = 10 + 21

y= 31

Check:

x+y=10

-21 + 31= 10

10 = 10 correct!

Hope this helped!

Vector v can be written as the sum of two vector components, \(v_x = -5i\)  and \(v_y = 6j\)

To write vector v as the sum of two vector components, we need to decompose it into two vectors along different directions.

v = -5i + 6j

w = 3i + 4j

We can decompose vector v into two components, one along the x-axis and the other along the y-axis.

Let's denote the component along the x-axis as \(v_x\) and the component along the y-axis as \(v_y.\)

The x-component, \(v_x,\) represents the projection of vector v onto the x-axis and can be found using the dot product of v and the unit vector i:

\(v_x = v . i = (-5i + 6j) . i = -5i . i + 6j . i = -5\)

Similarly, the y-component, \(v_y,\) represents the projection of vector v onto the y-axis and can be found using the dot product of v and the unit vector j:

\(v_y = v . j = (-5i + 6j) . j = -5i . j + 6j . j = 6\)

Now, we can write vector v as the sum of its components:

\(v = v_x + v_y = -5i + 6j\)

So, vector v can be written as the sum of two vector components: -5i along the x-axis and 6j along the y-axis.

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One possible solution from the graph(attached at the end of the solution) is selling 9 bags of popcorn and 5 bags of candies.

As per the given data:

The number of popcorn bags is x

The number of candies is y

The cost of each popcorn bag is $7

The cost of each candy is $4

She has $80 to spend on them

Inequality is the nonequal comparison of two or more variables and numbers.

So the first inequality is

7 x + 4 y ≤ 80

Since she must buy no less than 16 bags of popcorn and candies

The second inequality is

x + y ≥ 16

The red area represents the 1st inequality

The blue area represents the 2nd inequality

The solutions lie in the common part of red and blue colors

One possible solution is ordered pair (9,5)

(Graph attached at the end of solution)

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Answer:

A, B, and E. Hope this helps!

The system that has linear inequalities has the point (3, -2) in its solution set is y > -3; y ≥ 2/3x - 4.

We know that

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities (makes true both inequalities)

In option B, we have

y > -3 ----> inequality A

y ≥ 2/3x - 4 ----> inequality B

In both inequality, change the values of x and y at the point (3, -2) and then compare the outcomes.

Inequality A

y > -3 ----> is true

Inequality B

y ≥ 2/3x - 4 ----> is true

Therefore

The ordered pair is a solution of the system B

As a result, the point (3, -2) in the solution set of the system with linear inequalities is y > -3; y 2/3x - 4.

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Answer:

I will give you three, but the answer is the number that has a 50

Step-by-step explanation:

52

654

7657

50 is the numbers that have a 5 in 10 times the value of the 13,725.

Place value is the value of each digit in a number.

For example, the 5 in 350 represents 5 tens, or 50; however the 5 in 5006 represents 5 thousands, or 5000.

Now the given number is,

13,725

10 times of the number is givens as,

13,725*10

or, 137,250

here place value of 5 in 137,250 is 5 tens or 50.

Therefore, 50 is the numbers that have a 5 in 10 times the value of the 13,725.

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To find the value of (f o g)' at a given value, you first need to understand the concept of composite functions and the chain rule of differentiation. Let's break it down step by step.

To find the value of (f o g)' at a given value, you need to evaluate g(x) and f(x), find their derivatives, and use the chain rule to find the derivative of (f o g) at the given value. It is important to understand the concepts of composite functions and the chain rule to be able to solve problems involving these concepts.

What are composite functions? Composite functions are functions that are formed by composing two or more functions. The notation used to denote composite functions is (f o g)(x), which means that the output of function g is used as the input for function f. In other words, we first evaluate g(x), and then use the result as the input for f(x).

What is the chain rule of differentiation? The chain rule of differentiation is a method used to find the derivative of composite functions. It states that if a function is composed of two or more functions, then its derivative can be found by taking the derivative of the outer function and multiplying it by the derivative of the inner function.

To find the value of (f o g)' at a given value, we need to follow these steps:1. Find g(x) and f(x)2. Find g'(x) and f'(x)3. Evaluate g(x) at the given value4. Use the chain rule to find (f o g)' at the given value

step 1: Find g(x) and f(x)Let's say that we have two functions: g(x) = x^2 + 3x + 1 and f(x) = sqrt(x). To find (f o g)(x), we first need to evaluate g(x) and then use the result as the input for f(x). Therefore, (f o g)(x) = f(g(x)) = sqrt(x^2 + 3x + 1)

Step 2: Find g'(x) and f'(x)To find g'(x), we need to take the derivative of g(x) using the power rule and the sum rule. Therefore, g'(x) = 2x + 3To find f'(x), we need to take the derivative of f(x) using the power rule and the chain rule. Therefore, f'(x) = 1/2(x)^(-1/2)

Step 3: Evaluate g(x) at the given valueSuppose we want to find (f o g)' at x = 2. To do this, we need to first evaluate g(x) at x = 2. Therefore, g(2) = 2^2 + 3(2) + 1 = 11

Step 4: Use the chain rule to find (f o g)' at the given value now we can use the chain rule to find (f o g)' at x = 2. Therefore, (f o g)'(2) = f'(g(2)) * g'(2) = 1/2(11)^(-1/2) * (2)(3) = 3/sqrt(11)

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The ends of the points of the latus rectum of the parabola are (x₁, y₁) = (0, 1) and (x₂, y₂) = (0, - 3), respectively.

In this question we find the equation of a parabola, whose axis of symmetry is parallel with the x-axis. From which we need to determine the points of the latus rectum, that is, the ends of the line that contains the focus of the parabola. First, write the entire equation of the parabola:

(y + 1)² = 4 · (x + 1)

Second, find the y-coordinates of the ends of the latus rectum:

(y + 1)² = 4 · (x + 1)

(y + 1)² = 4 · (0 + 1)

(y + 1)² = 4

y + 1 = ± 2

y = - 1 ± 2

y = 1 or y = - 3

Third, write the coordinates of the ends of the latus rectum:

(x₁, y₁) = (0, 1) and (x₂, y₂) = (0, - 3)

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Answer:

To prove that none of the terms of the sequence 2, 5, 8 are perfect squares, we need to show that no integer exists such that its square equals 2, 5, or 8.

First, let's consider 2. We can start by assuming that there exists an integer n such that n^2 = 2. However, this leads to a contradiction, because the square of any integer is always non-negative, so n^2 cannot be equal to a negative number like 2. Therefore, there is no integer n such that n^2 = 2, and 2 is not a perfect square.

Similarly, let's consider 5. Again, assume that there exists an integer n such that n^2 = 5. However, this leads to another contradiction, because the square of any integer is always either a perfect square or a number that is slightly larger than a perfect square (for example, 4^2 = 16 and 6^2 = 36). Since 5 is not a perfect square, it cannot be equal to the square of any integer, and there is no integer n such that n^2 = 5.

Finally, let's consider 8. We can again assume that there exists an integer n such that n^2 = 8. However, this leads to a contradiction, because the square of any even integer is always a multiple of 4 (for example, (2n)^2 = 4n^2). Since 8 is not a multiple of 4, it cannot be equal to the square of any integer, and there is no integer n such that n^2 = 8.

Therefore, we have shown that none of the terms of the sequence 2, 5, 8 are perfect squares.

Answer:

You can make equivalent fractions by multiplying or dividing both top and bottom by the same amount. You only multiply or divide, never add or subtract, to get an equivalent fraction. Only divide when the top and bottom stay as whole numbers

Answer:

Step-by-step explanation:

By using multiplication or division. Multiplying the numerator and denominator or simplifying which is also division.

Answer

If its times then its 48 if its divide then its 1.3 if its adding then its 14, if its subtract then its 2

Step-by-step explanation:

Use a calculator

Answer:

24

Step-by-step explanation:

Answer:

\(y=-\frac{2}{3}x-\frac{10}{3}\)

Step-by-step explanation:

Hi there!

What we need to know:

1) Determine the slope (m)

\(y =-\frac{2}{3} x +4\)

In the given equation, \(-\frac{2}{3}\) is in the place of m, making it the slope. Because parallel lines always have the same slope, the slope of the line we're currently solving for is \(-\frac{2}{3}\). Plug this into \(y=mx+b\):

\(y=-\frac{2}{3}x+b\)

2) Determine the y-intercept (b)

\(y=-\frac{2}{3}x+b\)

Plug in the given point (-2,-2) to solve for b

\(-2=-\frac{2}{3}(-2)+b\\-2=\frac{4}{3}+b\)

Subtract \(\frac{4}{3}\) from both sides to isolate b

\(-2-\frac{4}{3}=\frac{4}{3}+b-\frac{4}{3}\\-\frac{10}{3} =b\)

Therefore, the y-intercept of the line is \(-\frac{10}{3}\). Plug this back into \(y=-\frac{2}{3}x+b\):

\(y=-\frac{2}{3}x-\frac{10}{3}\)

I hope this helps!

Answer:

24 units

Step-by-step explanation:

Each line plot equals 2, so the length is 8 and the hight is 4. 8+8+4+4 = 16 + 8 = 24 units.

Answer:

-10

Step-by-step explanation:

\(5(a-b)\\\\5(-4-(-2))\\\\5(-4+2)\\\\5(-2)\\\\\boxed{-10}\)

Hope this helps!