I need help on this asap!!!

If the weekly salary is reduced by 9.5% for CCSS and the salary is 50000, then they would pay $45250.

The percent reduction in the weekly salary is = 9.5% = 0.095,

So, the amount of the reduction is;

⇒ reduction = 9.5% of 50,000

⇒ reduction = 0.095 x 50,000

⇒ reduction = 4750

So, the reduction in the weekly salary is 4750,

Now, we subtract the reduction from the original salary to get the new salary after the reduction,

⇒ new salary = 50000 - 4750

⇒ new salary = 45250

Therefore, the new salary after the 9.5% reduction for CCSS is $45250.

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A solution to the given system of linear inequalities is (-6, -1).

In order to to graph the solution to the given system of inequalities on a coordinate plane, we would use an online graphing calculator to plot the given system of inequalities and then take note of the point of intersection;

2x + 7y ≤ -14     .....equation 1.

-3x + 5y > 5    .....equation 2.

Based on the graph (see attachment), we can logically deduce that the solution to the given system of inequalities is the shaded region below the solid and dashed line, and the point of intersection of the lines on the graph representing each, which is given by the ordered pair (-6, -1).

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The values of the variables x, y, and z obtained by using Cramer's rule are 83/105, -277/105, and -99/105, respectively.

To find the values of the variables x, y, and z using Cramer's rule, we need to find the determinant of the coefficient matrix and the determinants of the matrices obtained by replacing the first, second, and third columns of the coefficient matrix with the constant terms.
The coefficient matrix is:
|−3 0 −6|
| 8 −2 3|
| 2 −1 −4|
The determinant of the coefficient matrix is:
|−3 0 −6| = (−3)(−2)(−4) − (0)(3)(2) − (−6)(8)(−1) − (−3)(−1)(3) − (0)(−4)(8) − (−6)(−2)(2)
= −24 − 0 − 48 − 9 − 0 − 24
= −105
The matrix obtained by replacing the first column with the constant terms is:
|11 0 −6|
|17 −2 3|
| 3 −1 −4|
The determinant of this matrix is:
|11 0 −6| = (11)(−2)(−4) − (0)(3)(3) − (−6)(17)(−1) − (11)(−1)(3) − (0)(−4)(17) − (−6)(−2)(3)
= 88 − 0 − 102 − 33 − 0 − 36
= −83
The matrix obtained by replacing the second column with the constant terms is:
|−3 11 −6|
| 8 17 3|
| 2  3 −4|
The determinant of this matrix is:
|−3 11 −6| = (−3)(17)(−4) − (11)(3)(2) − (−6)(8)(3) − (−3)(3)(3) − (11)(−4)(2) − (−6)(17)(2)
= 204 − 66 − 144 − 9 + 88 + 204
= 277
The matrix obtained by replacing the third column with the constant terms is:
|−3 0 11|
| 8 −2 17|
| 2 −1  3|
The determinant of this matrix is:
|−3 0 11| = (−3)(−2)(3) − (0)(17)(2) − (11)(8)(−1) − (−3)(−1)(17) − (0)(3)(8) − (11)(−2)(2)
= 18 − 0 + 88 − 51 − 0 + 44
= 99
Now we can use Cramer's rule to find the values of the variables x, y, and z:
x = (−83)/(-105) = 83/105
y = (277)/(-105) = -277/105
z = (99)/(-105) = -99/105
83/105, -277/105, and -99/105 are the values of x,y and z respectively.

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1. The positive x-intercept of the function f(x)= -1201(x-88)^2 + 151058 is 88 feet.

2. The y-intercept of the function f(x) = -1201(x-88)^2 + 151058 is 151058 feet.

3. The maximum height that the football reaches is 151058 feet, and the horizontal distance from Patrick Mahomes when this happens is 88 feet.

4. Yes, Travis Kelce will be able to catch the football, since it reaches a height of 10 feet or lower at the horizontal distance of 160 feet from Patrick Mahomes.

5. The horizontal distance of the football from Patrick Mahomes when it first reaches a height of 40 feet is 176 feet.

1. The positive x-intercept represents the horizontal distance from Patrick Mahomes at which the football is initially released.

2. The y-intercept of the function represents the height of the football when it is released from Patrick Mahomes.

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The Zeros of the equation would be x = -5/2.

An equation is an expression that shows the relationship between two or more numbers and variables.

A mathematical equation is a statement with two equal sides and an equal sign in between. An equation is, for instance, 4 + 6 = 10. Both 4 + 6 and 10 can be seen on the left and right sides of the equal sign, respectively.

We are given the equation as;

[tex]f(x)=-8x^3-20x^2[/tex]

We can factor;

4x^2 ( 2x+5)  

Using the zero product property

2x = 0

2x + 5 = 0

x = -5/2

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AMSAA growth model

The MTTF at the conclusion of the test cycle is 232.59 hours. 22423.54 more hours of test time will be necessary to achieve and MTTF of 3000 hours. 52.15 items must now be placed on test to obtain another 10 failures with 5,000 hours of test time available. If the testing continues for only 600 more hours, 1.75 is the expected number of failures.

A) To fit the AMSAA growth model, we first need to calculate the cumulative number of failures (C) and the logarithm of the test time (lnT).

| Test Time (T) | C | lnT |
|---------------|---|-----|
| 28            | 1 | 3.33|
| 146           | 2 | 4.98|
| 258           | 3 | 5.55|
| 426           | 4 | 6.05|
| 521           | 5 | 6.26|
| 1027          | 6 | 6.93|
| 1273          | 7 | 7.15|

Next, we can use linear regression to estimate the parameters of the AMSAA model, β and η:

C = βlnT - η

Using linear regression, we get β = 1.11 and η = 3.82.

To estimate the MTTF at the conclusion of the test cycle, we can use the formula:

MTTF = (T/C)^(1/β)

Plugging in the values for T (1273), C (7), and β (1.11), we get:

MTTF = (1273/7)^(1/1.11) = 232.59 hours

B) To achieve an MTTF of 3000 hours, we need to solve for T in the equation: 3000 = (T/C)^(1/β)

Plugging in the values for C (7), β (1.11), and rearranging the equation, we get:

T = 3000^(β) * C = 3000^(1.11) * 7 = 23696.54 hours

Since we have already tested for 1273 hours, we need an additional 23696.54 - 1273 = 22423.54 hours of test time to achieve an MTTF of 3000 hours.

C) To obtain another 10 failures with 5000 hours of test time available, we can use the formula:

C = βlnT - η

Plugging in the values for C (10), β (1.11), η (3.82), and T (5000), and rearranging the equation, we get:

10 = 1.11ln(5000) - 3.82

ln(5000) = (10 + 3.82)/1.11 = 12.47

5000 = e^(12.47) = 260753.13

Therefore, we need to place 260753.13/5000 = 52.15 items on test to obtain another 10 failures with 5000 hours of test time available.

D) If the testing in (c) continues for only 600 more hours, we can use the formula:

C = βlnT - η

Plugging in the values for β (1.11), η (3.82), and T (600), and rearranging the equation, we get:

C = 1.11ln(600) - 3.82 = 1.75

Therefore, the expected number of failures in 600 more hours of testing is 1.75.

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There is a significant relationship between the variables (H-value = 2 and P-value = 0.994)

The test for significant relationship is used to determine whether there is a statistically significant relationship between two or more variables. In this case, the test can be used to determine whether there is a statistically significant relationship between the number of daily passenger trips (in thousands) for subways and commuter rail service. Based on the H-value and P-value, the result of this test indicates that there is a significant relationship between the variables (H-value = 2 and P-value = 0.994). Therefore, option (a) is correct and option (d) is incorrect.

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Keegan's error is that he did not provide enough information to fully determine the result of the given composition of transformations, namely (R90•D2)(Triangle ABC).

To be able to graph the image of Triangle ABC under the composition of R90 (a 90-degree counterclockwise rotation) and D2 (a dilation with center the origin and scale factor 2), we need to know the center of rotation for the rotation R90.

The reason for this is that the composition of a rotation and a dilation is not commutative, meaning that the order of the transformations matters. Specifically, the result of the composition depends on whether the dilation is applied before or after the rotation. If the dilation is applied before the rotation, then the center of dilation becomes the center of rotation for the rotation. If the rotation is applied before the dilation, then the center of dilation is not affected by the rotation.

Therefore, without knowing the center of rotation for the rotation R90, we cannot determine the exact result of the composition (R90•D2)(Triangle ABC), and thus we cannot graph it accurately.

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Answer:

Let's assume that the amount of saving is x.

According to the problem, the ratio of expenditure to saving is 4:1, so the amount of expenditure can be expressed as 4x.

The total income of the family is Rs. 500000, and it can be expressed as the sum of expenditure and saving:

Expenditure + Saving = 500000

Substituting the values of expenditure and saving, we get:

4x + x = 500000

Simplifying this equation, we get:

5x = 500000

Dividing both sides by 5, we get:

x = 100000

Therefore, the amount of saving is Rs. 100000, and the amount of expenditure is 4 times this value, which is Rs. 400000.

The rate of Bill's boat in still water is 33 mph and the speed of the current is 3 mph.

To find the rate of Bill's boat and the speed of the current, we can use the distance formula, which states that distance = rate × time. Since we know the distance and time for both the upstream and downstream trips, we can set up two equations and solve for the rate of the boat and the speed of the current.

Let x = the rate of the boat in still water and y = the rate of the current.

For the upstream trip:
90 = (x - y) × 3

For the downstream trip:
90 = (x + y) × 2.5

Simplifying the equations gives us:
90 = 3x - 3y
90 = 2.5x + 2.5y

Multiplying the first equation by 2.5 and the second equation by 3 gives us:
225 = 7.5x - 7.5y
270 = 7.5x + 7.5y

Adding the two equations together eliminates the y variable:
495 = 15x

Solving for x gives us:
x = 33

Substituting x back into the first equation gives us:
90 = (33 - y) × 3
90 = 99 - 3y
3y = 9
y = 3

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Answer:

it is 2,475

Step-by-step explanation:

Answer: 21 degrees

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The general solution of a fifth order linear ODE with one real root r and two complex roots a ± bi and c ± di can be determined by using the fact that the general solution of a linear ODE is the sum of the solutions corresponding to each of the roots.

For the real root r, the solution is given by:
y1 = C1 * e^(r*x)

For the complex roots a ± bi, the solution is given by:
y2 = C2 * e^(a*x) * cos(b*x) + C3 * e^(a*x) * sin(b*x)

For the complex roots c ± di, the solution is given by:
y3 = C4 * e^(c*x) * cos(d*x) + C5 * e^(c*x) * sin(d*x)

The general solution of the fifth order linear ODE is the sum of these solutions:
y = y1 + y2 + y3
= C1 * e^(r*x) + C2 * e^(a*x) * cos(b*x) + C3 * e^(a*x) * sin(b*x) + C4 * e^(c*x) * cos(d*x) + C5 * e^(c*x) * sin(d*x)

This is the general solution of the fifth order linear ODE with one real root r and two complex roots a ± bi and c ± di.

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Answer:12km

Step-by-step explanation:

She ran 2km on Monday, and three times as much on Tuesday so we do 2 x 3 which equals 6. So 6km, plus the original 2km is 8km (6+2) is she wants to run 20km this week, then she needs to subtract what shes already ran to find out how many more km she needs to run. 20-8=12 so 12km.

Answer: I think putting Prop of ║lines might work. I haven't done this in a long time, so I'm most likely wrong, but that might help. I'm so sorry.

40 adult tickets were sold and 80 student tickets were sold.

What is an equation?

An equation is a mathematical statement that asserts that two expressions are equal. It typically consists of variables, constants, and mathematical operations.

Let's use algebra to solve this problem.

Let's call the number of student tickets sold "s" and the number of adult tickets sold "a".

From the problem, we know two things:

The total number of tickets sold was 120:

s + a = 120

The total amount of money made from ticket sales was $1,400:

10s + 15a = 1400

Now we have two equations with two variables, so we can solve for "a" and "s".

Let's start by solving for "s" in the first equation:

s + a = 120

s = 120 - a

Now we can substitute this expression for "s" into the second equation:

10s + 15a = 1400

10(120 - a) + 15a = 1400

1200 - 10a + 15a = 1400

5a = 200

a = 40

So 40 adult tickets were sold.

Now we can substitute this value of "a" into the first equation to find "s":

s + a = 120

s + 40 = 120

s = 80

So 80 student tickets were sold.

Therefore, 40 adult tickets were sold and 80 student tickets were sold.

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\( \operatorname{det}\left(2 A^{-1} B^{2}\right) = \frac{18}{12} = \frac{3}{2} \)Explanation: We are given that \( \operatorname{det} A=12 \) and \( \operatorname{det} B=3 \). We need to find the determinant of \( 2 A^{-1} B^{2} \). We can use the properties of determinants to simplify the expression. Recall that \( \operatorname{det}(cA) = c^n \operatorname{det}(A) \) for an \( n \times n \) matrix \( A \) and a scalar \( c \), and that \( \operatorname{det}(AB) = \operatorname{det}(A)\operatorname{det}(B) \). Using these properties, we can write:\( \operatorname{det}\left(2 A^{-1} B^{2}\right) = \operatorname{det}(2) \operatorname{det}(A^{-1}) \operatorname{det}(B^{2}) \)\( = 2^3 \operatorname{det}(A^{-1}) \operatorname{det}(B)^2 \)\( = 8 \cdot \frac{1}{\operatorname{det}(A)} \cdot (\operatorname{det}(B))^2 \)\( = 8 \cdot \frac{1}{12} \cdot (3)^2 \)\( = \frac{18}{12} = \frac{3}{2} \)Therefore, \( \operatorname{det}\left(2 A^{-1} B^{2}\right) = \frac{3}{2} \).

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To solve the given equations, we need to follow these steps:

1. Isolate the radical on one side of the equation.
2. Square both sides of the equation to eliminate the radical.
3. Solve the resulting equation for the variable.
4. Check for extraneous solutions by plugging the solution back into the original equation.
5. If there are rational exponents, use the same steps but raise both sides of the equation to the reciprocal of the exponent.

Let's apply these steps to the sample problems:
1. V4x + 1 - Vx+ 10 = 2
First, isolate the radical on one side:
V4x + 1 = Vx+ 10 + 2
V4x + 1 = Vx+ 12
Next, square both sides to eliminate the radical:
(4x + 1) = (x+ 12)^2
4x + 1 = x^2 + 24x + 144
Solve for x:
0 = x^2 + 20x + 143
Using the quadratic formula:
x = (-20 ± √(20^2 - 4(1)(143)))/(2(1))
x = (-20 ± √(400 - 572))/2
x = (-20 ± √(-172))/2
x = (-20 ± √172i)/2
There are no real solutions for this equation.

2. V2x+30 = (x+3)
Isolate the radical:
V2x+30 = x+3
Square both sides:
2x + 30 = (x+3)^2
2x + 30 = x^2 + 6x + 9
Solve for x:
0 = x^2 + 4x - 21

Using the quadratic formula:
x = (-4 ± √(4^2 - 4(1)(-21)))/(2(1))
x = (-4 ± √(16 + 84))/2
x = (-4 ± √100)/2
x = (-4 ± 10)/2
x = 3 or x = -7
Check for extraneous solutions by plugging the solutions back into the original equation:
V2(3)+30 = (3+3)
V36 = 6
6 = 6 (True)
V2(-7)+30 = (-7+3)
V16 = -4
4 = -4 (False)

Therefore, the only solution is x = 3.

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1) A  perpendicular bisector in kite ABCD is; BD

2) An isosceles triangle in kite ABCD is; ΔABC

3) A right triangle in kite ABCD is; ΔABM

In geometry, a kite is defined as a quadrilateral with reflection symmetry across a diagonal.

here, we have,

1) A perpendicular bisector is defined as a line segment which bisects another line segment at 90 degrees.

Looking at the diagram, Line BD bisects Line AC and as such BD is the perpendicular bisector.

2) An Isosceles triangle is defined as a a triangle in which two sides have the same length.

In this case, in Triangle ABC, AB and BC have the same length and as such Triangle ABC is the Isosceles Triangle.

3) A right triangle is defined as a triangle, that has one of its interior angles equal to 90 degrees or any one angle is a right angle.

In this case if Mis the intersection of BD and AC, then the right angle triangle is ABM

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The area of the trapezoid in simplest radical form is 78 feet².

Given a trapezoid.

Length of the bases are 10 feet and 16 feet.

We have to find the height.

Consider the smaller right triangle formed by the height of the trapezoid.

Triangle base length or one leg = 16 - 10 = 6 feet

Since one of the angle is 45°, the other angle in the right triangle is also 45°.

Since it is isosceles, opposite sides for 45° angles are same.

Other leg = 6 feet, which is the height.

Area of the trapezoid = 1/2 (a + b)h, where a and b are bases and h is the height.

A = 1/2 (10 + 16) 6

   = 78 feet²

Hence the correct option is b.

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Answer:

1278in^3

Step-by-step explanation:

21×8=168

168×6=1008in^3

5×9=45

45×6=270in^3

1008+270=1278in^3

Answer:

Step-by-step explanation:

The least common denominator for 1/6 + 3/8 is 24

first find the multiples of 6 and 8

6= 6, 12, 18, 24, 30

8 = 8, 16, 24, 32

Next circle the number that appear in both

That number will be 24 which is the least common denominator.

The percentage of yellow paint that has to be used is given as 0.45

The amount of seafoam paint is given as 1.5 quartz

The percentage of yellow paint in the seafoam paint is 30 percent

Hence the amount that would be in it that would be made of yellow paint is given as

1.5 x 30%

1.5 x 0.30

= 0.45

Hence we would conclude by saying that the amount of yellow paint that has to be used in order to make the paint should be 0.45

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Curt and Melanie are mixing blue and yellow paint to make seafoam green paint. 1.5 quarts of seafoam paint is made of 70% blue, 30% yellow. Use the percent equation to find how much yellow paint they should use.

Answer:

see explanation

Step-by-step explanation:

(a)

• the opposite sides of a rectangle are congruent

then

4x + 1 = 2x + 12

(b)

solving

4x + 1 = 2x + 12 ( subtract 2x from both sides )

2x + 1 = 12 ( subtract 1 from both sides )

2x = 11 ( divide both sides by 2 )

x = 5.5

(c)

the perimeter (P) is the sum of the 4 sides of the rectangle , that is

P = 4x + 1 + x + 2x + 12 + x ( collect like terms )

  = 8x + 13 ( substitute x = 5.5 )

  = 8(5.5) + 13

  = 44 + 13

  = 57

Answer:

a) Because two opposite sides of the rectangle have the same sizes

B)

[tex]x = \frac{11}{2} [/tex]

c)

[tex]57[/tex]

Step-by-step explanation:

b)

[tex]4x + 1 = 2x + 12 \\ 4x - 2x = 12 - 1 \\ 2x = 11 \\ \frac{2x}{2} = \frac{11}{2} \\ x = \frac{11}{2} [/tex]

c) Perimeter

[tex]p = 2(x + y) \\ 2( \frac{11}{2} + 23) \\ 2( \frac{57}{2} ) \\ 57[/tex]

23 comes from

[tex]2x + 12 \\ x = \frac{11}{2} \\ 2( \frac{11}{2} ) + 12 \\ 11 + 12 \\ \\ 23[/tex]

The real zeroes of the function f(x)=x^(6)+8x^(5)+9x^(4)-2x^(2)-7x+4 are -4, -1, and 1.

The real zeroes of the function f(x)=x^(6)+8x^(5)+9x^(4)-2x^(2)-7x+4 can be found by using the Rational Zero Theorem and synthetic division.

The Rational Zero Theorem states that the possible rational zeroes of a polynomial are the factors of the constant term divided by the factors of the leading coefficient. In this case, the constant term is 4 and the leading coefficient is 1, so the possible rational zeroes are ±1, ±2, and ±4.

We can use synthetic division to test these possible zeroes and find the actual zeroes of the function. Synthetic division is a method of dividing a polynomial by a linear factor of the form x-a. The result of synthetic division is a quotient and a remainder, and if the remainder is 0, then x-a is a factor of the polynomial and a is a zero of the function.

Using synthetic division, we find that the real zeroes of the function are -4, -1, and 1. These are the values of x that make the function equal to 0. There are no repeated zeroes in this case.

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To determine the value(s) of a such that [1 a], [a a+2] are linearly independent, we need to find the values of a that make the determinant of the matrix non-zero. The determinant of a 2x2 matrix is given by:

|1 a|
|a a+2| = (1)(a+2) - (a)(a) = a + 2 - a^2

To make the determinant non-zero, we need to solve the equation:

a + 2 - a^2 ≠ 0

Rearranging the equation, we get:

a^2 - a - 2 ≠ 0

Factoring the equation, we get:

(a - 2)(a + 1) ≠ 0

Therefore, the values of a that make the determinant non-zero are a ≠ 2 and a ≠ -1. These are the values of a that make the vectors [1 a], [a a+2] linearly independent.

So, the value(s) of a such that [1 a], [a a+2] are linearly independent are a ≠ 2 and a ≠ -1.

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The age of fluffy is 30 in dog years. The solution has been obtained by using linear equation.

One is the degree of the linear equation. The absence of variables in linear equations with exponents greater than one is obvious. The graph's equation results in a straight line.

We are given that combined age of 3 dogs in dog years is 82.

Let Fluffy's age be 'f'.

Age of Spot = (f - 14)

Age of Shampy = (f + 6)

From this, we get

f + (f - 14) + (f + 6) = 82

On solving this, we get

⇒f + f - 14 + f + 6 = 82

⇒3f - 8 = 82

⇒3f = 90

⇒f = 30

Hence, the age of fluffy is 30 in dog years.

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As per the concept of integer, the best proof of her conjecture is Let 2n + 1 be an odd number; (2n + 1) + (2n + 1) = 4n + 2. Because 4(n + 2) is divisible by 2, the sum of 2n + 1 and itself is even. (option b)

Now, let's look at the given choices to find the best proof for Gemma's conjecture.

Choice B) Let 2n + 1 be an odd number; (2n + 1) + (2n + 1) = 4n + 2. Because 4(n + 2) is divisible by 2, the sum of 2n + 1 and itself is even.

This choice provides a proof by using algebraic equations. It begins by defining an odd integer as 2n + 1, where n is any integer. Then, it uses algebraic manipulation to show that the sum of this odd integer and itself results in an even integer. Specifically, (2n + 1) + (2n + 1) simplifies to 4n + 2, which is equal to 2(2n + 1). This proves that the sum of an odd integer and itself is always even.

This choice is a tautology, which means that it's always true, but it doesn't provide any evidence or proof to support Gemma's conjecture.

The best proof for Gemma's conjecture is choice B.

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Complete Question:

Gemma makes a conjecture that the sum of an odd integer and itself is always an even integer. Which choice is the best proof of her conjecture?

A) Look at these different examples: 7 + 7 = 14, 13 + 13 = 26, 23 + 23 = 46. So the sum of an odd number and itself must be even

B) Let 2n + 1 be an odd number; (2n + 1) + (2n + 1) = 4n + 2. Because 4(n + 2) is divisible by 2, the sum of 2n + 1 and itself is even.

C) Let n represent an odd number, and let n + n be an even number. Therefore, n + n = n + n, which shows that the sum of an odd number and itself is even.

D) Every time you add an odd number and itself, the sum is an even number.

The distance between any point on the parabola and the directrix is equal to the distance between that point and the focus. This property is what defines a parabola.

To draw a parabola given the directrix and focus, follow these steps:

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