suppose that f(x)=1x−2 and g(x)=5x 1. if we were to add these two functions together to create a new function h(x) then what is the domain of the new function h(x)?

The probability of rolling an even number is the sum of the probabilities of rolling 2 and 4 and 6, which is:

1/12 + 1/12 + 3/12 = 5/12

Let p be the probability of rolling each of the numbers 1, 2, 3, and 4. Since 5 and 6 are three times as likely to come up as each of the other numbers, the probabilities of rolling 5 and 6 are 3p each. The sum of all probabilities must be equal to 1, so we have:

p + p + p + p + 3p + 3p = 1

Simplifying this equation, we get:

12p = 1

p = 1/12

Therefore, the probability distribution is:

Outcome 1 2 3 4 5 6

Probability 1/12 1/12 1/12 1/12 3/12 3/12

The probability of rolling an even number is the sum of the probabilities of rolling 2 and 4 and 6, which is:

1/12 + 1/12 + 3/12 = 5/12

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Answer:

$0.16

Step-by-step explanation:

$4.88 is how much 8 apple juice boxes cost

to find out how many 1 costs we divide by 8

$4.88÷8=$0.61

so 1 apple juice box costs $0.61

The inner product of u and v is (-15).

In this problem, we are given two vectors, u and v, and asked to compute their inner product. The first step in calculating the inner product is to write the vectors in component form. We are given that

u = (1, -3) and v = (6, 4).

The next step is to compute the product of the corresponding components and sum them up. This gives us:

u · v = (1)(6) + (-3)(4) = 6 - 12 = -6

Therefore, the inner product of u and v is (-6).

Inner product is an important concept in linear algebra and has many applications in fields such as physics, engineering, and computer science. It is a way to measure the similarity between two vectors and can be used to find angles between vectors, project one vector onto another, and solve systems of linear equations.

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Among the given expressions and equations, two are equations represented by triangles (A), while the remaining three are expressions represented by boxes().

The first equation, "2x + 9 = 2," is represented by a triangle (A) because it contains an equal sign, indicating that both sides are equal. The second expression, "32 + 3 x 9) = 59," is represented by a box () as it does not have an equal sign, making it an arithmetic expression rather than an equation.

The third equation, "3k + 7 = 34," is an equation and represented by a triangle (A) because it has an equal sign, signifying an equality between two expressions. The fourth expression, "5 (b + 28) = 150," is an expression and represented by a box () because it lacks an equal sign. It involves arithmetic operations but does not establish an equality.  

Finally, the fifth equation, "9a + 7 =," is an equation and represented by a triangle (A). Although it appears incomplete, it still contains an equal sign, indicating that the expression on the left side is equal to an unknown value on the right side.  

In summary, two equations are represented by triangles (A) because they contain equal signs and establish equalities between expressions, while the remaining three are expressions represented by boxes () as they lack equal signs and do not create equalities.

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The given integral can be expressed in terms of simpler integrals as:

[tex]\int (−3x^2 + 5x + 6x cos(x)) dx = -x^3 + (5/2)x^2 + 6x sin(x) + 6 cos(x) + C[/tex](

To express the given integral in terms of simpler integrals, we can use the properties of the indefinite integral, including the linearity property and integration by parts.

We can first break down the integrand using linearity:

[tex]\int (−3x^2 + 5x + 6x cos(x)) dx = \int (-3x^2) dx + \int (5x) dx + \int (6x cos(x)) dx[/tex]

Now, we can integrate each term separately:

[tex]\int (-3x^2) dx = -x^3 + C1[/tex] (where C1 is the constant of integration)

[tex]\int (5x) dx = (5/2)x^2 + C2[/tex] (where C2 is another constant of integration)

To integrate ∫(6x cos(x)) dx, we can use integration by parts with u = 6x and dv = cos(x) dx:

∫(6x cos(x)) dx = 6x sin(x) - ∫(6 sin(x)) dx

= 6x sin(x) + 6 cos(x) + C3 (where C3 is another constant of integration)

Putting everything together, we have:

[tex]\int (−3x^2 + 5x + 6x cos(x)) dx = -x^3 + C1 + (5/2)x^2 + C2 + 6x sin(x) + 6 cos(x) + C3[/tex]

So the given integral can be expressed in terms of simpler integrals as:

[tex]\int (−3x^2 + 5x + 6x cos(x)) dx = -x^3 + (5/2)x^2 + 6x sin(x) + 6 cos(x) + C[/tex](where C = C1 + C2 + C3 is the overall constant of integration)

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The degrees of freedom for this two-mean non pooled hypothesis test is 15.

To find the degrees of freedom for a two-mean nonpooled hypothesis test, we use the following formula:

df = (s1^2/n1 + s2^2/n2)^2 / ( (s1^2/n1)^2 / (n1 - 1) + (s2^2/n2)^2 / (n2 - 1) )

Substituting the given values, we get:

df = (3^2/17 + 7^2/24)^2 / ( (3^2/17)^2 / (17 - 1) + (7^2/24)^2 / (24 - 1) )

= 14.97

Rounding to the nearest integer, we get:

df = 15

Therefore, the degrees of freedom for this two-mean non pooled hypothesis test is 15.

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The length of the enlarged pencil will be 29.75 units.The original length of the pencil is 7 units, and the width is 1.5 units. The scale factor is 425%.

We need to find the new length of the pencil after it is enlarged by the given scale factor of 425%.

The formula for calculating the new length of the pencil is:New Length of Pencil = Original Length × Scale Factor/100 Adding the given values in the above formula,

To find the length of the enlarged pencil, we need to multiply the original length by the scale factor.

The scale factor is given as 425%, which can be written as a decimal as 4.25.

Length of enlarged pencil = Original length * Scale factor

= 7 units * 4.25

= 29.75 units

Therefore, the length of the enlarged pencil will be 29.75 units.

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The conditional statement "T → (8<5)" is true because the antecedent "T" is false, and by the truth table of a conditional statement, a conditional with a false antecedent is always true, regardless of the truth value of the consequent.

what is antecedent?

In logic, an antecedent is the first part of a conditional statement (if-then statement) that precedes the word "if." It is the statement that implies or asserts the truth of the consequent. For example, in the conditional statement "If it is raining, then I will stay inside," the antecedent is "it is raining."

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(a) The change is -5.403 million enplaned passengers.

The number of enplaned passengers on the given airline decreased from 98.175 million in 2007 to 92.772 million in 2008, resulting in a decrease of 5.403 million enplaned passengers.

(b) The percentage change is -5.51%.

The percentage change is calculated using the formula: ((new value - old value) / old value) x 100%. In this case, the percentage change is ((92.772 - 98.175) / 98.175) x 100% = -5.51%. This indicates a 5.51% decrease in the number of paying passengers on the given airline between 2007 and 2008.

(c) The average rate of change is -2.702 million enplaned passengers per year.

The average rate of change is calculated by dividing the total change in the number of enplaned passengers by the number of years between 2007 and 2008. In this case, the average rate of change is (-5.403 / 2) = -2.702 million enplaned passengers per year.

This means that the number of paying passengers on the given airline decreased by an average of 2.702 million per year between 2007 and 2008.

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The slope for the regression equation is given by:

b = sp / ssx

where sp is the sum of products of deviations, and ssx is the sum of squared deviations of x scores.

Substituting the given values, we get:

b = 55 / 21

b ≈ 2.62

Rounding to 2 decimal places, we get the slope as 2.62.

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The derivative of the function f(x) = 0 to x sec(7t) dt is sec^2(7x) * tan(7x).

The derivative of the function f(x) = 0 to x sec(7t) dt is sec(7x).

To see why, we use part one of the fundamental theorem of calculus, which states that if F(x) is an antiderivative of f(x), then the definite integral from a to b of f(x) dx is F(b) - F(a).

Here, we have f(x) = sec(7t), and we know that an antiderivative of sec(7t) is ln|sec(7t) + tan(7t)| + C, where C is an arbitrary constant of integration.

So, using the fundamental theorem of calculus, we have:

f(x) = 0 to x sec(7t) dt = ln|sec(7x) + tan(7x)| + C

Now, we can take the derivative of both sides with respect to x, using the chain rule on the right-hand side:

f'(x) = d/dx [ln|sec(7x) + tan(7x)| + C] = sec(7x) * d/dx [sec(7x) + tan(7x)] = sec(7x) * sec(7x) * tan(7x) = sec^2(7x) * tan(7x)

Therefore, the derivative of the function f(x) = 0 to x sec(7t) dt is sec^2(7x) * tan(7x).

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The answer of the given question based on the triangle is , - 15.75 ,  this is not possible as the length cannot be negative.

We are given:

In ΔCDE, the measure of ∠E = 90°, CD = 9.2 feet, and DE = 8.3 feet.

To find:

The measure of ∠C to the nearest tenth of a degree.

Solution:

In ΔCDE, applying Pythagoras theorem:

CE² + CD² = DE²CE² + (9.2)² = (8.3)²

CE² = (8.3)² - (9.2)²CE²

= 68.89 - 84.64CE²

= - 15.75

This is not possible as the length cannot be negative.

Hence, the given values are not possible.

So, there is no such triangle ΔCDE, which satisfies the given conditions.

Hence, we cannot find the measure of ∠C.

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The area enclosed by the given parametric curve and the y-axis is -4.5 square units.

To find the area enclosed by the given parametric curve and the y-axis, we can use the formula for calculating the area bounded by a parametric curve:

A = ∫ |x(t) dy/dt| dt

In this case, the parametric equations are:

x = t^2 - 3t

y = t

To calculate the derivative dy/dt, we differentiate y = t with respect to t:

dy/dt = 1

Now we can substitute the values into the area formula:

A = ∫ |(t^2 - 3t)(1)| dt

A = ∫ |t^2 - 3t| dt

To calculate the integral, we need to split it into two parts based on the absolute value:

A = ∫ (t^2 - 3t) dt (for t ≥ 0)

A = ∫ -(t^2 - 3t) dt (for t < 0)

Evaluating the integrals:

For t ≥ 0:

A = (1/3)t^3 - (3/2)t^2 + C1

For t < 0:

A = -(1/3)t^3 + (3/2)t^2 + C2

To find the specific bounds of integration, we need to determine the range of t that corresponds to the area enclosed by the curve and the y-axis. This can be done by finding the points where the curve intersects the y-axis.

Setting x = 0, we have:

0 = t^2 - 3t

t(t - 3) = 0

t = 0 or t = 3

Therefore, the bounds of integration will be from t = 0 to t = 3.

Substituting these bounds into the area formula, we get:

A = [(1/3)(3)^3 - (3/2)(3)^2] - [(1/3)(0)^3 - (3/2)(0)^2]

A = [(1/3)(27) - (3/2)(9)] - 0

A = 9 - 13.5

A = -4.5

The area enclosed by the given parametric curve and the y-axis is -4.5 square units. Note that the negative sign indicates that the curve is below the x-axis for part of the interval.

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The statement is false. Every quantitative data set has a median, which is the middle value when the data is arranged in ascending or descending order. If there is an even number of data points, the median is the average of the middle two values.

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The result "100" in the simulation simulates that exactly one person in a group of three randomly chosen people who buy movie tickets also buys popcorn.


In the simulation, Lindsey generated three random numbers for each trial to represent the behavior of three people at the movie theater. According to the given rules, any number from 1 through 6 represents a person who buys a movie ticket and popcorn, while any number from 7 through 9 or 0 represents a person who buys only a movie ticket.

To estimate the probability that exactly two in a group of three people selected randomly at a movie theater will buy both a movie ticket and popcorn, Lindsey needed to run multiple trials of the simulation. In one of the trials, the result was "100", which means that one of the three randomly-chosen people bought both a movie ticket and popcorn, while the other two only bought a movie ticket.

Therefore, the result "100" in the simulation simulates that exactly one person in a group of three randomly-chosen people who buy movie tickets also buys popcorn.


Based on the simulation results, Lindsey can estimate the probability of exactly two people buying both a movie ticket and popcorn out of a group of three randomly chosen people who buy movie tickets at the theater. By analyzing all 30 trials of the simulation, Lindsey can calculate the relative frequency of this event and use it as an estimate of the probability.

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The product as a sum or difference is:

1) 16 sin(28x) sin(11x) = 8[cos(17x) - cos(39x)]
2) sin(-x) sin(9x) = ([tex]\frac{1}{2}[/tex])[cos(-10x) - cos(8x)]

1) 16 sin(28x) sin(11x)
We can use the Product-to-Sum identity: sin(A)sin(B) = (1/2)[cos(A-B) - cos(A+B)]
So, 16 sin(28x) sin(11x) can be rewritten as:
8[cos(28x - 11x) - cos(28x + 11x)] = 8[cos(17x) - cos(39x)]
2) sin(-x) sin(9x)
Again, we use the Product-to-Sum identity: sin(A)sin(B) = ([tex]\frac{1}{2}[/tex])[cos(A-B) - cos(A+B)]
So, sin(-x) sin(9x) can be rewritten as:
([tex]\frac{1}{2}[/tex])[cos(-x - 9x) - cos(-x + 9x)] = ([tex]\frac{1}{2}[/tex])[cos(-10x) - cos(8x)]

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To determine whether the matrix is in echelon form, reduced echelon form, or neither, let's first write the given matrix clearly:

[1 0 5 4]
[0 1 -5 -3]
[0 0 0 0]
[0 0 0 0]

Now, let's analyze its form:

a) Echelon form requires:
1. All nonzero rows are above any rows of all zeros.
2. The leading coefficient (pivot) of a nonzero row is always to the right of the pivot of the row above it.

This matrix satisfies both conditions, so it is in echelon form.

b) Reduced echelon form requires:
1. The matrix is in echelon form.
2. The pivot in each nonzero row is 1.
3. Each pivot is the only nonzero entry in its column.

This matrix fulfills the first two conditions, but the third condition is not met due to the presence of '5' in the first row and the same column as the pivot '1' in the second row.

Therefore, the matrix is in echelon form (option b) but not in reduced echelon form.

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Answer:

To eliminate x, you need a positive coefficient in front of x for one equation and its negative counterpart in front of the other equation as a positive number plus its negative opposite equals 0 (e.g., -4 + 4 = 0 and -80 + 80 = 0)

Step 1:  Therefore, we can eliminate x by first determining the least common multiple (LCM) between 5 and 4.  We know that 5 * 4 = 20 and 4 * 5, so the LCM between 5 and 4 is 20.

Step 2:  In order to have 20 as coefficient for x in one equation and -20 for x as a coefficient in the other equation, we can multiply the entire first equation by 4 and the entire second equation by -5:

Equation 1 multiplied by 4:  4 * (5x - 2y = 10) = 20x - 8y = 40

Equation 2 multiplied by -5:  -5* (4x - 3y = 15) = -20x + 15y = -75

Step 3:  Adding the two equations shows that the xs cancel as 20x - 20x = 0, leaving us with 15y - 8y = 40 - 75, which simplifies to 7y = -35

Answer: See below.

Step-by-step explanation:

       First, we are already given these equations in standard form.

5x − 2y = 10

4x − 3y = 15

       Next, we need to make the coefficients of the x variables opposites (as in 5 and -5, etc), since we want to eliminate the x's. To do this, we will find a common multiple (here, the Lowest Common Multiplb is 20). Then, we will multiply every term by the number that makes the coefficient of x our common multiple.

       We will make the first equation with a coefficient of 20 for the x and the second with a coefficient of -20 for the x.

       See this visually below.

5x − 2y = 10 ➜ 4(5x) − 4(2y) = 4(10) ➜ 20x - 8y = 40

4x − 3y = 15 ➜ -5(4x) − -5(3y) = -5(15) ➜ -20x + 15y = -75

       Lastly, add these two equations together. The x's are eliminated. This also will let us solve for y.

      20x - 8y = 40

+   -20x + 15y = -75

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

7y = -35

y = -5

Ryan needs to contribute $1000.07 per month.

To determine the monthly contribution needed, we will use the formula for monthly payment [tex]FV = P * [(1 + r)^n - 1] / r,[/tex]

Plugging values:

[tex]208,000 = P * [(1 + 0.078/12)^{11*12} - 1] / (0.078/12).\\208,000 = P * [1.0065^{132} - 1] / 0.0065.[/tex]

Rearranging to solve for P

[tex]P = 208,000 * 0.0065 / [1.0065^{132} - 1].[/tex]

P = 208,000 * 0.0065 / 1.35190003004

P = 1000.07394775

P = $1000.07

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The second boxplot is correct representation of the number of customers at each Slurpee Sam's Spaghetti Shop.

Given the data set is,

24, 25, 29, 30, 31, 31, 32, 34, 34

Hence, Minimum value = 24

Maximum value = 34

And, First quartile (Q1) = 1/4(n+1)th term

Q1 = 1/4 x 10 = 10/4 = 2.5th term = (25+29)/2 = 27

Q2 = 1/2(n+1)th term = 1/2(10) = 5th = 31

Q3 = 3/4(n+1)th term = 3/4(10) = 30/4 = 7.5th term = (7th +8th) term = (32 + 34) / 2 = 66/2 = 33

Hence, correct boxplot should have :

Minimum value = 24

Q1 = 27

Q2 = 31

Q3 = 33

Maximum value = 34

Thus, The second boxplot is correct.

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Let's assume that the marked price of the article is "M" dollars. The marked price of the article is approximately $4941.18.

According to the problem statement, the dealer gives a discount of 10%, so the selling price (S) of the article is:

S = M - 0.10M = 0.90M

Now, the GST of 12% is applied on the marked price, so the amount of GST paid is:

GST = 0.12M

Therefore, the total amount paid by the consumer (C) is:

C = S + GST

C = 0.90M + 0.12M

C = 1.02M

We are given that the consumer pays $5040, so we can set up the equation:

1.02M = 5040

Solving for M, we get:

M = 5040 / 1.02

M ≈ 4941.18

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The triangle in the image is a right triangle. We are given a side and an angle, and asked to find another side. Therefore, we should use a trigonometric function.

Trigonometric Functions: SOH-CAH-TOA

---sin = opposite/hypotenuse, cosine = adjacent/hypotenuse, tangent = opposite/adjacent

In this problem, looking from the angle, we are given the adjacent side and want to find the opposite side. This means we should use the tangent function.

tan(40) = x / 202

x = tan(40) * 202

x = 169.498

x (rounded) = 169 meters

Answer: the tower is 169 meters tall

Hope this helps!

(a) The expected value of Y is :

E(Y) = (n+1)/3

(b) The value of E(Y^2) = (2n^2+5n+1)/6

(c) The variance of Y = (2n^2+5n+1)/6 - [(n+1)/3]^2

(d) P(X+Y=2) = 1/n

a) To find the expected value of Y, we use the law of total probability:

E(Y) = ∑ P(X=k)E(Y|X=k) for k=1 to n

Since Y is uniformly distributed on {1,2,...,X}, we have E(Y|X=k) = (k+1)/2.

Therefore,

E(Y) = ∑ P(X=k)(k+1)/2 for k=1 to n

To find P(X=k), note that X can take on any value from 1 to n with equal probability, so P(X=k) = 1/n for k=1 to n. Thus,

E(Y) = ∑ (k+1)/2n for k=1 to n

E(Y) = [1/2n ∑ k] + [1/2n ∑ 1] for k=1 to n

E(Y) = [1/2n (n(n+1)/2)] + [1/2n n]

E(Y) = (n+1)/3

b) To find E(Y^2), we use the law of total probability again:

E(Y^2) = ∑ P(X=k)E(Y^2|X=k) for k=1 to n

Since Y is uniformly distributed on {1,2,...,X}, we have E(Y^2|X=k) = (k^2+3k+2)/6. Therefore,

E(Y^2) = ∑ P(X=k)(k^2+3k+2)/6 for k=1 to n

Using the same values of P(X=k) as before, we get:

E(Y^2) = ∑ (k^2+3k+2)/6n for k=1 to n

E(Y^2) = [1/6n ∑ k^2] + [1/2n ∑ k] + [1/6n ∑ 1] for k=1 to n

E(Y^2) = [1/6n (n(n+1)(2n+1)/6)] + [1/2n (n(n+1)/2)] + [1/6n n]

E(Y^2) = (2n^2+5n+1)/6

c) The variance of Y is given by Var(Y) = E(Y^2) - [E(Y)]^2. Therefore,

Var(Y) = (2n^2+5n+1)/6 - [(n+1)/3]^2

d) To find P(X+Y=2), we note that X+Y=2 if and only if X=1 and Y=1. Since X is uniformly distributed on {1,2,...,n}, we have P(X=1) = 1/n. Since Y is uniformly distributed on {1,2,...,X}, we have P(Y=1|X=1) = 1. Therefore,

P(X+Y=2) = P(X=1)P(Y=1|X=1) = 1/n

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The predicted clutch engagement time is approximately 1.81 seconds when the starting speed is 18 m/s, the maximum drive torque is 17 N.m, the system inertia is 0.006 kg•m2, and the applied force rate is 10 kN/s.

The given regression model for predicting clutch engagement time (y) based on four predictor variables (x1, x2, x3, x4) is:

[tex]y = -0.83 + 0.017x1 + 0.0895x2 + 42.771x3 + 0.027x4 - 0.0043x2x4[/tex]

To predict the clutch engagement time when x1 = 18 m/s, x2 = 17 N.m, x3 = 0.006 kg•m2, and x4 = 10 kN/s, we simply substitute these values into the regression equation:

[tex]y = -0.83 + 0.017(18) + 0.0895(17) + 42.771(0.006) + 0.027(10) - 0.0043(17)(10)\\y = -0.83 + 0.306 + 1.5215 + 0.256626 + 0.27 - 0.731[/tex]

y = 1.809126

Therefore, the predicted clutch engagement time is approximately 1.81 seconds when the starting speed is 18 m/s, the maximum drive torque is 17 N.m, the system inertia is 0.006 kg•m2, and the applied force rate is 10 kN/s.

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The correct equation to represent this situation is D-95=16.50, where D represents the amount Ms. Park had to start with.

Let's assume that Ms. Park had D dollars to start with. After depositing $95, her balance would be D + $95.

But now, her balance is in debt by $16.50. So we can set up an equation:D + $95 = $16.50

The left side of the equation represents the amount Ms. Park has after depositing $95, while the right side of the equation represents the amount she owes.

We can then isolate D by subtracting $95 from both sides:D + $95 - $95 = $16.50 - $95D = $78.50

Therefore, Ms. Park had $78.50 to start with.

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Answer:

x=9.3

Step-by-step explanation:

use SohCahToa

in this case u use cos

cos(41°)=7/x

x=7/cos(41)

x=9.275090953

x=9.3

The general antiderivative of n(x) = x⁸ + 5x⁴ + x⁵ is N(x) = (1/9)x⁹ + (1/5)x⁵ + (1/6)x⁶ + C.

To find the antiderivative of n(x) = x⁸ + 5x⁴ + x⁵, we apply the power rule for integration, which states that ∫x^n dx = (xⁿ⁺¹)/(n+1) + C, where C is the constant of integration.

1. For the first term, x⁸, integrate using the power rule: ∫x⁸ dx = (1/9)x⁹ + C₁.
2. For the second term, 5x⁴, integrate: ∫5x⁴ dx = 5(1/5)x⁵ + C₂ = x⁵ + C₂.
3. For the third term, x⁵, integrate: ∫x⁵ dx = (1/6)x⁶ + C₃.

Now, add the results of each integration and combine the constants: N(x) = (1/9)x⁹ + x⁵ + (1/6)x⁶ + (C₁ + C₂ + C₃). Since the constants are arbitrary, we can represent them as a single constant, C: N(x) = (1/9)x⁹ + (1/5)x⁵ + (1/6)x⁶ + C.

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Here are the masses of some planets and moons expressed in both standard and scientific notation:

Planet Mass in Standard NotationMass in Scientific Notation:

Venus = 4,870,000,000,000,000,000,000,000 kg4.87 × 10²⁴ kg

Earth = 5,970,000,000,000,000,000,000,000 kg5.97 × 10²⁴ kg

Mars = 6,420,000,000,000,000,000,000,000 kg6.42 × 10²⁴ kg

Jupiter = 1,898,000,000,000,000,000,000,000,000 kg1.90 × 10²⁷ kg

Saturn = 568,000,000,000,000,000,000,000,000 kg5.68 × 10²⁶ kg

Uranus = 86,800,000,000,000,000,000,000 kg8.68 × 10²⁵ kg

Neptune = 102,000,000,000,000,000,000,000 kg1.02 × 10²⁶ kg

Moon = 7,340,000,000,000,000,000 kg7.34 × 10²² kg

Io = 8,930,000,000,000,000,000 kg8.93 × 10²² kg

Ganymede = 1,480,000,000,000,000,000,000 kg1.48 × 10²³ kg

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If `f(x)` is a polynomial, then `f(x²)` is also a polynomial. Polynomials are mathematical expressions that consist of variables and coefficients with only the operations of addition, subtraction, multiplication, and non-negative integer exponents. We can prove this statement using the definition of a polynomial. Definition of a polynomial polynomial is an expression that can be written as follows:$$f(x)= a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdot\cdot\cdot +a_1x+a_0$$where `a0, a1, …, an` are constants, and `n` is a non-negative integer. This definition of the polynomial can be used to show that `f(x²)` is also a polynomial. Using the definition of a polynomial, we can write:$$f(x²)= a_n(x²)^n+a_{n-1}(x²)^{n-1}+a_{n-2}(x²)^{n-2}+\cdot\cdot\cdot +a_1(x²)+a_0$$Simplifying the terms of the expression, we get:$$f(x²)= a_nx^{2n}+a_{n-1}x^{2(n-1)}+a_{n-2}x^{2(n-2)}+\cdot\cdot\cdot +a_1x^2+a_0$$This proves that `f(x²)` is also a polynomial. Therefore, if `f(x)` is a polynomial, then `f(x²)` is also a polynomial.

Yes, if f(x) is a polynomial, then f(x²) is also a polynomial.

A polynomial is a mathematical expression consisting of variables, coefficients, and non-negative integer exponents. It can include addition, subtraction, and multiplication operations. The terms in a polynomial can be in the form of axⁿ, where a is the coefficient, x is the variable, and n is a non-negative integer exponent.

When we substitute x² into f(x), each occurrence of x in the polynomial f(x) is replaced by x². Since x² is still a variable with a non-negative integer exponent, the resulting expression f(x²) remains a polynomial. The coefficients and exponents may change, but the essential structure of a polynomial is preserved.

Therefore, if f(x) is a polynomial, then f(x²) is also a polynomial.

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