What is the surface area of the cone? Use 3.14 for π and round to the nearest hundredth.

About 5.23 square centimeters

About 18.84 square centimeters


About 75.36 square centimeters


About 37.68 square centimeters

We have shown that h₁ and h₂ are normalized and orthogonal, which means they form a valid set of hybrid orbitals appropriate for sp³ hybridization.

To show that two of the set of four equivalent orbitals appropriate for sp³ hybridization, h₁ = ½(-s + px + py + pz) and h₂ = ½(-s - px - py + pz), are normalized and orthogonal, we need to compute their inner product and confirm that it equals zero.

The normalization condition for an orbital is:

[tex]\int |\psi|^2 d\tau = 1[/tex]

where Ψ is the wavefunction and dτ is the volume element. For a normalized orbital, the integral of the square of the wavefunction over all space is equal to 1.

To compute the inner product of h₁ and h₂, we need to integrate their product over all space:

[tex]\int h_1\times h_2 d\tau = \dfrac{1}{4}\int (s^2 + p_x^2 + p_y^2 + p_z^2) - (sp_x + sp_z + sp_y + p_xp_z + p_xp_y + p_yp_z) d\tau[/tex]

We can evaluate this integral by using the orthogonality of the atomic orbitals, which means that the cross terms involving different atomic orbitals vanish, leaving only the diagonal terms. Therefore, the integral simplifies to:

[tex]\int h_1\times h_2 d\tau = \dfrac{1}{4}\int (s^2 + p_x^2 + p_y^2 + p_z^2) - (sp_x + sp_z + sp_y + p_xp_z + p_xp_y + p_yp_z) d\tau = 0[/tex]

This shows that h₁ and h₂ are orthogonal.

To check that h₁ and h₂ are normalized, we need to evaluate the integral of their square over all space:

[tex]\int h_1\times h_1 d\tau = \dfrac{1}{4}\int (s^2 + p_x^2 + p_y^2 + p_z^2) - (sp_x + sp_z + sp_y + p_xp_z + p_xp_y + p_yp_z) d\tau = 1[/tex]

[tex]\int h_2\times h_2 d\tau = \dfrac{1}{4}\int (s^2 + p_x^2 + p_y^2 + p_z^2) - (sp_x + sp_z + sp_y + p_xp_z + p_xp_y + p_yp_z) d\tau = 1[/tex]

These integrals are equal to 1, which shows that h₁ and h₂ are normalized.

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The greatest possible value of 1/x when the hundredths digit of x is 8 is 12.5.

To find the greatest possible value of 1/x when the hundredths digit of x is 8, we need to find the smallest possible value of x.

Since the hundredths digit is 8, we can express x as a decimal as follows:

x = k + 0.08,

where k is an integer representing the remaining digits of x.

To maximize 1/x, we need to minimize x.

Therefore, we need to find the smallest possible value of k. Since k is an integer, the smallest possible value of k is 0. Thus, x = 0.08.

Now, we can find 1/x:

1/x = 1/(0.08) = 12.5.

Therefore, the greatest possible value of 1/x when the hundredths digit of x is 8 is 12.5.

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The minimum probability of error achievable for a signal set with white Gaussian noise, denoted as "n0/2," is an important concept in digital communication systems. To determine this probability, we can use the concept of "minimum Euclidean distance" between the signal points in the signal space. The larger the minimum Euclidean distance, the better the system can differentiate between the signals, and the lower the probability of error.

In the presence of white Gaussian noise, the minimum probability of error is determined using the following formula:

Pe = Q(sqrt(2 * Eb / n0))

Here, Pe represents the probability of error, Q() is the Q-function (a well-known function in statistics representing the tail probability of a Gaussian distribution), Eb is the energy per bit of the signal, and n0 is the noise spectral density.

This formula shows that the minimum probability of error is a function of the signal-to-noise ratio (Eb / n0). As the ratio increases, the probability of error decreases, indicating better signal detection and communication performance. To improve the system's performance, one can either increase the energy per bit (Eb) or decrease the noise spectral density (n0).

In summary, the minimum probability of error achievable with a signal set in the presence of white Gaussian noise (n0/2) can be determined using the Q-function and the signal-to-noise ratio (Eb / n0). By increasing the signal-to-noise ratio, the system can achieve a lower probability of error, resulting in better communication performance.

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Answer:

Below.

Step-by-step explanation:

We multiply each term in the parentheses by the term outside.

36x - 27y - 6

12x - 18y + 2

4y - 4/5 x - 13/5

-8x + 4

8x + 6y -28

The Maclaurin series for given function f(x) = x cos(8x) is f(x) = x - 8x^2 - 85.333x^3 + ...

The Maclaurin series expansion for a function f(x) can be obtained by finding the derivatives of f(x) at x=0 and evaluating them at x=0. The general formula for the Maclaurin series is:

f(x) = f(0) + f'(0)x + f''(0)x^2/2! + f'''(0)x^3/3! + ...

To find the Maclaurin series for f(x) = x cos(8x), we need to take the derivatives of f(x) and evaluate them at x=0.

f(x) = x cos(8x)

f'(x) = cos(8x) - 8x sin(8x)

f''(x) = -16 sin(8x) - 8x cos(8x)

f'''(x) = -128 cos(8x) + 24x sin(8x)

Evaluating at x=0, we get:

f(0) = 0

f'(0) = 1

f''(0) = -16

f'''(0) = -128

Plugging these into the Maclaurin series formula, we get:

f(x) = 0 + 1x - 16x^2/2! - 128x^3/3! + ...

f(x) = x - 8x^2 - 85.333x^3 + ...

Therefore, the Maclaurin series for f(x) = x cos(8x) is:

f(x) = x - 8x^2 - 85.333x^3 + ...

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Complete question is:

Use a Maclaurin series to obtain the Maclaurin series for the given function

f(x) = x cos(8x)

The Lagrange polynomial basis L = {l0(t), l1(t),..., ln(t)} is a set of polynomials with specific properties: li(t_i) = 1 for all i = 0, 1,..., n, and li(t_j) = 0 for all i, j = 0, 1,..., n, where i ≠ j. This basis allows you to represent any polynomial p(t) in Pr by a linear combination of the Lagrange polynomials in the basis L.

Polynomials are mathematical expressions consisting of variables and coefficients, often used in algebra and calculus. In this context, we are considering polynomials of degree n or less, denoted by Pr.

Lagrange polynomials are a type of polynomial that form a basis for Pr. This means that any polynomial in Pr can be expressed as a linear combination of the Lagrange polynomials. The Lagrange polynomial basis is formed from a set of n + 1 distinct points in R, denoted {to, t1, ..., tn}.

Each Lagrange polynomial, denoted li(t), has the following properties:
- li(ti) = 1 for all i = 0, 1,..., n
- li(tj) = 0 for all i, j = 0, 1,..., n, i + j.

In other words, each Lagrange polynomial is equal to 1 at its corresponding point ti, and is equal to 0 at all other points. These properties ensure that the Lagrange polynomials are linearly independent and form a basis for Pr.

Hope this helps! Let me know if you have any further questions.
Hi! I'd be happy to help you with your question. Polynomials are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. The Lagrange polynomial is a specific type of polynomial that provides a way to interpolate a polynomial function using a set of n + 1 distinct points in R, {t0, t1, ..., tn}.

The Lagrange polynomial basis L = {l0(t), l1(t),..., ln(t)} is a set of polynomials with specific properties: li(t_i) = 1 for all i = 0, 1,..., n, and li(t_j) = 0 for all i, j = 0, 1,..., n, where i ≠ j. This basis allows you to represent any polynomial p(t) in Pr by a linear combination of the Lagrange polynomials in the basis L.

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True. In the context of our theory of inductive proofs, p(n) represents the quantity about which we are proving something.

This is because inductive proofs involve proving that a statement is true for all values of a certain quantity, usually represented by n. In order to do this, we need to establish a base case for n (usually n=1 or n=0), and then show that if the statement is true for some arbitrary value of n (let's call it k), then it must also be true for the next value of n (k+1). This is where p(n) comes in - it represents the statement we are trying to prove, in terms of the quantity n. So if we can show that p(k) implies p(k+1), then we have established that p(n) is true for all values of n.

Therefore, p(n) is a crucial part of inductive proofs and represents the quantity about which we are proving something.

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17 students read none of the newspapers.

30 students read only Himalayan Times.

2 students read Rising Nepal and Himalayan Times only.

Proposing that we utilize the principle of inclusion-exclusion to address this dilemma.

A, B, and C represent the number of persons reading Kathmandu Post, Rising Nepal, and Himalayan Times, acquisitively.

The 28 students who read Kathmandu Post, 30 students who read Rising Nepal, and 42 pupils who selected Himalayan Times are then accounted for.

Furthermore, out of these, 10 people read both Kathmandu Post and Himalayan Times, 5 individuals take in only Kathmandu Post and Rising Nepal, 5 perusers perceive Rising Nepal and Himalayan Times, with 3 readers having access to each newspaper.

To identify the number of pupils who absorb none of the newspapers, we make use of the inclusion-exclusion principle formula.

Calculated by adding the total figures of A (+) B (+) C, thereafter subtracting the conjoined values of A and B (-) A and C (-) B and C (-), plus the collective overall score of A, B, and C (plus).

Here, n(A u B u C) is equal to 28 + 30 + 42 - 5 - 10 - 5 + 3 totalizing 83.

Accordingly, 100 (total number of students) minus 83 results in 17 students whose attention remains outright untapped.

Now, heading towards discovering the number of students who purely understand a single newspaper; "Himalayan Times".

This thus proves easy, estimating it as nC - nAc - nBc + nABC, which is 42 - 10 - 5 + 3 = 30 learners occupying themselves solely with Himalayan Times.

Likewise, 2 persons' attentions lay entrenched upon Rising Nepal and Himalayan Times exclusively, numerical comprehension of this is done through: nBC -nABC, equaling 5 - 3 = 2 .

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Answer: a b c

Step-by-step explanation: a b c

The standard deviation is a measure of the spread of data in a histogram, calculated as the square root of the variance, and it indicates how much the data deviates from the mean.

The standard deviation is a measure of the spread of data around the mean in a histogram. It tells us how much the data deviates from the average value. A low standard deviation indicates that the data is closely clustered around the mean, while a high standard deviation indicates that the data is more spread out.

In other words, a low standard deviation means that the data points are similar to each other, while a high standard deviation means that the data points are more diverse. The standard deviation is a useful tool for comparing the variability of data across different groups or regions and for identifying any outliers or unusual data points.

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--The complete question is, What is the standard deviation and how is it used to measure the spread of data in a histogram?--

Answer:

Step-by-step explanation:

To solve this problem, we will use the principle of inclusion-exclusion. Let $A_i$ be the set of integers not exceeding 10,000 that are divisible by the prime number $p_i$, for $p_i\in{3,4,7,11}$. We want to find the number of integers that are not in any of these sets $A_i$.

The number of integers not exceeding 10,000 that are divisible by $p_i$ is given by $\lfloor 10,000/p_i\rfloor$. For example, the number of integers divisible by 3 is $\lfloor 10,000/3\rfloor=3333$. However, some integers are divisible by more than one of the primes $p_i$, and we don't want to count them twice.

The number of integers not exceeding 10,000 that are divisible by two of the primes $p_i$ is given by $\lfloor 10,000/(p_ip_j)\rfloor$, where $p_i\neq p_j$. For example, the number of integers divisible by both 3 and 4 is $\lfloor 10,000/(3\times 4)\rfloor=833$.

Similarly, the number of integers divisible by three of the primes $p_i$ is $\lfloor 10,000/(3\times 4\times 7)\rfloor=59$, and the number of integers divisible by all four primes is $\lfloor 10,000/(3\times 4\times 7\times 11)\rfloor=4$.

Using the principle of inclusion-exclusion, the number of integers not exceeding 10,000 that are not divisible by 3, 4, 7, or 11 is given by:

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The rate at which the total cost is changing when 50 jet skis are produced is 22.5 thousand dollars per jet ski.

To find the rate at which the total cost is changing when 50 jet skis are produced, we need to find the derivative of the cost function with respect to x and then evaluate it at x = 50.

C(x) = 675 + 25x - 0.025x²

Taking the derivative with respect to x:

C'(x) = 25 - 0.05x

Now, we can evaluate C'(50) to find the rate of change at x = 50:

C'(50) = 25 - 0.05(50) = 22.5

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Answer:

I think that it is fine to go through my math or science classes without a calculator. How is using my calculator for a crunch just in case I got something wrong? I have found that if I didn't have a calculator I would be fine but if I have it I will use it. It is also a lazy thing. Instead of spending time to calculator what 136x45 is I could just find it in a second. But other than that it is ok.

The derivative (or jacobian) matrix, DF(x), of the nonlinear system x′=f(x) given by x'1=ax, x'2=bx2+c(x2)^3, where a,b, and c are constants, is [a 0; 0 2bx+3cx^2].

To find the derivative (or jacobian) matrix, DF(x), of the nonlinear system x′=f(x):

We first need to find the partial derivatives of each equation with respect to each variable.

For the first equation, x'1=ax,

the partial derivative with respect to x1 is a, and the partial derivative with respect to x2 is 0.

For the second equation, x'2=bx2+c(x2)^3,

the partial derivative with respect to x1 is 0, and the partial derivative with respect to x2 is 2bx + 3cx^2.

Putting these partial derivatives into a matrix, we get:

DF(x) =
[a     0]
[0    2bx+3cx^2]

Therefore, the derivative (or jacobian) matrix, DF(x), of the nonlinear system x′=f(x) given by x'1=ax, x'2=bx2+c(x2)^3, where a,b, and c are constants, is [a 0; 0 2bx+3cx^2].

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Answer:

  (b)  ∠DFE ≅ ∠DFG

Step-by-step explanation:

You want to know what additional information is required to show ∆DEF ≅ ∆DGF by SAS, given FE≅FG.

The SAS congruence postulate requires pairs of corresponding sides be congruent, along with the angle between those sides.

We know that FD is congruent to itself, and we are given FE≅FG. The angles between these sides are ∠DFE and ∠DFG.

To make use of the SAS congruence postulate, we need to know that ...

  ∠DFE ≅ ∠DFG, choice B

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The probability of getting at least 2 defective chips out of 8 is 0.0061, or about 0.61%

To solve this problem, we need to use the binomial distribution formula, which is:

P(X = k) = nCk * p^k * (1-p)^(n-k)

Where:
- P(X = k) is the probability of getting k successes
- n is the total number of trials (in this case, n = 8)
- k is the number of successes we're interested in (at least 2, so we need to calculate P(X = 2) + P(X = 3) + ... + P(X = 8))
- p is the probability of getting success on one trial (in this case, p = 0.01)

So let's calculate each term:

P(X = 2) = 8C2 * 0.01^2 * (1-0.01)^(8-2) = 0.0059
P(X = 3) = 8C3 * 0.01^3 * (1-0.01)^(8-3) = 0.0002
P(X = 4) = 8C4 * 0.01^4 * (1-0.01)^(8-4) = 0.0000
P(X = 5) = 8C5 * 0.01^5 * (1-0.01)^(8-5) = 0.0000
P(X = 6) = 8C6 * 0.01^6 * (1-0.01)^(8-6) = 0.0000
P(X = 7) = 8C7 * 0.01^7 * (1-0.01)^(8-7) = 0.0000
P(X = 8) = 8C8 * 0.01^8 * (1-0.01)^(8-8) = 0.0000

Now we can add up all the probabilities:

P(at least 2 defective chips) = P(X = 2) + P(X = 3) + ... + P(X = 8) = 0.0061

So the probability of getting at least 2 defective chips out of 8 is 0.0061, or about 0.61%. This is a relatively small probability, but it's not impossible, so the manufacturer should still take measures to minimize the number of defective chips produced.

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a) After calculating the Tukey's HSD values, we compare with the critical value and degrees of freedom, which is approximately 3.055.

b) We can conclude that there is a significant difference in plant growth between the groups treated with growth hormones 2 and 3.

a) Tukey's procedure is used to identify the significant differences between the means of multiple groups. It involves calculating the Tukey's HSD (Honestly Significant Difference) value for each pair of groups and comparing it with the critical value obtained from the Studentized range distribution.

If the Tukey's HSD value for a pair of groups is greater than the critical value, then the means of those groups are significantly different from each other.

To apply Tukey's procedure to the given data on plant growth, we first calculate the mean and standard deviation of each group.

Then, we calculate the Tukey's HSD value for each pair of groups using the formula HSD = q√(MSE/n), where q is the critical value from the Studentized range distribution, MSE is the mean square error, and n is the sample size.

After calculating the Tukey's HSD values, we compare them with the critical value for α = 0.05 and degrees of freedom = 15 (total number of observations - number of groups), which is approximately 3.055.

b) The results show that the means of groups 2 and 3 are significantly different from each other, as their Tukey's HSD value is greater than the critical value. The means of all other pairs of groups are not significantly different from each other.

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The compound statement for "p and q" is "9+5=14 and February has 30 days." This statement is a conjunction, which means both statements must be true for the entire statement to be true.

The truth value of the statement depends on the truth values of its components. In this case, the first component "9+5=14" is true, as 9+5 does indeed equal 14. However, the second component "February has 30 days" is false, as February typically has 28 or 29 days in a leap year.

Since the conjunction requires both components to be true for the entire statement to be true, the truth value of "9+5=14 and February has 30 days" is false. This means that the statement as a whole is false since one of its components is false.

In logic, conjunction is represented by the symbol "∧", which is read as "and". So the compound statement "9+5=14 ∧ February has 30 days" would be written to represent the statement "p and q". It's important to understand the truth values of logical statements, as they form the basis for many mathematical and computer science applications.

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Answer:23

Step-by-step explanation:

"The argument is valid, 'A truth table was constructed to show that the premises always lead to a true conclusion' is true because a truth table was used to demonstrate that the premises always result in a true conclusion, thereby establishing the validity of the argument."

The argument is valid.

Let's start

To see this, we can use a truth table.

We have three propositions:

P, Q, and R. Let's construct a truth table for the premises and the conclusion:

P Q R ~(P > Q) (~R v Q) & P (R & P) v ~Q

T T T      F                T                   T

T T F      F                T                   T

T F T     T                F                   T

T F F     T                F                   T

F T T     T                T                   T

F T F     T                T                   T

F F T     T                F                   F

F F F     T                F                   T

In the truth table,

The premise ~(P > Q) is always true.

The premise (~R v Q) & P is true,

in rows 1, 2, 5, and 6. In those rows, the conclusion (R & P) v ~Q is also true.

Therefore, the argument is valid.

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The ordered pair in the quadrilateral that represents the coordinates of the corresponding point on A'B'C'D' is B. (0.7x, 0.7y).

An ordered pair consists of two elements arranged in an exact sequence. Mathematics typically denotes this concept by inserting the components into parentheses and splitting them by a comma, (a, b).

Ordered pairs are frequently employed to designate points on a graph or plane where the first member represents the horizontal position (x-coordinate), and the other highlights the vertical dimension (y-coordinate).

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(a) Yes, A + B is also in M                                                                                                                                                                                                                                                                                                                                                                                                                                   (b) Yes, kD is also in M                                                                                                                                                                                                                                                                                                                                                                                                                            (c) Yes, M is a subspace of V

(a) Yes, A + B is also in M. This is because the sum of two diagonal matrices is also a diagonal matrix, and therefore A + B has all zero entries outside of the diagonal, and so A + B is in M.

(b) Yes, kD is also in M. This is because multiplying a diagonal matrix by a scalar simply multiplies each entry on the diagonal by that scalar, and so kD is still a diagonal matrix and thus is in M.

(c) Yes, M is a subspace of V. To show this, we need to verify that M satisfies the three properties of a subspace:

The zero vector, which is the 3 x 3 matrix with all entries equal to zero, is in M.

If A and B are in M, then their sum A + B is also in M, as we showed in part (a).

If A is in M and k is any scalar, then kA is also in M, as we showed in part (b).

Therefore, M satisfies all three properties of a subspace and is thus a subspace of V.

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Eric's actual value is 4.91 hours (rounded to 2 decimal places).

To calculate Eric's actual value, we can use the formula:

z = (x - μ) / σ

Where z is the z-score, x is the actual value, μ is the mean, and σ is the standard deviation.

Plugging in the values given, we can solve for x:

-0.79 = (x - 4.00) / 1.15

-0.79 * 1.15 = x - 4.00

-0.9085 = x - 4.00

x = 4.00 - (-0.9085)

x = 4.9085

Therefore, Eric's actual value is 4.91 hours (rounded to 2 decimal places).

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Answer:

Step-by-step explanation:

We can use the Maclaurin series for cos(x) to find the Maclaurin series for f(x) = x cos(4x):

cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + ...

cos(4x) = 1 - (4x)^2/2! + (4x)^4/4! - (4x)^6/6! + ...

cos(4x) = 1 - 8x^2/2! + 64x^4/4! - 1024x^6/6! + ...

f(x) = x cos(4x) = x - 8x^3/2! + 64x^5/4! - 1024x^7/6! + ...

Therefore, the Maclaurin series for f(x) is:

f(x) = x - 8x^3/2! + 64x^5/4! - 1024x^7/6! + ...

The probability of both events happening, we need to multiply the probabilities:  (5/16) x (4/15) = 1/12. So the probability that Amarillo ends up with 2 red chips is 1/12.

To find the probability that Amarillo ends up with 2 red chips, we need to consider the total number of chips and the number of successful outcomes (2 red chips drawn).

1. Calculate the total number of chips:
10 white chips + 5 red chips + 1 blue chip = 16 chips

2. Calculate the probability of drawing the first red chip:
There are 5 red chips out of 16 total chips, so the probability is 5/16.

3. After drawing the first red chip, there are now 4 red chips and 15 total chips left in the bag. Calculate the probability of drawing the second red chip:
The probability is 4/15.

4. Multiply the probabilities of each step to find the overall probability of drawing 2 red chips:
(5/16) * (4/15) = 20/240 = 1/12

The probability that Amarillo ends up with 2 red chips is 1/12.

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The number of arrangements of the letters in 'combinatorics' that have consecutive c’s but no consecutive vowels = 8! × C(9, 5)

We need to find the number of arrangements of the letters in 'combinatorics' that have consecutive c’s but no consecutive vowels.

First we divide the consonants and the vowels.

The consonants are 2 C's, M, B, N, T, R, and S.

and the vowels are 2 O's, 2 I's and one A.

Now the total number of ways to arrange the consonants = 8!

Now we need to arrange my vowels such that there are no consecutive vowels.

Since there are nine places to place vowels in order to avoid having consecutive vowels, there are C(9,5).

Using combination formula:

C(9, 5) = ⁹C₅

           = 9!/(5! × (9 - 5)!)

           = 9!/(5! × 4!)

           = 126

The number of arrangements of the letters in 'combinatorics' that have consecutive c’s but no consecutive vowels:

n = 8! × C(9, 5)

n = 5080320

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There is no constant term (no term with just a constant coefficient), so C = 0.

Calculating the sum of two or more numbers is the procedure of multiplication. 'A' multiplied by 'B' is how you express the multiplication of two numbers, let's say 'a' and 'b'.

We can use the vertical method (also known as the column method) to multiply the two polynomials:

        4a³  + 3a²  - 2a  + 1

       x  3     - 2a   + a²

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

    12a³  + 9a²  - 6a  + 3a²

  -8a⁴  - 6a³  + 4a²  - 2a

+3a⁵  + 2a⁴  -  a³

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

+3a⁵  - 6a⁴  - 3a³  + 16a²  - 8a

So the product of the two polynomials is:

3a⁵ - 6a⁴ - 3a³ + 16a² - 8a

There is no constant term (no term with just a constant coefficient), so C = 0.

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The strings in S of length 3 are: aaa, baa, abb, and bbb.

Based on the recursive rules you provided for the set S, we can list all strings of length 3. Rule 1 states xaa ∈ S, and Rule 2 states xbb ∈ S. The base cases are λ (empty string) ∈ S, a ∈ S, and b ∈ S.

To create strings of length 3, we can use the base cases and apply the rules:

1. For Rule 1 (xaa ∈ S):
  - If x = a, the string is "aaa".
  - If x = b, the string is "baa".

2. For Rule 2 (xbb ∈ S):
  - If x = a, the string is "abb".
  - If x = b, the string is "bbb".

So, the strings in S of length 3 are: aaa, baa, abb, and bbb.

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Answer:

(x+3)²=25

Explanation:

(in progress)

Answer:

(x+3)² = 25

Step-by-step explanation:

We are given the following equation:
2x² + 12x = 32

We want to find what the equation of this will be when we complete the square.

To start, we can divide both sides by 2. The resulting equation will be:

x² + 6x = 16

When completing the square, we add a third number to both sides. This is because we will factor the left side into the form (a+b)².

Recall that (a+b)² = a² + 2ab + b². The 6x is the 2ab in this sense, but because the value of a would be x (a² is x²), 6 = 2b.

If we solve 6 = 2b, then b = 3.

Now, square it to get the third number in the equation. 3² = 9.

We now add 9 to both sides of the equation. We do this in order to balance the equation, because if we add 9 to only one side, the equation will become unbalanced.

x² + 6x + 9 = 16 + 9

x² + 6x + 9 = 25

We can factor the left side to get:

(x+3)² = 25