What is the area of a rectangle with vertices at (−3, −1) , (1, 3) , (3, 1) , and (−1, −3) ?

Enter your answer in the box. Do not round any side lengths.

Answer: Each face would have an area of 22 * 22 = 484 square inches.

Since a cube has 6 faces, then the total area = 6 * 484 = 2,904 square inches.

Step-by-step explanation:

Answer:

The statement is false.

Step-by-step explanation:

To begin an indirect proof, you assume the converse of what you intend to prove is true. This is false.

Rather the correct answer is that to begin an indirect proof, you assume the inverse of what you intend to prove is true.

The converse can be either true or false, depending on what the original statement is, so assuming the converse is pointless.

Answer: 5/9 will be the fraction of the space will be for storage and baking.

Step-by-step explanation:

Since area of bakery will be used as our storefront, where customers come to visit= 100/225

Thus, the total area of bakery = 225  

And, the area will be used our storefront, where customers come to visit= 100

So, the remaining area which is used for storage and baking = 225 - 100 =125

So the part of area which is used for storage and baking= the remaining area which is used for storage and baking/ total area

=125/225= 5/9

which is the simplest form.

The area of the front cover of the square notebook is A = 400 cm²

A square is a four sided quadrilateral with all of the side having equal lengths. The total internal angle of a square is 360° with each side perpendicular to each other.

A square is a 4 sided polygon where all the sides are of equal length.

The area of the square is the square of any one of its side

Let a be the side of the square

Area of the square = a²

Given data ,

Let the area of the square be represented as A

Now , the side length of the notebook a = 20 cm

And , the area of the front cover of the notebook = area of square

On simplifying , we get

The area of front cover A = a²

The area of the front cover of the notebook A = 20 x 20

The area of the front cover of the notebook A = 400 cm²

Hence , the area of the square notebook is 400 cm²

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In 24 ways the first, second, third, and fourth place be assigned

A permutation is an arrangement of objects in a definite order. .

Given that, for a spelling contest awards four prizes Tim, Sam, Jen, and Kate all made it to the final round.

We need to find in how many ways can first, second, third, and fourth place be assigned.

This can be arranged in 4! ways.

4!=4×3×2×1

4!=24

Hence in 24 ways the first, second, third, and fourth place be assigned.

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The distance covered by the vehicle in 6 hours will be 600 km.

The distance covered by the particle or the body in an hour is called speed. It is a scalar quantity. It is the ratio of distance to time.

We know that the speed formula

Speed = Distance/Time

If you drive at 100km/hr for 6 hours.

Then the distance covered by the vehicle in 6 hours will be given as,

100 = Distance / 6

Distance = 100 x 6

Distance = 600 km

The distance covered by the vehicle in 6 hours will be 600 km.

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Answer:

Option A) EF = 15 units.  

Step-by-step explanation:

We are given the following information in the question:

ΔABC ≅ ΔDEF

AB = 12, BC = 15, and AC = 17

If two triangles are congruent then they have the following properties:

AB = DE

BC = EF

AC = DF

[tex]\angle A = \angle D \\\angle B = \angle E \\\angle C = \angle F[/tex]

Hence, by congruency:

BC = EF

EF = 15 units.

Thus, option A is the correct answer.

Answer: A(11x=55)

Step-by-step explanation:

in the equation 2x + 3y = 55, subsitute 3y for 3x, because y = 3x. this will result in the equation 2x + 3(3x) which equals 11x. so 11x=55

the answer is A x=15 and y =12

Answer: "Use the straightedge to draw a line through points X and Y." is the right answer.

Step-by-step explanation:

To perpendicular bisector of line segment AB. There are following steps:

1) Draw arcs from points A and B on the both sides of AB.

2) Name the intersection points as X and Y.

3) Use the straightedge to draw a line through points X and Y.

4) Name the point as O

hence we have construct perpendicular bisector XY of AB which bisects at O.

The scale factor of the two rectangular prisms is 3/2 of the given problem the volumes of two similar rectangular prisms are 40 mm and 135 mm

To find the scale factor between two similar rectangular prisms, we can compare their volumes. The scale factor is the ratio of the cube root of the volumes.

Let's denote the volume of the first rectangular prism as V1 and the volume of the second rectangular prism as V2.

[tex]V1 = 40 mm^3\\V2 = 135 mm^3[/tex]

The scale factor (SF) can be calculated as follows:

[tex]SF = (V2)^{(1/3)} / (V1)^{(1/3)[/tex]

Plugging in the given values:

[tex]SF = (135)^{(1/3)} / (40)^{(1/3)[/tex]

To simplify the calculation, we can rewrite the values as prime factorization:

[tex]135 = 3^3 * 5\\40 = 2^3 * 5[/tex]

Now, let's calculate the cube roots:

[tex]SF = (3^3 * 5)^{(1/3) }/ (2^3 * 5)^{(1/3)}[/tex]

Using the property of exponents [tex](a^{(m/n) }= (a^m}^{(1/n)})[/tex]:

[tex]SF = (3^{(3/3)} * 5^{(1/3)}) / (2^{(3/3)} * 5^{(1/3)})[/tex]

Simplifying the fractions:

[tex]SF = 3 * 5^{(1/3)} / 2 * 5^{(1/3)[/tex]

The [tex]5^{(1/3)}[/tex] terms cancel out:

SF = 3/2

Therefore, the scale factor of the two rectangular prisms is 3/2 of the given problem the volumes of two similar rectangular prisms are 40 mm and 135 mm.

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Answer:

CPCTC; JP≅MI; SAS

Step-by-step explanation:

For the first question, we use the congruence statement.  When we are using the congruence statement to match corresponding pieces, the justification is that Congruent Parts of Congruent Triangles are Congruent, or CPCTC.

For the second question, we again use CPCTC and the congruence statement.  P matches T and B matches I, so PB = TI and B = I.  J matches M also, so BJP = IMT.  However, P does not match I, so JP does not equal MI.

For the third question, we have two sides (AM matching ZD and BD matching itself) and an included angle (AMD and ZDM).   This is Side-Angle-Side, or SAS.

The formula for the distance Lorrae swims at a constant speed for 2 hours is d = 2s, where d is distance and s is her speed.

The question is asking for a formula to calculate the distance that Lorrae swims. Since distance (d) is equal to speed (s) multiplied by time (t), and we know that Lorrae swims at a constant speed, s, for 2 hours, a formula for her distance would be d = s * t.

In this case, the time, t, is 2 hours. So, we can substitute t with 2 in our formula. Hence, the formula for the distance Lorrae swims is d = s * 2 or d = 2s. This is the equation you can use to compute the distance Lorrae swims for any given speed at a constant period of 2 hours.

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To find the number Wilma is thinking of, we create an equation and solve for x.

To solve this problem, we need to create an equation based on the information given. Let's suppose the number Wilma is thinking of is x. According to the problem, if we add 4 to x and then multiply that result by 3, we get 30. We can write this as the equation 3(x + 4) = 30. To find the value of x, we need to isolate it. First, distribute the 3 to both terms inside the parentheses: 3x + 12 = 30. Next, isolate x by subtracting 12 from both sides of the equation: 3x = 18. Finally, solve for x by dividing both sides of the equation by 3: x = 6. Therefore, Wilma is thinking of the number 6.

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The Interest is $12.595 while the value of the Proceeds is $307.405

The amount of loan; A = $320

The discount rate = 16%=0.16

Time = 90 days = 90/365 year = 0.246 year

The formula of interest is;

Interest = Amount × Discount rate × Time

Thus;

Interest = 320 × 0.16 × 0.246

Interest = $12.595

Proceed = Amount - Interest

Proceed = $320 - $12.595

Proceeds = $307.405

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Answer:

12.80

307.20

Step-by-step explanation:

In modified boxplots, a data value is a potential outlier if it is outside the range of Q3 + (1.5)(IQR) or Q1 - (1.5)(IQR). The IQR is the spread of the middle 50% of the data, calculated as Q3 - Q1.

In modified​ boxplots, a data value is​ a potential outlier if it is above Q3 + (1.5)(IQR) or below Q1 - (1.5)(IQR). The interquartile range (IQR) is crucial in this determination, representing the spread of the middle 50 percent of the data. It is found by subtracting the first quartile (Q1) from the third quartile (Q3). For example, if Q1 is 2 and Q3 is 9, the IQR is 7. Any data value greater than 9 + (1.5)(7) or less than 2 - (1.5)(7) would be considered a potential outlier.

Constructing a box plot involves identifying these five key values: minimum value, first quartile, median, third quartile, and maximum value. The box plot visually represents the data's spread, with whiskers extending to the most extreme values within the acceptable range, and separate marks for potential outliers.

Answer:

20

Step-by-step explanation:

:D

The value of the x is 20.

We have given that,

[tex]81^5 = 3^x[/tex]

We have to determine the value of x.

3^4 = 81

Exponentiation is a mathematical operation, written as bⁿ, involving two numbers, the base b, and the exponent or power n, and pronounced as b raised to the power of n.

exponent 4 x 5 = 20

so, 3^20 = 81^5

Therefore the value of the x is 20.

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