Here is Triangle A. Lin created a scaled copy of Triangle A with an area of 72 square units. a. How many times larger is the area of the scaled copy compared to that of Triangle A? b. What scale factor did Lin apply to the Triangle A to create the copy?

Answer:

x≤−9

Step-by-step explanation:

x+5≤−4

Subtract 5 from each side

x+5-5≤−4-5

x≤−9

The height of the rectangle is

⇒ Height = 1/2 ft

A rectangle is a 2 - dimension figure with 4 sides, 4 corners and 4 right angles. And, Opposite sides of the rectangle are equal and parallel to each other.

We have to given that;

Length of the rectangle = 5/3 ft

Area of the rectangle = 5/6 ft

We know that;

Area of rectangle = Length x Height

Hence, We get;

5/6 ft = 5/3 ft x Height

5/6 x 3/5 = Height

Height = 1/2 ft

Thus, The height of the rectangle is;

⇒ Height = 1/2 ft

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Answer:

$3623.85

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Equality Properties

Algebra I

Compounded Interest Formula: \(A=P(1 + \frac{r}{n} )^{nt}\)

Step-by-step explanation:

Step 1: Identify Variables

P = 3000

r = 0.065

n = 1

t = 3

Step 2: Find Final Amount A

Answer:

cp=(SP * 100 ) / ( 100 – percentage loss)

6786*100/100-13

7800

Answer: 2

Step-by-step explanation:

theres a chance i could be right

Answer:

2(y-3)=7+2y-3

Step-by-step explanation:

2y-6=7+2y-3

-6=7-3

-6=4

Answer:

I think B

Step-by-step explanation:

Answer:

C

Step-by-step explanation:

If you follow rise/run

The rise over run is positive 2/3

If you are dealing with a data set that consists of sales with a consistent percentage increase, a multiplicative regression model would be the best method to use.

Multiplicative models are useful when the relationship between variables is not linear, but rather exponential or logarithmic. In the case of sales with a consistent percentage increase, the relationship between sales and time would be multiplicative, because each successive time period's sales would be a multiple of the previous period's sales. Multiplicative models are also useful for dealing with trends that increase or decrease at an accelerating rate, rather than at a consistent rate. In contrast, a simple moving average or a random walk additive regression model would not be appropriate for a data set with a consistent percentage increase, as they assume linear relationships between variables and would not account for the exponential or logarithmic relationship between sales and time.

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Answer : 1.5 mi

To calculate the distance closer to the shortest distance, we will be apply the pythagora's theorem

Answer:

x = -2

x = 3

The two x's resemble to roots of the parabola.

Step-by-step explanation:

In the following, I uploaded an image with the steps to the equation.

Answer: A) 87

The median value is the second quartile (Q₂), which is 87.

Answer:

A: 87

Step-by-step explanation:

Sorry for this very late response. But if anyone is not sure if answer A is correct, it is. I can confirm this because I just took the test. I hope I could help! (:

The equation of the tangent line to the circle x^2 + y^2 = 37 at the point (6,1) is y = -6x + 37.

The equation of the tangent line to the circle x^2 + y^2 = 37 at the point (6,1) can be found by first finding the slope of the tangent line at that point and then using point-slope form to write the equation of the tangent line.

To find the slope of the tangent line at (6,1), we need to take the derivative of the circle equation with respect to x:

2x + 2y dy/dx = 0

Solving for dy/dx, we get:

dy/dx = -x/y

At the point (6,1), the slope of the tangent line is:

dy/dx = -(6)/(1) = -6

Now we can use point-slope form to write the equation of the tangent line:

y - y1 = m(x - x1)

where (x1,y1) is the point (6,1) and m is the slope we just found.

Substituting the values, we get:

y - 1 = -6(x - 6)

Simplifying the equation, we get:

y = -6x + 37

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42.57 x 46630= 75769 =3794874 =37489438790=3474936

The null hypothesis can be rejected at the 0.05 or 5 % level of significance according to Decision Rule.

The null hypothesis is a type of hypothesis that explains the population parameter and is used to examine if the provided experimental results are reliable. Depending on whether the population or sample under consideration is viable, this hypothesis is either rejected or not. Or to put it another way, the null hypothesis is a hypothesis that assumes that the sample observations are the product of chance. It is claimed to be a claim made by surveyors who wish to look at the data. The symbol for it is H0.

Given :  n = 8          

\(\large \bar{X}=105.20 \\\\ \larg\frac{\sigma}{\sqrt{n}}=1.77 \\ \\ \large \alpha=0.05 \large \mu_0=100\)

a )  We want to find the 95% confidence interval for mean

Therefore ,

\(\large (105.20-Z_{0.05}*1.77,105.20+Z_{0.05}*1.77)\\\\\large (105.20-1.64*1.77,105.20+1.64*1.77)\\\\\large (105.20-2.9028,105.20+2.9028)\)

(102.2972,108.1028)

b )  Hypothesis :  

\(\large H_0:\mu=100 \\ \\ \large H_1:\mu\neq 100\)

The test statistic under H is given by ,

\(\large Z\rightarrow N(0,1)\\\\ \large Z =\frac{105.20-100}{1.77}\\\\ \large =\frac{5.20}{1.77}\\\\\large =2.9379\)

\(P value \large =P(Z > |Z_{cal}|)\)

=P(Z>2.9379)

=0.001652

Decision Rule :  If P value \(< \large \alpha\)  then reject  at  \(\large \alpha\) % level of significance accept otherwise

Here , P value = 0.001652 < \large \alpha = 0.05

Therefore , reject H at 5% level of significance.

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Weighing using a tared container is not appropriate due to the potential for errors and inaccuracies. This method can impact the results by introducing uncertainties in the measurements.

Using a tared container involves placing the substance to be weighed on a container that has already been weighed and then subtracting the weight of the container to obtain the weight of the substance alone. While this method is commonly used for qualitative analysis or when the accuracy requirements are not strict, it is not suitable for quantitative preparation where precise measurements are essential.

The use of a tared container introduces several potential sources of error. First, the accuracy of the tare weight might not be exact, leading to uncertainties in subsequent measurements. Additionally, the tare weight may change over time due to factors like evaporation or contamination, further affecting the accuracy of subsequent measurements. Moreover, the process of transferring the substance to the tared container introduces the risk of loss or gain of material, leading to errors in the final measurements.

Overall, relying on weighing with a tared container for quantitative preparation can result in inaccurate quantities of the substance being weighed, compromising the reliability and reproducibility of experimental results. Therefore, more precise weighing techniques, such as using calibrated weighing balances or analytical techniques, should be employed for quantitative preparations.

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The integers s and t that satisfy the equation 330s + 156t = gcd(330, 156), are s = 9 and t = -19.

To find integers s and t such that 330s + 156t = gcd(330, 156), we can use the Extended Euclidean Algorithm. This algorithm allows us to find the greatest common divisor (gcd) of two numbers and express it as a linear combination of the two numbers.

Step 1: Apply the Euclidean Algorithm.

Dividing 330 by 156, we get:

330 = 2 * 156 + 18

Dividing 156 by 18, we get:

156 = 8 * 18 + 12

Dividing 18 by 12, we get:

18 = 1 * 12 + 6

Dividing 12 by 6, we get:

12 = 2 * 6 + 0

Step 2: Backward substitution.

Starting with the last equation:

6 = 18 - 1 * 12

Substituting 12 with the previous equation:

6 = 18 - 1 * (156 - 8 * 18) = -1 * 156 + 9 * 18

Substituting 18 with the previous equation:

6 = -1 * 156 + 9 * (330 - 2 * 156) = 9 * 330 - 19 * 156

Therefore, s = 9 and t = -19. So, 330s + 156t = gcd(330, 156) becomes:

330 * 9 + 156 * (-19) = gcd(330, 156)

The integers s and t that satisfy the equation are s = 9 and t = -19.

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The requried area of the board is 110 square feet, option D is correct.

From the figure,

The length of the board is 11 feet
The width of the board is 10 feet

The area of the board is given as:

Area = length x width
Area = 11 × 10
Area = 110 square feet.

Thus, the requried area of the board is 110 square feet, and option D is correct.

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Answer: That would be a vertical pair and X is 38

Step-by-step explanation:

Answer: A

Step-by-step explanation: I would say A because the angle is greater than 90 degrees

Answer:

We have supplementary angles.

76 + 3x + 2 = 180

3x + 78 = 180

3x = 102

x = 34

No, a fraction is not a term. The distributive property can be applied to fractions because it is a general mathematical principle.

A fraction is not considered a term in the traditional sense. It is a mathematical expression that represents division. However, the distributive property can still be applied to fractions because the property itself is a fundamental rule of arithmetic that extends beyond specific types of expressions.

The distributive property states that for any real numbers a, b, and c:

a × (b + c) = (a × b) + (a × c).

When working with fractions, we can apply the distributive property as follows:

Let's consider the expression: a × (b/c).

We can rewrite this as: (a × b)/c.

Now, let's distribute the 'a' to 'b' and 'c':

(a × b)/c = (a/c) × b.

In this step, we applied the distributive property to the fraction (a/c) by treating it as a whole.

Although fractions represent division, we can still use the distributive property because it is a general mathematical principle that allows for manipulating expressions involving addition, subtraction, multiplication, and division.

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Answer:

1st blank = Shorter

2nd blank = shorter

3rd blank = will not

Step-by-step explanation:

Base of the box

a² + b² = c²

Lets input the numbers

32² + 8² = c²

1024 + 64 = c²

1088 = c²

Sq rt c = 32.9

32.9<34 so it will not fit.

Interior of the box

32² + 6² = c²

1024 + 36 = c²

1060 = c²

Sq root = 32.6

If my answer is incorrect, pls correct me!

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-Chetan K

The correct answer is the minimal cover of F is:F = {AB→C, AB→S, AB→E, ES→S}

Given, relation schema R(ABC,DE) with functional dependencies F:AB→SCDE→ABAB→SEES→C.

The candidate key of a relation R is a minimal set of attributes that can uniquely identify each tuple in R. To find the candidate key, we can compute the closure of each subset of attributes in R. Let's begin by finding the closure of AB, the subset of R. We can use the functional dependency AB→SC to compute the closure of AB as follows: AB+ = ABSC (by AB→SC)

Next, let's compute the closure of DE, the other subset of R. We can use the functional dependency DE→AB to compute the closure of DE as follows: DE+ = DEABSE (by DE→AB and AB→SE)

Now we have two potential candidate keys, AB and DEABSE. However, we need to check if they are minimal. To do this, we check if any subset of these candidate keys can also function as a candidate key. We can see that neither AB nor DEABSE has any proper subset that can function as a candidate key.

Therefore, both AB and DEABSE are candidate keys for the relation R. Next, let's compute the minimal cover of F. The minimal cover of a set of functional dependencies F is a minimal set of functional dependencies that is equivalent to F.

To find the minimal cover of F, we can use the following steps:

Step 1: Eliminate redundant dependencies. We can eliminate the dependency AB→SC since it is implied by AB→C.

Step 2: Decompose dependencies. We can decompose the dependency AB→SE into two dependencies, AB→S and AB→E, since S and E are not a subset of any other attribute set in F.

Step 3: Eliminate extraneous attributes. We can eliminate the attribute C from the dependency ES→C since C is already determined by AB→C.

Therefore, the minimal cover of F is:F = {AB→C, AB→S, AB→E, ES→S}

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The variables on the horizontal and vertical axes of a scatter plot are always continuous.Corresponding: When two sets of data are plotted on a scatter plot, each data point in the two sets has a corresponding point in the other set.

When you plot two sets of data on a scatter plot, you should use different colors or symbols for the data points from each set.The horizontal axis represents one variable while the vertical axis represents the other. Each point on the graph represents one set of data that corresponds to both variables, with the x-value and y-value corresponding to the respective data points being plotted.Scatterplot versus: A scatter plot has two axes that correspond to two variables. In a scatter plot, we can determine how one variable changes as the other variable changes. The variables on the horizontal and vertical axes of a scatter plot are always continuous.Corresponding: When two sets of data are plotted on a scatter plot, each data point in the two sets has a corresponding point in the other set. When you plot two sets of data on a scatter plot, you should use different colors or symbols for the data points from each set.

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The vectors are:

To find the vectors:

Use the graph to solve the problem:

We can add and subtract the vectors:

-To find b - a subtract a from b:

2a means multiply vector a by 2:

Now let's find the resulting vector 2a - b.

- Subtract b from 2a:

The resulting vector 2a - b is <8, -9>

Therefore, the vectors are:

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The complete question is given below:

The graph shows vectors a and b the resulting vector for b - a is blank, blank >, and the resulting vector for 2a -b is < blank, blank >.

Answer:

AB = √(4^2 + 1^2) = √(16 + 1) = √17

GH = √(4^2 + 4^2) = √(16 + 16) = √32

= 4√2

The inequality to show the values of [w] that will not exceed the weight limit of the elevator is w + 1643 ≤ 2000. On solving the inequality, we get w ≤ 357. The graph of the inequality is attached.

        ≤ : less than or equal to

        ≥ : greater than or equal to

        < : less than

        > : greater than

        ≠ : not equal to

Given is the maximum load for a certain elevator is 2000 pounds. The total weight of the passengers on the elevator is 1400 pounds. A delivery man who weighs 243 pounds enters the elevator with a crate of weight [w].

        1400 + 243 + w ≤ 2000

        w + 1643 ≤ 2000

        w + 1643 ≤ 2000

        w ≤ 2000 - 1643

        w ≤ 357

Therefore, the inequality to show the values of [w] that will not exceed the weight limit of the elevator is w + 1643 ≤ 2000. On solving the inequality, we get w ≤ 357. The graph of the inequality is attached.

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- det(2A) = 8
- det(-A) = 1/2
- det(A^2) = 1/4
- det(A^-1) = 2

To find the determinants of the given matrices, follow these steps:

1. det(2A): Multiply the original determinant by the scalar raised to the power of the matrix size.
det(2A) = 2^n * det(A) = 2^4 * (1/2) = 16 * (1/2) = 8.

2. det(-A): Multiply the original determinant by the scalar raised to the power of the matrix size.
det(-A) = (-1)^n * det(A) = (-1)^4 * (1/2) = 1 * (1/2) = 1/2.

3. det(A^2): Raise the original determinant to the power of 2.
det(A^2) = (det(A))^2 = (1/2)^2 = 1/4.

4. det(A^-1): Invert the original determinant.
det(A^-1) = 1/det(A) = 1/(1/2) = 2.

In summary:
- det(2A) = 8
- det(-A) = 1/2
- det(A^2) = 1/4
- det(A^-1) = 2

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