# 4.08 quiz: factoring difference of squares question 1 (20 points) factor: x^2 - 36 a.) (x - 6)(x + 6)

###### Question:

question 1 (20 points)

factor: x^2 - 36

a.) (x - 6)(x + 6)

b.) (x - 6)^2

c.) (x - 36)(x + 36)

d.) (x + 36)^2

question 2 (10 points)

factor: 4x^4 - 64

a.) (4x - 8)(4x + 8)

b.) (2x^2 -8)2

c.) (2x^2 - 32)(2x^2 + 32)

d.) (2x^2 + 8)(2x^2 - 8)

## Answers

you from k12? me too if so :)

i already took this and C and A are the answers

11. Ans: (D)

Since all the vertices and the foci lie along the y axis, therefore, we would need the following equation for vertical hyperbola:

[tex]\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1[/tex]

Since (h,k) = (0,0)

Therefore, the above equation becomes,

[tex]\frac{(y)^2}{a^2} - \frac{(x)^2}{b^2} = 1[/tex]

Now the distance between the vertices is:

2a = 12

=> a = 6

And the distance between the foci is:

2c = 18

=> c = 9

Since,

[tex]c^2 = a^2 + b^2[/tex]

=> [tex]b^2 = 45[/tex]

Hence, the equation becomes,

[tex]\frac{(y)^2}{36} - \frac{(x)^2}{45} = 1[/tex] (Option D:y squared over 36 minus x squared over 45 = 1)

12. Ans: (B)

The hyperbola's standard form is(as it is a vertical):

[tex]\frac{y^2}{16} - \frac{x^2}{b^2} = 1[/tex] -- (X)

=> [tex]y^2 = ( \frac{16}{b^2})*(b^2 + x^2)[/tex]

=> y = ± [tex]( \frac{4}{b} ).x[/tex] --- (A)

Since asymptotes at y = ± [tex]( \frac{1}{4} ).x[/tex]. --- (B)

Compare (A) and (B), you would get,

[tex]\frac{4}{b} = \frac{1}{4}[/tex]

=> b=16

The equation (X) would become:

[tex]\frac{y^2}{16} - \frac{x^2}{256} = 1[/tex] (Option-B)

13. Ans: (A) [tex]y = x^{2} + 6x + 14[/tex]

Equations given:

x = t - 3 --- (equation-1)

y = [tex]t^{2}[/tex] + 5 --- (equation-2)

From equation-1,

t = x + 3

Put the value of t in (equation-2),

[tex]y = (x+3)^{2} + 5[/tex]

[tex]y = x^2 + 9 + 6x + 5[/tex]

[tex]y = x^2 + 6x + 14[/tex]

Hence, the correct option is (A)

14. Ans: (A)

The polar coordinates given: [tex](3, \frac{2 \pi }{3} )[/tex] = (r, θ)

Since,

x = r*cosθ,

y = r*sinθ

Plug-in the values of r, and θ in the above equations:

x = (3) * cos(120°); since [tex]\frac{2 \pi }{3}[/tex] = 120°

=> x = [tex]- \frac{3}{2}[/tex]

y = (3) * sin(120°);

=> y = [tex]\frac{3 \sqrt{3} }{2}[/tex]

Ans: (x,y) = [tex](- \frac{3}{2} ,\frac{3 \sqrt{3} }{2})[/tex] (Option A)

15. Ans: (D)

The general forms of finding all the polar coordinates are:

1) When r >= 0(meaning positive): (r, θ + 2n [tex]\pi[/tex]) where, n = integer

2) When r < 0(meaning negative): (-r, θ + (2n+1) [tex]\pi[/tex]) where, n = integer

Since r is not mentioned in the question, but in options every r slot has the value r=1, therefore, I would take r = +1, -1(plus minus 1)

θ(given) = [tex]\frac{- \pi }{6}[/tex]

When r = +1(r>0):

(1, [tex]\frac{- \pi }{6}[/tex] + 2n[tex]\pi[/tex])

When r = -1(r<0):

(-1, [tex]\frac{- \pi }{6}[/tex] + (2n+1)[tex]\pi[/tex])

Therefore, the correct option is (D): (1, negative pi divided by 6 + 2nπ) or (-1, negative pi divided by 6 + (2n + 1)π)

16. Ans: (B)

In polar coordinates,

[tex]r = \sqrt{x^{2} + y^{2}}[/tex]

Since x = 4, y=4; therefore,

[tex]r = \sqrt{16 + 16} = 4 \sqrt{2}[/tex]

To find the angle,

tanθ = y/x = 4/4 = 1

=> θ = 45° (when [tex]r =4 \sqrt{2}[/tex])

If r = -[tex]r =4 \sqrt{2}[/tex], then,

θ = 45° + 180° = 225°

Therefore, the correct option is (B) (4 square root 2 , 45°), (-4 square root 2 , 225°)

17. Ans: (B)

(Question-17 missing Image is attached below) The general form of the limacon curve is:

r = b + a cosθ

If b < a, the curve would have inner loop. As you can see in the image attached(labeled Question-17), the limacon curve graph has the inner loop. Therefore, the correct option is (B) r = 2 + 3 cosθ, since b = 2, and a = 3; and the condition b < a (2 < 3) is met.

18. Ans: (C)

Let's find out!

1. If we replace θ with -θ, we would get:

r = -2 + 3*cos(-θ )

Since, cos(-θ) = +cosθ, therefore,

r = -2 + 3*cos(θ)

Same as the original, therefore, graph is symmetric to x-axis.

2. If we replace r with -r, we would get:

-r = -2 + 3*cos(θ )

r = 2 - 3*cos(θ)

NOT same as original, therefore, graph is NOT symmetric to its origin.

3. If we replace θ with -θ and r with -r, we would get:

-r = -2 + 3*cos(-θ )

Since, cos(-θ) = +cosθ, therefore,

r = 2 - 3*cos(3θ)

NOT same as original, therefore, graph is NOT symmetric to y-axis.

Ans: The graph is symmetric to: x-axis only!

19. (Image is attached below) As the question suggests that it is a horizontal ellipse, therefore, the equation for the horizontal ellipse is:

[tex]\frac{x^{2}}{a^{2}} + \frac{y_{2}}{b_{2}} = 1[/tex] -- (A)

Since, x = 8f,

y = 18ft,

b = 54ft,

[tex]a^{2}[/tex] = ?

Plug-in the values in equation (A),

(A)=> [tex]\frac{64}{a^{2}} + \frac{324}{2916} = 1[/tex]

=> [tex]a^{2}[/tex] = 72

Therefore, the equation becomes,

Ans: [tex]\frac{x^{2}}{72} + \frac{y_{2}}{2916} = 1[/tex]

20. Ans: x-axis only

Let's find out!

1. If we replace θ with -θ, we would get:

r = 2*cos(-3θ )

Since, cos(-θ) = +cosθ, therefore,

r = +2*cos(3θ) = Same as original

Therefore, graph is symmetric to x-axis.

2. If we replace r with -r, we would get:

-r = 2*cos(3θ )

r = -2*cos(3θ) = Not same

3. If we replace θ with -θ and r with -r, we would get:

-r = 2*cos(-3θ )

Since, cos(-θ) = +cosθ, therefore,

r = -2*cos(3θ) = Not Same

Ans: The graph is symmetric to: x-axis only!

[tex]8.08, part 2 11. find an equation in standard form for the hyperbola with vertices at (0, ±6) and fo[/tex]

[tex]8.08, part 2 11. find an equation in standard form for the hyperbola with vertices at (0, ±6) and fo[/tex]

11. Ans: (D)

Since all the vertices and the foci lie along the y axis, therefore, we would need the following equation for vertical hyperbola:

[tex]\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1[/tex]

Since (h,k) = (0,0)

Therefore, the above equation becomes,

[tex]\frac{(y)^2}{a^2} - \frac{(x)^2}{b^2} = 1[/tex]

Now the distance between the vertices is:

2a = 12

=> a = 6

And the distance between the foci is:

2c = 18

=> c = 9

Since,

[tex]c^2 = a^2 + b^2[/tex]

=> [tex]b^2 = 45[/tex]

Hence, the equation becomes,

[tex]\frac{(y)^2}{36} - \frac{(x)^2}{45} = 1[/tex] (Option D:y squared over 36 minus x squared over 45 = 1)

12. Ans: (B)

The hyperbola's standard form is(as it is a vertical):

[tex]\frac{y^2}{16} - \frac{x^2}{b^2} = 1[/tex] -- (X)

=> [tex]y^2 = ( \frac{16}{b^2})*(b^2 + x^2)[/tex]

=> y = ± [tex]( \frac{4}{b} ).x[/tex] --- (A)

Since asymptotes at y = ± [tex]( \frac{1}{4} ).x[/tex]. --- (B)

Compare (A) and (B), you would get,

[tex]\frac{4}{b} = \frac{1}{4}[/tex]

=> b=16

The equation (X) would become:

[tex]\frac{y^2}{16} - \frac{x^2}{256} = 1[/tex] (Option-B)

13. Ans: (A) [tex]y = x^{2} + 6x + 14[/tex]

Equations given:

x = t - 3 --- (equation-1)

y = [tex]t^{2}[/tex] + 5 --- (equation-2)

From equation-1,

t = x + 3

Put the value of t in (equation-2),

[tex]y = (x+3)^{2} + 5[/tex]

[tex]y = x^2 + 9 + 6x + 5[/tex]

[tex]y = x^2 + 6x + 14[/tex]

Hence, the correct option is (A)

14. Ans: (A)

The polar coordinates given: [tex](3, \frac{2 \pi }{3} )[/tex] = (r, θ)

Since,

x = r*cosθ,

y = r*sinθ

Plug-in the values of r, and θ in the above equations:

x = (3) * cos(120°); since [tex]\frac{2 \pi }{3}[/tex] = 120°

=> x = [tex]- \frac{3}{2}[/tex]

y = (3) * sin(120°);

=> y = [tex]\frac{3 \sqrt{3} }{2}[/tex]

Ans: (x,y) = [tex](- \frac{3}{2} ,\frac{3 \sqrt{3} }{2})[/tex] (Option A)

15. Ans: (D)

The general forms of finding all the polar coordinates are:

1) When r >= 0(meaning positive): (r, θ + 2n [tex]\pi[/tex]) where, n = integer

2) When r < 0(meaning negative): (-r, θ + (2n+1) [tex]\pi[/tex]) where, n = integer

Since r is not mentioned in the question, but in options every r slot has the value r=1, therefore, I would take r = +1, -1(plus minus 1)

θ(given) = [tex]\frac{- \pi }{6}[/tex]

When r = +1(r>0):

(1, [tex]\frac{- \pi }{6}[/tex] + 2n[tex]\pi[/tex])

When r = -1(r<0):

(-1, [tex]\frac{- \pi }{6}[/tex] + (2n+1)[tex]\pi[/tex])

Therefore, the correct option is (D): (1, negative pi divided by 6 + 2nπ) or (-1, negative pi divided by 6 + (2n + 1)π)

16. Ans: (B)

In polar coordinates,

[tex]r = \sqrt{x^{2} + y^{2}}[/tex]

Since x = 4, y=4; therefore,

[tex]r = \sqrt{16 + 16} = 4 \sqrt{2}[/tex]

To find the angle,

tanθ = y/x = 4/4 = 1

=> θ = 45° (when [tex]r =4 \sqrt{2}[/tex])

If r = -[tex]r =4 \sqrt{2}[/tex], then,

θ = 45° + 180° = 225°

Therefore, the correct option is (B) (4 square root 2 , 45°), (-4 square root 2 , 225°)

17. Ans: (B)

(Question-17 missing Image is attached below) The general form of the limacon curve is:

r = b + a cosθ

If b < a, the curve would have inner loop. As you can see in the image attached(labeled Question-17), the limacon curve graph has the inner loop. Therefore, the correct option is (B) r = 2 + 3 cosθ, since b = 2, and a = 3; and the condition b < a (2 < 3) is met.

18. Ans: (C)

Let's find out!

1. If we replace θ with -θ, we would get:

r = -2 + 3*cos(-θ )

Since, cos(-θ) = +cosθ, therefore,

r = -2 + 3*cos(θ)

Same as the original, therefore, graph is symmetric to x-axis.

2. If we replace r with -r, we would get:

-r = -2 + 3*cos(θ )

r = 2 - 3*cos(θ)

NOT same as original, therefore, graph is NOT symmetric to its origin.

3. If we replace θ with -θ and r with -r, we would get:

-r = -2 + 3*cos(-θ )

Since, cos(-θ) = +cosθ, therefore,

r = 2 - 3*cos(3θ)

NOT same as original, therefore, graph is NOT symmetric to y-axis.

Ans: The graph is symmetric to: x-axis only!

19. (Image is attached below) As the question suggests that it is a horizontal ellipse, therefore, the equation for the horizontal ellipse is:

[tex]\frac{x^{2}}{a^{2}} + \frac{y_{2}}{b_{2}} = 1[/tex] -- (A)

Since, x = 8f,

y = 18ft,

b = 54ft,

[tex]a^{2}[/tex] = ?

Plug-in the values in equation (A),

(A)=> [tex]\frac{64}{a^{2}} + \frac{324}{2916} = 1[/tex]

=> [tex]a^{2}[/tex] = 72

Therefore, the equation becomes,

Ans: [tex]\frac{x^{2}}{72} + \frac{y_{2}}{2916} = 1[/tex]

20. Ans: x-axis only

Let's find out!

1. If we replace θ with -θ, we would get:

r = 2*cos(-3θ )

Since, cos(-θ) = +cosθ, therefore,

r = +2*cos(3θ) = Same as original

Therefore, graph is symmetric to x-axis.

2. If we replace r with -r, we would get:

-r = 2*cos(3θ )

r = -2*cos(3θ) = Not same

3. If we replace θ with -θ and r with -r, we would get:

-r = 2*cos(-3θ )

Since, cos(-θ) = +cosθ, therefore,

r = -2*cos(3θ) = Not Same

Ans: The graph is symmetric to: x-axis only!

[tex]8.08, part 2 11. find an equation in standard form for the hyperbola with vertices at (0, ±6) and fo[/tex]

[tex]8.08, part 2 11. find an equation in standard form for the hyperbola with vertices at (0, ±6) and fo[/tex]

11. The technique to answering this problem quickly is by examining the choices. We know that the denominator for y and x should be equal to the square of the foci (from the fact that [tex]c^{2}=a^{2}+b^{2}[/tex]). This would mean that the denominators should add up to 81. From the choices, only two satisfy this and those are options A and D. We know it should be D since the smaller denominator always has to be in the first term.

ANSWER: D. y squared over 36 minus x squared over 45 = 1.

12. You can also quickly identify the equation of the hyperbola given the vertices and the asymptotes. The square root of the denominator of the FIRST TERM in the equation is the numerator of the asymptote while the square root of the denominator of the SECOND TERM is the denominator of the asymptote. HOWEVER, we have to consider that the vertices are at (0,4) and (0,-4) so the asymptote must be in the lowest term. Considering the vertices, we can arrive at the asymptote [tex]y=+- \frac{4}{16}x[/tex]. This means that the first term will have a denominator of 16 while the second term will have a denominator of 256. This equation is option B.

ANSWER: B. y squared over 16 minus x squared over 256 = 1.

13. To eliminate the parameter, t, we just need to equate both equations such that t is equal.

[tex]t=x+3[/tex]

[tex]t=\sqrt{y-5}[/tex]

[tex]x+3=\sqrt{y-5}[/tex]

[tex]x^{2}+6x+9=y-5[/tex]

[tex]y=x^{2}+6x+14[/tex]

ANSWER: A. [tex]y=x^{2}+6x+14[/tex]

14. To find the x coordinate we just need to multiply r and cos θ while for the y coordinate we would need to multiply r and sin θ.

[tex]x=(3)[cos(\frac{2\pi}{3})]=-\frac{3}{2}[/tex]

[tex]y=(3)[sin(\frac{2\pi}{3})]=\frac{3\sqrt{3}}{2}[/tex]

ANSWER: A. ordered pair negative 3 divided by 2 comma 3 square root 3 divided by 2.

15. The same polar coordinate as point P will be arrived if: (1) a full rotation of 2nπ is performed or (2) a rotation of (2n+1)π is performed and r is negated. Among the choices, we can see exactly one option with coordinates that follow these two rules, and it is choice D.

ANSWER: D. (1, negative pi divided by 6 + 2nπ) or (-1, negative pi divided by 6 + (2n+1)π)

16. To convert the coordinate (4, 4) to polar, we just let x be rcosθ and y be rsinθ. We can get r by solving [tex]r^{2}=x^{2}+y^{2}[/tex] while we can get θ by solving for [tex]\theta=arctan(\frac{y}{x})[/tex].

[tex]r^{2}=(4)^{2}+(4)^{2}[/tex]

[tex]r=4sqrt(2)[/tex]

[tex]\theta=arctan(\frac{4}{4})=45degrees[/tex]

Therefore, one pair of polar coordinates would be (4 square root of 2, 45 degrees) and another one would be (-4 square root of 2, 225 degrees) [note the rule we stated in number 15].

ANSWER: B. (4 square root of 2, 45 degrees), (-4 square root of 2, 225 degrees)

17. Based on your definition that the graph is "circular" with an inner loop on the left, we can only deduce that the limacon with the form r = a + bcosθ has a value of b that is greater than a. Looking at the choices, we only have one option following this criteria, thus we can be sure that it is the correct answer.

ANSWER: B. r = 2 + 3cos θ

18. To test for symmetry about the x-axis, we replace the variables r and θ with r and -θ respectively or -r and π-θ. The equation will be symmetric if it will be unchanged (i.e. the same points will still satisfy the new equation).

[tex]r=-2+3cos\theta[/tex]

[tex]r=-2+3cos(-\theta)[/tex]

[tex]-r=-2+3cos(\pi-\theta)[/tex]

If you examine closely, the same set of points will satisfy the equation above. Therefore, the equation is symmetric about the x axis.

For symmetry about the y-axis, we replace r and θ with -r and -θ.

[tex]r=-2+3cos\theta[/tex]

[tex]-r=-2+3cos(-\theta)[/tex]

Unfortunately, the equation was changed upon substitution therefore we know that it is not symmetric about the y-axis.

For symmetry about the origin, we just replace r and θ with -r and θ respectively.

[tex]r=-2+3cos\theta[/tex]

[tex]-r=-2+3cos\theta[/tex]

These two equations are not also similar so no symmetry about the origin is exhibited.

ANSWER: C. x-axis only

19. Let's consider the general form of the ellipse: [tex]\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1[/tex].

From the problem we know that a is equal to 54 since we are given the fact that the height of the tunnel is 54ft. To find b, we use the fact that the point (8,18) is a point on the ellipse as stated in the problem.

[tex]\frac{x^{2}}{b^{2}}+\frac{y^{2}}{54^{2}}=1[/tex]

[tex]\frac{8^{2}}{b^{2}}+\frac{18^{2}}{54^{2}}=1[/tex]

[tex]\frac{8^{2}}{b^{2}}+\frac{1}{9}=1[/tex]

[tex]64=(\frac{8}{9})b^{2}[/tex]

[tex]b^{2}=72[/tex]

ANSWER: [tex]\frac{x^{2}}{72}+\frac{y^{2}}{2916}=1[/tex]

20. For this item, we follow similar rules stated in item #18. To test for symmetry we need to examine if the equation will remain unchanged after performing substitutions.

[tex]r=2cos3\theta[/tex]

x-axis:

[tex]r=2cos(-3\theta)[/tex]

[tex]-r=2cos[3(\pi-\theta)][/tex]

y-axis:

[tex]-r=2cos(-3\theta)[/tex]

origin:

[tex]-r=2cos3\theta[/tex]

If you examine the equations, you'll see that it is only symmetric at the x-axis.

ANSWER: The equation is symmetric about the x-axis.

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5 sayo ehemmm

11. Equation of hyperbola having vertices at (0, ±6) and foci at (0, ±9).

is given by [tex]\frac{y^2}{a^2}- \frac{x^2}{b^2}=1[/tex]

also b²= c²-a²

b²=81-36

b²=45

So, Equation becomes , [tex]\frac{y^2}{36}- \frac{x^2}{45}=1[/tex]

Option (D) is correct.

12.Vertices (0, ± 4), Asymptotes = ±1/4.x

equation of asymptote is given by, [tex]x = \pm y \frac{b}{a}[/tex]

[tex]\frac{4}{b}=\frac{1}{4}[/tex]

So a=4 and b=16,

So , equation becomes [tex]\frac{y^2}{16}- \frac{x^2}{256}=1[/tex]

Option (B) is correct.

13. x= t-3, and y = t²+ 5

Replace t by x+3, we get

y= (x+3)² +5

y=x²+ 6x +14

Option (A) is correct.

14. Polar coordinates is given by (r,∅)

Polar coordinates of point is (3, 2π/3)

So, r =3, ∅ =2π/3

x= r cos∅ and y = r sin∅

x=3 Cos (2π/3) and y= 3 Sin (2π/3)

x= -3/2 and y=3√3/2

Option (A) is correct.

14. Polar coordinate is given by (r,∅)

Here , r=1, and ∅ = -π/6

x= r Cos ∅ and y =r Sin∅

x= 1 Cos (-π/6) and y= 1 Sin (-π/6)

x =√3/2 and y =1/2

Option (B) which is B) (1, negative pi divided by 6 + 2nπ) or (-1, negative pi divided by 6 + 2nπ) is correct.

16. Two pairs of polar coordinates for the point (4, 4) with 0° ≤ θ < 360°.

is option (B) which is B) (4 square root 2 , 45°), (-4 square root 2 , 225°).

17. A circular graph with inner loop on the left of a limacon curve is given by

r = a + b Cos∅.

In this case a> b.

So , Option (D) r = 4 + cos θ as well as (A) r = 3 + 2 cos θ looks correct.

18. Equation of limacon curve is given by r = -2 + 3 cos θ, here , a<b

So it is symmetric about y-axis only.Option (B) is correct.

Explanation:

I CAN'T UNDERSTAND YOUR QUESTION. SORRY

Periodic table large-ar.svg

الجدول الدوري ترتيب مجدول للعناصر الكيميائية، مرتبة حسب عددها الذري، والتوزيع الإلكتروني، والخواص الكيميائية المتكررة، والذي يُظهر هيكله اتجاهات دورية. بشكل عام، تكون العناصر في الصف واحد (الدورة) فلزات باتجاه اليسار، ولا فلزات باتجاه اليمين، بحيث توضع العناصر التي لها سلوكيات كيميائية مماثلة في نفس العمود. تسمى صفوف الجدول عادةً بالدورات وتسمى الأعمدة بالمجموعات. وتمتلك ستة مجموعات أسماء بالإضافة إلى الأرقام المخصصة: على سبيل المثال، عناصر المجموعة 17 هي الهالوجينات؛ والمجموعة 18 هي الغازات النبيلة. كما أنه يُعرض في شكل أربع مناطق مستطيلة بسيطة أو مستويات فرعية مرتبطة بملء المدارات الذرية المختلفة.

يمكن استخدام تنظيم الجدول الدوري لاشتقاق العلاقات بين خواص العناصر المختلفة، وأيضًا الخصائص والسلوكيات الكيميائية المتوقعة للعناصر غير المكتشفة أو المركَّبة حديثًا. كان الكيميائي الروسي ديمتري مندلييف أول من نشر جدولًا دوريًا معروفًا في عام 1869، وقد تم تطويره بشكل أساسي لتوضيح الاتجاهات الدورية للعناصر المعروفة آنذاك. كما توقع بعض خصائص العناصر غير المحددة التي كان من المتوقع أن تملأ الفجوات داخل الجدول. ثبتت صحة معظم توقعاته. وقد تم توسيع فكرة مندلييف ببطء وصقلها مع اكتشاف أو توليف عناصر جديدة أخرى وتطوير نماذج نظرية جديدة لشرح السلوك الكيميائي. يوفر الجدول الدوري الحديث الآن إطارًا مفيدًا لتحليل التفاعلات الكيميائية، ولا يزال يستخدم على نطاق واسع في الكيمياء، والفيزياء النووية، والعلوم الأخرى.

تم اكتشاف أو تركيب جميع العناصر من العدد الذري 1 (هيدروجين) إلى 118 (أوغانيسون)، واستكمال الصفوف السبعة الأولى من الجدول الدوري.[1][2] توجد العناصر الـ 98 الأولى في الطبيعة، على الرغم من أن بعضها موجود فقط بكميات شحيحة وأن البعض الآخر تم تصنيعه في المختبرات قبل أن يتم العثور عليه في الطبيعة.[n 1] تم تركيب العناصر 99 إلى 118 فقط في المختبرات أو المفاعلات النووية.[3] ويجري حاليًا متابعة تجميع العناصر التي تحتوي على أعداد ذرية أعلى: تبدأ هذه العناصر في الصف الثامن، وقد اقترح العمل النظري مرشحين محتملين لهذا التمديد. كما أُنتجت العديد من النويدات المشعة الاصطناعية من العناصر الطبيعية في المختبرات.

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