Create 3 different graphs that model the three kinds of systems of equations solutions. Do not put them all on the same graph.
1. Graph of a system with one solution
2. Graph of a system with no solution
3. Graph of a system with infinite solutions
Rubric: Each of the 3 graphs must contain
● The two lines of the system
● The two equations that correspond to the two lines
● A label that marks the solution if there is one
● An x and y axis scale that is accurate for the equations on the graph
You may create the 3 graphs in any format you are comfortable with and can easily submit to me as long as the above requirements are met. Some format options include
● Desmos, screenshot your 3 graphs and equations to submit
● Desmos, export your 3 graphs and then add information on the equations to go with the graphs.
● Desmos, create a free account and share your saved graphs link with me
● Graph paper (NOT notebook paper), take a pic and submit
● Any other free graphing app, site or software you can easily access that will let you create graphs you can download, screen shot or take a pic of. But you must include information on the equations and the solutions as well.
● Desmos, screenshot your 3 graphs and equations to submit
● Desmos, export your 3 graphs and then add information on the equations to go with the graphs.
● Desmos, create a free account and share your saved graphs link with me
● Graph paper (NOT notebook paper), take a pic and submit
● Any other free graphing app, site or software you can easily access that will let you create graphs you can download, screen shot or take a pic of. But you must include information on the equations and the solutions as well.
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Task: Create 3 different graphs that model the three kinds of systems of equations solutions. Do not put them all on the same graph.
1. Graph of a system with one solution
2. Graph of a system with no solution
3. Graph of a system with infinite solutions

Rubric: Each of the 3 graphs must contain
● The two lines of the system
● The two equations that correspond to the two lines
● A label that marks the solution if there is one
● An x and y axis scale that is accurate for the equations on the graph

Submission: You may create the 3 graphs in any format you are comfortable with and can easily submit to me as long as the above requirements are met. Some format options include...
● Desmos, screenshot your 3 graphs and equations to submit
● Desmos, export your 3 graphs and then add information on the equations to go with the graphs.
● Desmos, create a free account and share your saved graphs link with me
● Graph paper (NOT notebook paper), take a pic and submit
● Any other free graphing app, site or software you can easily access that will let you create graphs you can download, screen shot or take a pic of. But you must include information on the equations and the solutions as well. Portfolio


Task: Create 3 different graphs that model the three kinds of systems of equations solutions. Do not put them all on the same graph.
1. Graph of a system with one solution
2. Graph of a system with no solution
3. Graph of a system with infinite solutions

Rubric: Each of the 3 graphs must contain
● The two lines of the system
● The two equations that correspond to the two lines
● A label that marks the solution if there is one
● An x and y axis scale that is accurate for the equations on the graph

Submission: You may create the 3 graphs in any format you are comfortable with and can easily submit to me as long as the above requirements are met. Some format options include...
● Desmos, screenshot your 3 graphs and equations to submit
● Desmos, export your 3 graphs and then add information on the equations to go with the graphs.
● Desmos, create a free account and share your saved graphs link with me
● Graph paper (NOT notebook paper), take a pic and submit
● Any other free graphing app, site or software you can easily access that will let you create graphs you can download, screen shot or take a pic of. But you must include information on the equations and the solutions as well.
● Desmos, screenshot your 3 graphs and equations to submit
● Desmos, export your 3 graphs and then add information on the equations to go with the graphs.
● Desmos, create a free account and share your saved graphs link with me
● Graph paper (NOT notebook paper), take a pic and submit
● Any other free graphing app, site or software you can easily access that will let you create graphs you can download, screen shot or take a pic of. But you must include information on the equations and the solutions as well.

Answer:

x ≥ 22/5

Step-by-step explanation:

I used a calculator called MathPapa

The greatest number of sides that can result from a cross section of a rectangular prism is 6.

To see why, consider a rectangular prism. The prism has 6 faces, which are all rectangles. If we take a cross section of the prism, the section will intersect some of the faces, resulting in a closed shape. The shape will have at most as many sides as the number of sides of the faces that the section intersects.

Since each face of the rectangular prism has 4 sides, the maximum number of sides that can result from a cross section is 4 x 6 = 24. However, this is only possible if the section intersects every face of the prism, which is unlikely to occur in general. In practice, a cross section of a rectangular prism is more likely to intersect just a few of the faces, resulting in a shape with fewer sides. Therefore, the greatest number of sides that can result from a cross section of a rectangular prism is 6, which occurs when the section intersects all 6 faces.

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Answer:

Larry has at least 9 books

Step-by-step explanation:

just add 6 and 3 to get 9. and since it says at least 3 more, you would know that even after adding them, he would have at least 9.

That was a bad explanation, but I hope you know what I mean, if you dont, just comment on my answer and ask any questions about it :)

Answer:

4308

Step-by-step explanation:

The population of fish at the end of 10 years will remain 4129 .

Depreciation is essentially a compound measure in the opposite direction, so decreasing the original value at a specific percentage at regular intervals. For depreciation you need to use the multiplier method for decreasing percentages and put this multiplier to the power of the number of time intervals in the question.

Formula of compound interest depreciation :

\(A= P(1-\frac{r}{100} )^{n}\)

Where,

A = final amount

P = initial principal

n = time in years

r = rate per annum

According to the question

A population of fish starts at (P)= 8000

decreases  per year (r) =  6%

n = 10

now , Applying Formula of compound interest depreciation :

\(A= P(1-\frac{r}{100} )^{n}\)

substituting the value

\(A= 8000(1-\frac{6.4}{100} )^{10}\)

\(A= 8000(0.936} )^{10}\)

A=8000 *  0.5161

A = 4,129.03 (approx.)

A = 4129 (round off to nearest whole no )

Hence,  the population of fish at the end of 10 years will remain 4129 .

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Answer:

The median is the same for both car

types, but Bandits have the lower

mean.

Step-by-step explanation:

From the boxplot attached, the median value for both car types can be clearly seen to be the same as the vertical line in between the box of both plots lie at the same point.

As for the mean which is the average of a set of data we could deduce from the plot that, the roadster data has a larger span than the bandit data with a far greater third percentile value. Hence, mean or average data for roadster will be higher Than that if bandits.

The absolute value of x minus one is equal to two. Therefore, x must be either one greater than two (x = 3) or one less than two (x = 1).

One less than x results in a value of two in absolute terms. As a result, x minus 1 has an absolute value of 2, which is a positive number. The separation of an integer from zero is its absolute value. As a result, x minus 1 is two units from zero in absolute terms. As x minus one has an absolute value of two, its real value must either be two or a negative two. That is to say, x must either be one more than two (x = 3) or one less than two (x = 1).In order to confirm this, let's look at a few examples. If x = 3, then |x - 1| = |2 - 1| = |1| = 1 which is not equal to two. Therefore, x = 3 is not the answer we are looking for. On the other hand, if x = 1, then |x - 1| = |1 - 1| = |0| = 0 which is not equal to two. Therefore, x = 1 is also not the answer we are looking for. The only two possible solutions for the equation |x - 1| = 2 are x = 3 and x = 1. Therefore, the result of |x - 1| = 2 is that x must be either one greater than two (x = 3) or one less than two (x = 1)

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The   statement that is true ,congruence of corresponding angles is preserved under rigid Transformations.

If a series of rigid transformations maps ∠F onto ∠C, where ∠F is congruent to ∠C, it implies that the two angles have the same measure or size. Given this information, the following statement is true:

The congruence of corresponding angles is preserved under rigid transformations.

Rigid transformations, such as translation, rotation, and reflection, preserve the size, shape, and angles of geometric figures. When ∠F is congruent to ∠C, it means that the measures of the angles are equal.

By applying a series of rigid transformations that map ∠F onto ∠C, the congruence between the two angles is maintained. This means that the resulting transformed angle, after the series of transformations, will still have the same measure as the original angle.

In other words, if ∠F and ∠C are congruent, the rigid transformations will preserve the equality of their measures. Therefore, the congruence between the corresponding angles is maintained throughout the series of rigid transformations.

It is important to note that congruent angles have equal measures, but their orientations or positions may differ. Rigid transformations do not change the measure of angles but can alter their positions or orientations in space.

Hence, the statement that is true in this context is:

The congruence of corresponding angles is preserved under rigid transformations.

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3 percent of n is (C) 300

We need to know about the percent theory first, the percent number is per 100. It can be determined by this formula

A% = A/100

From the text we know that :

0.03%. n = 3           A

3%. n =?                  B

Find the value of n to get 3 percent of n by using A

0.03%. n = 3

0.03/100 . n = 3

0.0003 n = 3

n = 3 / 0.0003

n = 10,000

Substitute n to B

3% . n = 3% . 10000

=3/100 x 10000

=300

Hence, 3 percent of n is 300.

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This situation does not contradict Rolle's Theorem because Rolle's Theorem requires the function to be continuous on a closed interval and differentiable on an open interval, which is not satisfied by f(x) = tan x in the interval (0, π).

To show that f(0) = f(π), we evaluate the tangent function at these points. At x = 0, tan(0) = 0, and at x = π, tan(π) = 0. Therefore, f(0) = f(π).

To investigate whether there exists a number c in the interval (0, π) such that f'(c) = 0, we need to find the derivative of f(x). The derivative of tan x is given by f'(x) = sec² x. However, the secant squared function is never equal to zero. Therefore, there is no c in the interval (0, π) where f'(c) = 0.

This situation does not contradict Rolle's Theorem because Rolle's Theorem requires certain conditions to be met. First, the function must be continuous on the closed interval [a, b], which is not satisfied by f(x) = tan x since it is not defined at x = π/2. Second, the function must be differentiable on the open interval (a, b), but f'(x) = sec^2 x is not defined at x = π/2. Thus, the requirements of Rolle's Theorem are not fulfilled, and its conclusion does not apply to f(x) = tan x in the interval (0, π).

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So to find any last digit of 3^2011 divide 2011 by 4 which comes to have 3 as remainder. Hence the number in units place is same as digit in units place of number 3^3. Hence answer is 7.

Answer:

Ari's is less than Davids

Ari=18.75

David=24.00

Step-by-step explanation:

Answer:

See below ~

Step-by-step explanation:

Given

Solving

Solution

Answer:

4 gigabytes

Step-by-step explanation:

$55-$35 =$20

Under $55 = less than (<) $55

According to given question we have,

$35 + $5 × g < 55

35 + 5g < 55

compare like terms and simplify we get,

5g < 55 - 35

5g < 20

g < 20/5

g < 4 gigabytes

Inequality: 5x + 35 <50

Answer: x < 3

I'll start first by finding the slope of the line at t = 2 mins up to t = 6 mins, t = 6 mins up to t = 8 mins, and t = 14 mins up to t = 20 mins.

For the first interval (2 mins to 6 mins), we have the coordinates (2, 7) and (6, 5). The slope of the line is

For the second interval (6 mins up to 8 mins), we have the coordinates (6, 5) and (8,0). The slope of the line is

For the last interval (14 mins up to 20 mins), we have the coordinates (14,0) and (20,9). The slope of the line is

The x-axis of the given graph pertains to time while its y-axis pertains to distance from home. Let's try to make a story about a person from work going home and will prepare something before going outside again.

For the first 2 mins, the person walks out of his office and will go to his car. Since he is still in the office, the distance from home does not change for the first two minutes.

For the next 4 mins (2 mins to 6 mins interval), he starts driving going home at a rate of 1/2 miles per minute. Because of traffic, he is driving slower than his usual driving speed. Upon passing away from the traffic, the person now travels at a rate of 5/2 miles per minute for 2 mins (6 to 8 mins interval). At the 8th minute mark, he is already home. He prepared something at home during the 8 min to 14 min interval time. After the preparation, he went again outside for some business trip, traveling at the speed of 9/6 miles per minute.

The third step in the data Modeling process when using a packaged data model typically involves customizing and refining the model to align with your specific business requirements.

Here's a breakdown of this step:

1. Identify unique business requirements: Understand the specific needs of your organization or project that are not addressed by the packaged data model's default settings.

2. Map out customizations: Determine which aspects of the packaged data model need to be adjusted or extended to accommodate your unique requirements. This may include adding or modifying entities, attributes, or relationships.

3. Document customizations: Keep a clear record of any changes made to the packaged data model. This will help maintain consistency across different team members and provide a reference point for future updates or modifications.

4. Implement customizations: Update the packaged data model with the required changes, following best practices for data modeling and ensuring the integrity of the overall structure.

5. Validate customizations: Test the updated data model to ensure that it accurately represents your business requirements and functions as expected. This may involve reviewing the model with stakeholders, running test queries, or using data validation tools.

6. Iterate as necessary: If any issues or further requirements are identified during validation, refine and update the data model as needed.

By customizing and refining the packaged data model, you can tailor it to better suit your organization's unique needs, ultimately leading to more accurate and useful insights from your data.

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(a) Purchase of the investment:

1 Available-for-Sale Securities $80,728

2 Cash (Payment for the investment) $80,728

(b) Receipt of annual interest and discount amortization:

1 Cash (Receipt of annual interest) $8,700

2 Discount on Bonds Payable (Amortization) $1,272

3 Investment Revenue (Interest Income) $7,428

(a) Journal entry for the purchase of the investment:

Date: January 1, 2020

Account Titles                  Debit     Credit

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

Available-for-Sale Investments  $80,728

Cash                            $80,728

(b) Journal entry for the receipt of annual interest and discount amortization:

Date: End of the year (Assuming December 31, 2020)

Account Titles               Debit   Credit

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

Cash                          $10,400

Interest Revenue              $7,728

Discount on Bonds Investment  $2,672

The cash received is the annual interest payment calculated as 10% of the face value of the bonds ($87,000 x 10%).

(c) Journal entry for the year-end fair value adjustment:

Date: End of the year (Assuming December 31, 2020)

Account Titles             Debit     Credit

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

Fair Value Adjustment                 $2,150

Unrealized Gain or Loss               $2,150

The fair value adjustment account is credited to increase its balance to reflect the increase in the fair value of the investment.

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The probability that all of the resulting pieces have length at least 2.5 cm is approximately 0.955.

Let's call the length of the first piece "x" and the length of the second piece "y". Since the stick has a total length of 41 cm, we know that

x + y = 41

We want to find the probability that both x and y are at least 2.5 cm. This means that

x >= 2.5 and y >= 2.5

We can rearrange the first equation to solve for one of the variables in terms of the other

y = 41 - x

Substituting this into the second equation, we get

x >= 2.5 and 41 - x >= 2.5

Simplifying the second inequality, we get

x <= 38.5

So the conditions for both x and y are

2.5 <= x <= 38.5 and 2.5 <= y <= 38.5

Now we need to find the probability of these conditions being met. We can visualize the possible values of x and y as a square with side length 38 (since the maximum value of x or y is 38.5). The area of this square is 38×38 = 1444.

The area of the region where both x and y are at least 2.5 can be found by subtracting the areas of the two triangles where x or y is less than 2.5 from the total area of the square

Area of region = 1444 - 2×(2.5×38) = 1378

So the probability that both x and y are at least 2.5 is

Probability = Area of region / Total area = 1378/1444 = 0.955

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The given question is incomplete, the complete question is:

Two points along a straight stick of length 41 cm are randomly selected. the stick is then broken at those two points. find the probability that all of the resulting pieces have length at least 2.5 cm?

According to the question the remainder when the polynomial is divided by (x + 1) is -5.

i. To find the value of (a), we know that (x + 2) is a factor of the polynomial. This means that when we substitute x = -2 into the polynomial, the result should be zero.

Substituting x = -2 into the polynomial:

(-2)^3 + a(-2)^2 - a(-2) - 10 = 0

-8 + 4a + 2a - 10 = 0

6a - 18 = 0

6a = 18

a = 3

Therefore, the value of (a) is 3.

ii. Now that we have the value of (a) as 3, we can find the remainder when the polynomial is divided by (x + 1). To do this, we can use the Remainder Theorem, which states that the remainder when a polynomial f(x) is divided by x - c is equal to f(c).

Substituting x = -1 into the polynomial:

(-1)^3 + 3(-1)^2 - 3(-1) - 10 = -1 + 3 + 3 - 10 = -5

So, the remainder when the polynomial is divided by (x + 1) is -5.

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Answer:

B 9

Step-by-step explanation:

We have 6 green socks, 18 purple socks, and 12 orange socks.

Adding more purple sock means 6 green socks, 18+x purple socks, and 12 orange socks.

We have a  probability of 60% of getting a purple sock.

P( purple) = number of purple socks / total

.60 = (18+x) / (6+18+x+12)

.60 = (18+x) / (36+x)

Multiply each side by 36+x

21.6 +.6x = 18+x

Subtract 18 from each side

3.6x +.6x = x

Subtract .6x from each side

3.6x = .4x

Divide each side by .4

9 =x

Jamal added 9 purple socks

Answer:

$5.00 per ounce

Step-by-step explanation:

18/3.6

5

From the given graph

The x-axis represents the number of hours

The y-axis represents the amount earned

The common solution of the two lines is the point of intersection

Since the point of intersection is (4, 40), then

They will earn the same amount of money $40 in 4 hours

After 4 hours they will the same amount of money

Answer:

b < 12.5

Step-by-step explanation:

2b + 5 < 30

1. Subtract 5 from both sides:

2b < 25

2. Divide both sides by 2:

b < 12.5

Final answer: b < 12.5

The value of b is any value that is less (smaller) than 12.5.

Hope this helps!

C. 61°

Angle ZVY is vertically opposite to angle WVX and all vertically opposite angles are equal to the same number.

Rewrite 13/5 as 26/10

Add the two pieces together:

26/10 + 33/10 = 59/10

Rewrite 7 meters as 70/10

Subtract the two pieces:

70/10 - 59/10 = 11/10 meter left

The range of x are;

1. x < -30

2. x > 8

3. x > 15

4. x < -2

A relationship between two expressions or values that are not equal to each other is called 'inequality.

1. -130 > 50x +20

-130-20> 50x

-150 > 50x

-150/50 > x

-30 > x

x < -30

2. -8(x-3) < -40

-8x +24< -40

collect like terms

-8x < -64

x > -64/-8

x > 8

3. 2x - 22 > 8

collect like terms

2x > 30

divide both sides by 2

x > 30/2

x > 15

4. -35 < -5(x+9)

-35 < -5x -45

collect like terms

10 < -5x

-2 > x

x < -2

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The 90th percentile of x is 70.

What is standard deviation?

A set of values' variance or dispersion is measured by the standard deviation. While a high standard deviation implies that the values are dispersed over a wider range, a low standard deviation shows that the values tend to be close to the mean of the collection.

Given:  Mean = 80, s = 5.

Computing values to standard deviation from mean,

Lower limit = mean - 2s

                  = 80 - 2(5)

                  = 80 - 10

Lower limit = 70

Hence, if x has a normal distribution with mean 80 and standard deviation 5 then the 90th percentile of x is 70.

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No, It is not possible to have a complex number in the form of a + bi when b = 0.

An expression is a way of writing a statement with more than two variables or numbers with operations such as addition, subtraction, multiplication, and division.

Example: 2 + 3x + 4y = 7 is an expression.

We have,

Complex number.

It is in the form of a + ib.

Where a and b are real numbers.

Now,

If b = 0 then we can not have a complex number.

Example:

a + bi

a = 1 and b = 0

1 + 0i = 1 which is not a complex number.

In order to have a complex number a can be any real number and 0 but b  can not be a zero.

Thus,

It is not possible to have a complex number in the form of a + bi when

b = 0.

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The value of x in the triangles in diagram A and B is 9.28 respectively.

Triangle is a plane shape that has three sides.

(A) To solve for x in the triangle in diagram A, we use the formula below

Formula:

Where:

Substitute these values into equation 1 and solve for x

(B) Similarly, to solve for x in the triangle in diagram B, we use the formula below

Formula:

Where:

Substitute these values into equation 2 and solve for x

Hence, the values of x in A and B is 9.28 respectively.

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The probability of there being children in a car involved in an auto accident is 0.3.

In the given scenario, the probability of there being children in a car involved in an auto accident is 0.3. If there are children in the car, the probability of the driver being 55 years or older is 0.1.

However, if there are no children in the car, the probability of the driver being 55 years or older is 0.25. These probabilities provide information about the likelihood of certain events occurring within the context of auto accidents.

It is important to note that these probabilities are based on the assumptions given and may not represent real-world statistics. Probability calculations help us understand the relative likelihood of different outcomes and can be useful in decision-making and risk assessment.

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