Andrea bought tacos from a food truck and left 25% tip of $2.00. What was the price of Andreas tacos, before tip?

The term 'GDP Gap' refers to the difference between the actual Gross Domestic Product (GDP) and the potential GDP of a country. It is a measure of the underutilization or overutilization of economic resources.

In the context of Australia's COVID-19 recession, the GDP Gap would indicate the extent to which the recession has impacted the country's economic output compared to its potential.

During a recession, the GDP Gap is usually negative, indicating a decline in economic activity. This gap arises due to factors such as decreased consumer spending, reduced investment, and increased unemployment. In the case of Australia, the COVID-19 pandemic triggered a significant economic plunge, leading to a confirmed recession.

The GDP Gap can be calculated by subtracting the actual GDP from the potential GDP. The potential GDP represents the level of economic output that could be achieved if all resources were fully employed. The larger the GDP Gap, the greater the underutilization of resources and the more severe the economic downturn.

The term 'GDP Gap' refers to the difference between actual GDP and potential GDP, and it is used to measure the extent of economic underutilization or overutilization. In the case of Australia's COVID-19 recession, the GDP Gap would indicate the negative impact of the recession on the country's economic output compared to its potential.

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The correct set of possible results would be (0, 1, 0, 0), (1, 2, 1, 0) and (1, 2, 0, 1).

Your approach seems to be correct, but there seems to be a minor mistake in your final list of possible solutions. Let's go through the steps to clarify.

Given the initial conditions z=t=0, you obtained a partial solution (0,1,0,0).

Next, you solved the homogeneous equations for z=1 and t=0, which resulted in a solution (1,1,1,0).

Similarly, solving the homogeneous equations for z=0 and t=1 gives another solution (1,1,0,1).

To find the general solution, you combine the partial solution with the solutions obtained in the previous step, using parameters a and b.

(0,1,0,0) + a(1,1,1,0) + b(1,1,0,1)

Expanding this expression, you get:

(0+a+b, 1+a+b, 0+a, 0+b)

Simplifying, you obtain the following set of solutions:

(0, 1, 0, 0)

(1, 2, 1, 0)

(1, 2, 0, 1)

Therefore, the correct set of possible results would be:

(0, 1, 0, 0)

(1, 2, 1, 0)

(1, 2, 0, 1)

Note that (0, 1, 1, 1) is not a valid solution in this case, as it does not satisfy the initial condition z = 0.

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Answer: Not 100% sure but I think it is p=7

Answer:

8+8p

Step-by-step explanation:

you use distributive property! Multiply 8 by 1 and 8 by p and add it!

(1+p)8

8+8p

Answer:

-17

Step-by-step explanation:

x=-8

-8+6=2

2 times -8 -1+-17

The $1314^\text{th}$ digit past the decimal point is 2.

To find the $1314^\text{th}$ digit past the decimal point in the decimal expansion of $\frac{5}{14}$, we can use long division to compute the decimal expansion of the fraction.

The long division of $\frac{5}{14}$ is as follows:

```

0.35    <-- Quotient

-----

14 | 5.00

   4.2    <-- Subtract: 5 - (14 * 0.3)

   -----

     80   <-- Bring down the 0

     70    <-- Subtract: 80 - (14 * 5)

     -----

     100   <-- Bring down the 0

      98    <-- Subtract: 100 - (14 * 7)

      -----

       20   <-- Bring down the 0

       14    <-- Subtract: 20 - (14 * 1)

       -----

        60   <-- Bring down the 0

        56    <-- Subtract: 60 - (14 * 4)

        -----

         40   <-- Bring down the 0

         28    <-- Subtract: 40 - (14 * 2)

         -----

         120   <-- Bring down the 0

         112    <-- Subtract: 120 - (14 * 8)

         -----

           80   <-- Bring down the 0

           70    <-- Subtract: 80 - (14 * 5)

           -----

           ...

```

We can see that the decimal expansion of $\frac{5}{14}$ is a repeating decimal pattern with a repeating block of digits 285714. Therefore, the $1314^\text{th}$ digit past the decimal point is the same as the $1314 \mod 6 = 0^\text{th}$ digit in the repeating block.

Since $1314 \mod 6 = 0$, the $1314^\text{th}$ digit past the decimal point in the decimal expansion of $\frac{5}{14}$ is the first digit of the repeating block, which is 2.

So, the $1314^\text{th}$ digit past the decimal point is 2.

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The given system of equations is:

5d + 20e - 10% = -85 …………………..(1)

-20d + 394e = 223 ………………...(2)

4d + 13e = -106 …………………….(3)

We can solve this system of equations using the Gaussian elimination method. In this method, we convert the system of equations into an equivalent system of equations that is easy to solve by eliminating one variable at a time.The elimination process starts by multiplying the equations so that one variable cancels out. We can do this by multiplying both sides of an equation with a constant or by adding or subtracting two equations.

Here, the first equation has a percentage sign which we need to remove.

We can do this by dividing both sides of equation (1) by

10.5d/10 + 20e/10 - 1 = -85/10

⇒ 0.5d + 2e - 1 = -8.5

⇒ 0.5d + 2e = -7.5 ……………………..(4)

Now, we can use equations (2) and (4) to eliminate d.

Multiplying equation (2) by 2 and adding it to equation (4), we get,

2(-20d + 394e = 223) + (0.5d + 2e = -7.5)

⇒ -40d + 788e + 0.5d + 2e = -7.5 + 446

⇒ -39.5d + 790e = 438 ……………………...(5)

Multiplying equation (2) by 5 and adding it to equation (3), we get,

5(-20d + 394e = 223) + (4d + 13e = -106)

⇒ -100d + 1970e + 4d + 13e = -106 + 1105

⇒ -96d + 1983e = 999 ……………………(6)

Now, we can use equations (5) and (6) to eliminate d.

Multiplying equation (5) by 2 and adding it to equation (6), we get,

2(-39.5d + 790e = 438) + (-96d + 1983e = 999)

⇒ -79d + 1563e + (-96d + 1983e) = 876 + 1998

⇒ -175d + 3546e = 2874 …………………..(7)

We can simplify equation (7) by dividing both sides by 7,-25d + 506e = 410 ……………………..(8)

Now, we can use equations (4) and (8) to solve for e.

Multiplying equation (4) by 506/4 and adding it to equation (8), we get,

506/4 × (0.5d + 2e = -7.5) + (-25d + 506e = 410)

⇒ 253d + 506e - 1895 - 25d + 506e = 410

⇒ 228d + 1012e = 2305 ……………………(9)

We can simplify equation (9) by dividing both sides by

4,57d + 253e = 576 ……………………(10)

Now, we can use equations (4) and (10) to solve for d.

Multiplying equation (4) by 57/2 and adding it to equation (10), we get,

57/2 × (0.5d + 2e = -7.5) + (57d + 253e = 576)

⇒ 28.5d + 114e - 214.125 + 57d + 253e = 576

⇒ 85.5d + 367e = 790.125 …………………….(11)

We can simplify equation (11) by dividing both sides by 17,5d + 23e = 46.4166667 …………………….(12)

Now, we can use equations (4) and (12) to solve for e.

Multiplying equation (4) by 23/2 and adding it to equation (12), we get,

23/2 × (0.5d + 2e = -7.5) + (5d + 23e = 46.4166667)

⇒ 11.5d + 46e - 86.625 + 5d + 23e = 46.4166667

⇒ 16.5d + 69e = 133.0416667 ……………………..(13)

We can simplify equation (13) by dividing both sides by 3,5.5d + 23e = 44.3472222 ……………………..(14)

Now, we can use equations (4) and (14) to solve for e.

Multiplying equation (4) by 23 and subtracting it from equation (14), we get,

23 × (0.5d + 2e = -7.5) - (5.5d + 23e = 44.3472222)

⇒ 11.5d - 48.5e = -212.0972222 ……………………..(15)

We can simplify equation (15) by dividing both sides by 1.5,7.6666667d - 32.3333333e = -141.3981481 ……………………..(16)

Now, we can use equations (4) and (16) to solve for e.

Multiplying equation (4) by 32 and adding it to equation (16), we get,

32 × (0.5d + 2e = -7.5) + (7.6666667d - 32.3333333e = -141.3981481)

⇒ 16d + 64e - 240 + 7.6666667d - 32.3333333e = -141.3981481

⇒ 23.6666667d + 31.6666667e = 98.60185185 …………………..(17)

We can simplify equation (17) by dividing both sides by 1.6666667,

14.2d + 19e = 59.1611111 ……………………..(18)

Now, we can use equations (4) and (18) to solve for e.

Multiplying equation (4) by 19 and subtracting it from equation (18), we get,

19 × (0.5d + 2e = -7.5) - (14.2d + 19e = 59.1611111)

⇒ 9.5d - 41e = -186.3111111 ……………………..(19)

We can simplify equation (19) by dividing both sides by 1.9,

5d - 21.57894737e = -98.05789474 ……………………..(20)

Now, we can use equations (4) and (20) to solve for e.

Multiplying equation (4) by 21.57894737 and adding it to equation (20), we get,

21.57894737 × (0.5d + 2e = -7.5) + (5d - 21.57894737e = -98.05789474)

⇒ 43.15789474d + 0e = -252.5263158

⇒ 43.15789474d = -252.5263158

⇒ d = -5.85416667

Substituting the value of d in equation (4), we get,

0.5d + 2e = -7.5

⇒ 0.5(-5.85416667) + 2e = -7.5

⇒ e = -6.5

Now, we have the values of d and e, and we can substitute them in any equation to get the value of the remaining variable.

We can use equation (2),

-20d + 394e = 223

⇒ -20(-5.85416667) + 394(-6.5) = 223

⇒ 117.0833334 + (-2561) = 223

⇒ -2443.916667 = 223

This is not true, which means there is no solution to this system of equations. Hence, the system is inconsistent.

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Hello!

Let's take this step-by-step:

The formula for the Area of the Triangle:

  \(Area = \dfrac{1}{2} *\text{Base} *\text{Height}\)

In this case:

Let's plug in those value's into the equation:

    \(\text{Area}=\dfrac{1}{2} *(2x +4)*(x+3)\\\\\\\text{Let's first multiple '2x + 4' and 'x+3'}\\\text{Area}=\dfrac{1}{2} *(2x^2+6x + 4x + 12)=\dfrac{1}{2} *(2x^2+10x + 12)\\\\\\\\\\\textNow multiple 1/2 in the function in the parentheses}\\\text{Area}=\dfrac{1}{2} *2x^2 + \dfrac{1}{2}*10x +\dfrac{1}{2} *12\\\\\text{Area}=x^2+5x + 6\)

Answer: \(x^2 + 5x +6\)

Hope that helps!

Answer:

2 and 6

Step-by-step explanation:

178 square meters is the area of the complex figure.

To find the area of this composite figure, we need to find the areas of the individual shapes and add them up.

The area of the first rectangle is:

16 m x 8 m = 128 m²

The area of the triangle is:

1/2 x base x height = 1/2 x 5 m x 8 m = 20 m²

The area of the second rectangle is:

6 m x 5 m = 30 m²

To find the total area, we add the areas of the three shapes:

128 m² + 20 m² + 30 m² = 178 m²

Therefore, the area of this complex figure is 178 square meters.

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It is well known that utilizing a zip file is an effective method for reducing the size of the document in order to boost storage and sharing possibilities. A zipped folder may be smoothly moved from one place (the transmitter) to another without any issues (the receiver).

One or more files can be compressed and stored together in one location using the ubiquitous ZIP file format. As a result, the file size is less and is simpler to transport and store. After transmission, the receiver can use the files in their original form by unzipping (or extracting) her ZIP file.

In order to increase storage and sharing capabilities by lowering the size of the document, it is known that using a zip file is an efficient technique. A zipped folder may be effortlessly transferred without any hiccups from one location (the transmitter) to another (the receiver). Using Zip files has the primary benefit of reducing the time required to send large files over email. The document's quality is unaffected and is identical to the original file when it is compressed using a zip file.

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Answer:

\((x-y)(x^2+xy++y^2)\)

Step-by-step explanation:

\(Proof :\\(x-y)(x^2+xy+y^2)\\=(x^3+x^2y+y^2x-x^2y-xy^2-y^3)\\=x^3-y^3\)

Answer:

It would be D

Step-by-step explanation:

Got it right on Edge 2021

Answer:

4 months

Step-by-step explanation:

\(\frac{22}{3}\)  = 7\(\frac{1}{3}\)

Hope this helps :-)

The value of y on the number line is 4.

A number line is a series of numbers placed on a straight line in order from least to greatest.

The number line is RT.

The point S is in between RT.

Therefore,

RT = RS + ST

Hence,

RT = 67

RS = 9y + 3

ST = 5y + 8

Therefore,

RT = RS + ST

67 = 9y + 3 + 5y + 8

67 = 9y + 5y + 11

67 = 14y + 11

subtract 11 from both sides of the equation

67 = 14y + 11

67 - 11 = 14y + 11 - 11

56 = 14y

divide both sides by 14

y = 56 / 14

Therefore,

y = 4

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The minimum score needed to be in the top 2% of the exam is 90.5, which is two standard deviations above the mean.

To calculate the minimum score needed to be in the top 2%, we need to find the value of two standard deviations above the mean. This can be done using the formula x = μ + (2σ), where μ represents the mean and σ represents the standard deviation. In this case, μ = 76.4 and σ = 6.1. We substitute these values into the formula to get x = 76.4 + (2*6.1), which equals 90.5. Therefore, the minimum score needed to be in the top 2% is 90.5. To further understand this calculation, it is important to note that the normal distribution curve is symmetrical. This means that the area under the curve that is two standard deviations above the mean is 2%, which is the same as the area under the curve that is two standard deviations below the mean. Therefore, the score that is two standard deviations above the mean is the minimum score needed to be in the top 2%.

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Answer:

a or b

Step-by-step explanation:

Answer:

B. 30

Step-by-step explanation:

3/4 = .75

120 * .75 = 90

Of those 90 students 1/3 wore sandals.

1/3 = approximately .33

90 * .33 = 29.7

Therefore, around 30 students wore jeans and sandals.

Answer:

E

Step-by-step explanation:

NO

Answer:

A group of seven friends each received one-half of a pound ofcandy. How much candy did they receive total?✓

0.5 pounds x 7= 3.5 poundsStep-by-step explanation:

The values of A and B that make AX + BY = C are 3 and 5 respectively.

The standard form of the equation is 3x +5y = 15

An intercept is the point where the a straight line or curve touches the x or y axis.

If the curve or line touches the x-axis the y-coordinate is zero and if touches the y-axis the x-coordinate is 0.

in the graph, the x-intercept is 5, as an ordered pair you have (5,0)

The y-intercept is 3 as an ordered pair you have (0,3)

The given equation is AX +BY = c, where C = 15

so the equation is AX +BY = 15

when x =5, y = 0

so that

5A = 15

A = 15/5 = 3

When x = 0, y = 3

so that you have,

3B = 15

B = 15/3 = 5

The values of A and B are 3 and 5 respectively.

Substituting A and B into the equation,

It becomes 3X +5Y = 15

In conclusion, the values of A and B are 3 and 5 respectively and the standard form of the equation is 3x + 5y = 15

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The type I error in this case is: The officer rejects a supplier who makes good quality products.

In hypothesis testing, a type I error occurs when the null hypothesis (H0) is true, but it is incorrectly rejected in favor of the alternative hypothesis (H1). In this scenario, the null hypothesis states that the supplier does not meet the quality standards (poor quality products). The alternative hypothesis states that the supplier does meet the quality standards (good quality products).

If the officer incorrectly rejects the null hypothesis (H0), it means they mistakenly conclude that the supplier does not meet the quality standards and, as a result, rejects the supplier. However, in reality, the supplier actually produces good quality products.

This decision is a type I error because the officer has made a false rejection based on incorrect evidence. The type I error in this case is the officer rejecting a supplier who makes good quality products.

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To find the absolute maximum and minimum of a function, determine the critical points and evaluate the function at those points and the endpoints of the interval.

To find the absolute maximum and minimum of a function, you need to follow these steps:

1. Determine the critical points of the function by finding where its derivative equals zero or is undefined. These points can be potential candidates for the maximum or minimum values.

2. Evaluate the function at the critical points and at the endpoints of the interval you are considering. The highest value among these points will be the absolute maximum, while the lowest value will be the absolute minimum.

Let's illustrate this with an example:

Consider the function f(x) = x^2 - 4x + 5 on the interval [0, 5].

1. To find the critical points, we need to find where the derivative is equal to zero or undefined. Taking the derivative of f(x),

we get f'(x) = 2x - 4.

Setting this equal to zero gives us 2x - 4 = 0, which implies x = 2. So, x = 2 is the only critical point.

2. Now, we evaluate the function at the critical point and the endpoints of the interval:
f(0) = 0^2 - 4(0) + 5 = 5
f(2) = 2^2 - 4(2) + 5 = 1
f(5) = 5^2 - 4(5) + 5 = -5

From these evaluations, we see that f(0) = 5 is the absolute maximum and f(5) = -5 is the absolute minimum.

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Answer:

you would make something like this:

Step-by-step explanation:

i cant understand your question define briefly

The maximum amount of rainfall that was ever recorded in a single day in Turramurra is 444 centimeters.

From the dot plot, we can see that the maximum rainfall recorded in a single day was 444 centimeters. This is the highest value plotted on the graph and is the maximum amount of rainfall that was ever recorded in Turramurra.

The dot plot reveals that the maximum amount of rain ever recorded in a single day in Turramurra is 444 centimeters. This is an unusually high amount of rain for the area, indicating that the day might have been particularly stormy or that the area was in the middle of a heavy rain season. It is interesting to note that the average rainfall for the area is about 50 centimeters, making this day an outlier in the 414141-year period.

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To solve this problem we need to use the formula for the surface area of a cylinder. So, the surface area of the given cylinder with base radius 3 and height 6 is 54π square units or approximately 169.65 square units.

The formula for the surface area of a cylinder is S=2πrh+2πr², where r is the radius and h is the height of the cylinder.

A cylinder has a base radius of 3 and a height of 6, therefore: S = 2πrh + 2πr²S = 2π(3)(6) + 2π(3)²

S = 36π + 18πS = 54π square units (exact answer in terms of π)

S ≈ 169.65 square units (approximate answer to two decimal places using π ≈ 3.14). Therefore, the surface area of the given cylinder with base radius 3 and height 6 is 54π square units or approximately 169.65 square units.

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Answer:

you have to look at the chart and use the chart to answer the questions at the botttom

Step-by-step explanation: