Find the area of the shape below.

The quadratic polynomial is x² - 6x + 31/4

A quadratic polynomial is a polynomial in which the highest power of the unknown is 2.

To form a quadratic polynomial whose zeroes are 3 + √5/2 and 3 - √5/2, we proceed as follows.

Since the zeroes are

Then its factors are

So, to get the quadratic polynomial p(x), we multiply the factors together.

So, we have that

p(x) =  [(x - 3) - √5/2][(x - 3) + √5/2]

= (x - 3)² - (√5/2)²

= x² - 6x + 9 - 5/4

= x² - 6x + (36 - 5)/4

= x² - 6x + 31/4

So, the polynomial is x² - 6x + 31/4

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The right answer is:

1. The amount of carbon dioxide in the atmosphere is increasing (or growing).

2. Political conflicts (disagreements) occur over control of these resources.

As the population continues to grow, the demand for energy will also increase, further exacerbating this problem.

The increase in human population has led to an increase in energy consumption, which is largely met by the burning of fossil fuels such as coal, oil, and gas.

As fossil fuels become increasingly scarce, there may be greater competition and conflicts over their control and distribution. This can lead to geopolitical tensions and instability in some regions.

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Probability of getting a 7 when a card is picked randomly is 1/3.

Here, given that

A card is picked at random.

Possible outcomes = {5, 6, 7}

Number of possible outcomes = 3

Favorable outcomes = {7}

Number of favorable outcomes = 1

Probability = Number of favorable outcomes / Total outcomes

                  = 1/3

Hence the required probability is 1/3.

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The limit of the sequence is 0.

To determine the limit of the sequence, we can use the Limit Laws and theorems. We will start by simplifying the expression:

cn=ln((5n-7)/(12n+4))

cn=ln(5n-7)-ln(12n+4)

Now we can use the Limit Laws:

lim n→∞ ln(5n-7) = ∞ (since ln(x) → ∞ as x → ∞)

lim n→∞ ln(12n+4) = ∞ (since ln(x) → ∞ as x → ∞)

Therefore, we have:

lim n→∞ cn = lim n→∞ (ln(5n-7)-ln(12n+4))

= lim n→∞ ln(5n-7) - lim n→∞ ln(12n+4)

= ∞ - ∞ (which is an indeterminate form)

To evaluate this limit, we can use L'Hopital's Rule:

lim n→∞ ln(5n-7) - ln(12n+4) = lim n→∞ [ln((5n-7)/(12n+4))]

= lim n→∞ [(5/12)/(5/n - 3/4n²)]

Since the denominator goes to ∞ and the numerator is constant, we have:

lim n→∞ [(5/12)/(5/n - 3/4n²)] = 0

Therefore, we have:

lim n→∞ cn = 0

So the limit of the sequence is 0.

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The order of a in (z/(pq))× is exactly (p-1)(q-1), as desired.

To prove that (z/(pq))× has order (p − 1)(q − 1), we need to show that the least positive integer n such that (z/(pq))×n = 1 is (p − 1)(q − 1).

First, let's define (z/(pq))× as the set of all integers a such that gcd(a,pq) = 1 (i.e., a is relatively prime to pq) and a mod pq is also relatively prime to pq.

Now, we know that the order of an element a in a group is the smallest positive integer n such that a^n = 1. Therefore, we need to find the order of an arbitrary element a in (z/(pq))×.

Let's assume that a is an arbitrary element in (z/(pq))×. Since gcd(a,pq) = 1, we know that a has a multiplicative inverse modulo pq, denoted by a^-1. Therefore, we can write:

a * a^-1 ≡ 1 (mod pq)

Now, let's consider the order of a. Since gcd(a,pq) = 1, we know that a^(p-1) is congruent to 1 modulo p by Fermat's Little Theorem. Similarly, we can show that a^(q-1) is congruent to 1 modulo q. Therefore, we have:

a^(p-1) ≡ 1 (mod p)
a^(q-1) ≡ 1 (mod q)

Now, we can use the Chinese Remainder Theorem to combine these congruences and get:

a^(p-1)(q-1) ≡ 1 (mod pq)

Therefore, we know that the order of a must divide (p-1)(q-1).

To show that the order of a is exactly (p-1)(q-1), we need to show that a^k is not congruent to 1 modulo pq for any positive integer k such that 1 ≤ k < (p-1)(q-1).

Assume for contradiction that there exists such a k. Then, we have:

a^k ≡ 1 (mod pq)

This means that a^k is a multiple of pq, which implies that gcd(a^k, pq) ≥ pq. However, since gcd(a,pq) = 1, we know that gcd(a^k, pq) = gcd(a,pq)^k = 1. This is a contradiction, and therefore our assumption must be false.

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The order of a in (z/(pq))× is exactly (p-1)(q-1), as desired.

To prove that (z/(pq))× has order (p − 1)(q − 1), we need to show that the least positive integer n such that (z/(pq))×n = 1 is (p − 1)(q − 1).

First, let's define (z/(pq))× as the set of all integers a such that gcd(a,pq) = 1 (i.e., a is relatively prime to pq) and a mod pq is also relatively prime to pq.

Now, we know that the order of an element a in a group is the smallest positive integer n such that a^n = 1. Therefore, we need to find the order of an arbitrary element a in (z/(pq))×.

Let's assume that a is an arbitrary element in (z/(pq))×. Since gcd(a,pq) = 1, we know that a has a multiplicative inverse modulo pq, denoted by a^-1. Therefore, we can write:

a * a^-1 ≡ 1 (mod pq)

Now, let's consider the order of a. Since gcd(a,pq) = 1, we know that a^(p-1) is congruent to 1 modulo p by Fermat's Little Theorem. Similarly, we can show that a^(q-1) is congruent to 1 modulo q. Therefore, we have:

a^(p-1) ≡ 1 (mod p)
a^(q-1) ≡ 1 (mod q)

Now, we can use the Chinese Remainder Theorem to combine these congruences and get:

a^(p-1)(q-1) ≡ 1 (mod pq)

Therefore, we know that the order of a must divide (p-1)(q-1).

To show that the order of a is exactly (p-1)(q-1), we need to show that a^k is not congruent to 1 modulo pq for any positive integer k such that 1 ≤ k < (p-1)(q-1).

Assume for contradiction that there exists such a k. Then, we have:

a^k ≡ 1 (mod pq)

This means that a^k is a multiple of pq, which implies that gcd(a^k, pq) ≥ pq. However, since gcd(a,pq) = 1, we know that gcd(a^k, pq) = gcd(a,pq)^k = 1. This is a contradiction, and therefore our assumption must be false.

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The probability of number of strings in two digits followed by a letter is 2600,

The probability of the mean contents of the 625 sample cans being less than 11.994 ounces can be calculated using the Z-score formula.

This formula takes into account the mean and standard deviation of the sample and the size of the sample.

The formula is Z = (x - μ) / (σ / √n),

Where,

x is the value we are looking for,

μ is the mean of the sample,

σ is the standard deviation of the sample and

n is the size of the sample.

In this case, x = 11.994, μ = 12, σ = 0.12, and n = 625.

The Z-score is then calculated to be -0.166, which corresponds to a probability of 0.106.

This means that there is a 0.106 probability that the mean contents of the 625 sample cans is less than 11.994 ounces.

The number of bit strings of length nine:

[tex]2^9[/tex] = 512 (Answer: 1)
The number of functions from a set with five elements to a set with four elements:

[tex]4^5[/tex] = 1024 (Answer: 2)
The number of one-to-one functions from a set with three elements to a set with eight elements:

8P3 = 8*7*6

= 336 (Answer: 3)
The number of strings of two digits followed by a letter:

10 X 10 X 26 = 2600 (Answer: 4)
So the correct matching is:
1 -> 1,
2 -> 2,
3 -> 3,
4 -> 4.

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Check the picture below.

so we have a cube with a pyramidical hole, so let's just get the volume of the whole cube and subtract the volume of the pyramid, what's leftover is the part we didn't subtract, the cube with the hole in it.

[tex]\textit{volume of a pyramid}\\\\ V=\cfrac{Bh}{3} ~~ \begin{cases} B=\stackrel{base's}{area}\\ h=height\\[-0.5em] \hrulefill\\ B=\stackrel{6\sqrt{2}\times 6\sqrt{2}}{72}\\ h=12 \end{cases}\implies V=\cfrac{72\cdot 12}{3}\implies 288 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{ \textit{\LARGE volumes} }{\stackrel{ cube }{12^3}~~ - ~~\stackrel{ pyramid }{288}}\implies \text{\LARGE 1440}~in^3[/tex]

Answer: (-7,4) is 11 units away.

Step-by-step explanation:

First we can see that (-7,4) is farther away on the coordinate plane.

Next, if we count the number of units from (-7,4) to (4,4) we count 11 units

There fore (-7,4) is 11 units away

To estimate the displacement of a moving object when given its velocity function, you can use the Definite Integral and Riemann Sum approaches.

Steps:

1. You're given the object's velocity function, which represents the rate of change of its position with respect to time.
2. To find the displacement, you need to calculate the total change in position over a given time interval. This can be done by finding the area under the velocity function curve within that time interval.
3. The Definite Integral approach allows you to find the exact area under the curve by integrating the velocity function over the specified time interval.
4. The Riemann Sum approach provides an approximation of the area under the curve by dividing the interval into smaller subintervals and summing up the areas of rectangles formed using the velocity function values at certain points within these subintervals.

Both of these approaches can help estimate the object's displacement, but the Definite Integral will give you a more accurate result, while the Riemann Sum provides an approximation that gets better as the number of subintervals increases. U-substitution is a method used for finding integrals, but it's not a direct approach to estimate displacement; it could be a part of the process if the velocity function requires it for integration.

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The given statement "The complement of the false positive rate is the sensitivity of a test" is true because the false positive rate is the proportion of negative instances incorrectly classified as positive, while sensitivity is the proportion of positive instances correctly identified.

False positive rate (FPR) is the proportion of negative instances incorrectly classified as positive, while sensitivity (also known as true positive rate or recall) is the proportion of positive instances correctly identified.

The complement of FPR is 1 - FPR, which is also known as specificity.

Specificity measures the proportion of negative instances correctly identified.

However, the statement would be false if it claimed that the complement of FPR is specificity.

The correct statement would be: the complement of the false positive rate is the specificity of a test.

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The given statement "The complement of the false positive rate is the sensitivity of a test" is true because the false positive rate is the proportion of negative instances incorrectly classified as positive, while sensitivity is the proportion of positive instances correctly identified.

False positive rate (FPR) is the proportion of negative instances incorrectly classified as positive, while sensitivity (also known as true positive rate or recall) is the proportion of positive instances correctly identified.

The complement of FPR is 1 - FPR, which is also known as specificity.

Specificity measures the proportion of negative instances correctly identified.

However, the statement would be false if it claimed that the complement of FPR is specificity.

The correct statement would be: the complement of the false positive rate is the specificity of a test.

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Step-by-step explanation:

Could you give a little more clearer explanation? I would be glad to help!

The outcomes with at least two tails are (T, T, H), (T, H, T), (H, T, T), and (T, T, T), which have a total of 4 outcomes.

The probability of getting at least two tails is 4/8 = 1/2. Therefore, the answer is (c) 1/2.

X can be expressed as a linear combination of v1 and v2 with the coordinates (a', b') in the basis B.  The coordinate vector [x]B:

[x]B = (a', b')

To determine whether the vector x is in the span V of vectors v1 and v2, we need to check if there exist scalar coefficients a and b such that:

x = a × v1 + b × v2

Given that x = [23 29], v1 = [46 58], and v2 = [61 67], the equation can be written as:

[23 29] = a × [46 58] + b × [61 67]

This equation can be represented in the form of a matrix:

| 46 61 | | a | = | 23 |
| 58 67 | | b | = | 29 |

We can now find the reduced row-echelon form of the augmented matrix to solve for a and b:

| 46 61 23 |
| 58 67 29 |

After row-reducing the matrix, we get:

| 1 0 a' |
| 0 1 b' |

Since the system has a unique solution, x is in the span V of vectors v1 and v2. We can now find the coordinates of x with respect to the basis B=(v1, v2) and write the coordinate vector [x]B:

[x]B = (a', b')

Therefore, x can be expressed as a linear combination of v1 and v2 with the coordinates (a', b') in the basis B.

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Joan should invest  $4720 annually in her sinking fund to have $250,000 when she retires

We can use the future value formula for an annuity to solve this problem:

FV = PMT * [(1 + r)^n - 1] / r

Where:

We want to find PMT, so we can rearrange the formula:

PMT = FV * r / [(1 + r)^n - 1]

Plugging in the values we know:

FV = $250,000

r = 0.04

n = 29

PMT = $250,000 * 0.04 / [(1 + 0.04)^29 - 1]

PMT = $250,000 * 0.04 / 22.718

PMT = $4720

So Joan should invest approximately $4720 annually in her sinking fund to have $250,000 when she retires in 29 years, assuming an interest rate of 4% compounded annually.

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The approximate probability that a randomly selected student spends at most 175 hours on the project is 0.7734, rounded to four decimal places.

To compute the approximate probability that a randomly selected student spends at most 175 hours on the project, we can use the normal approximation to the gamma distribution.

First, we need to find the mean and variance of the gamma distribution:

Mean = a×B = 40×4 = 160

Variance = a×B² = 40*4² = 640

Next, we can use the following formula to standardize the gamma distribution:

Z = (X - Mean) / √(Variance)

where X is the random variable following the gamma distribution.

For X <= 175 hours, we have:

Z = (175 - 160) / √(640) = 0.750

Using a standard normal distribution table or calculator, we can find the probability that Z is less than or equal to 0.750:

P(Z <= 0.750) = 0.7734

Therefore, the approximate probability that a randomly selected student spends at most 175 hours on the project is 0.7734, rounded to four decimal places.

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The measures of each term are; WS=39, WY=78, WZ=58, ZR=29, ZT=16 and YZ=32.

WE are given that Z is the centroid of triangle. Since centroid is the centre point of the object. The point in which the three medians of the triangle intersect is the centroid of a triangle.

Given WR=87 SY=39 and YT=48

WS=39

As WS=WR

WY=WS+SY

WY=39+39=78

WZ=58

Now, ZR=WR-WZ

ZR=87-48=29

ZT=16

Similalry;

YZ=YT-ZT

=48-16=32

YZ=32

Hence, the measures are; WS=39, WY=78, WZ=58, ZR=29, ZT=16 and YZ=32

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To find the radius of convergence, we can use the ratio test.

Let's apply the ratio test to the series:

sigma^[infinity]_n=1 4(−1)^n nx^n

The ratio test tells us that the series converges if the limit of the absolute value of the ratio of the (n+1)th term to the nth term is less than 1:

lim n -> infinity |a_n+1 / a_n| < 1

where a_n = 4*(-1)^n * n * x^n.

Let's compute the ratio of the (n+1)th term to the nth term:

|a_n+1 / a_n| = |4*(-1)^(n+1) * (n+1) * x^(n+1) / (4*(-1)^n * n * x^n)|

             = |(n+1) / n| * |x|

             = (n+1) / n * |x|

We want to find the values of x for which the above limit is less than 1. So we need to solve the inequality:

lim n -> infinity (n+1) / n * |x| < 1

Taking the limit as n approaches infinity, we get

lim n -> infinity (n+1) / n * |x| = |x|

lim n -> infinity (n+1) / n * |x| = |x|

So the inequality reduces to:

|x| < 1

Therefore, the radius of convergence R is 1.

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Step-by-step explanation:

See image

The conditional PDF of X given Y = y is a geometric distribution with parameter 1-p.

Let X be a geometric random variable with parameter p, and let Y be a uniform random variable on the interval [0,1], which means the PDF of Y is fY(y) = 1 for 0 ≤ y ≤ 1 and 0 otherwise. We want to find the conditional PDF of X given Y = y.

By Bayes' theorem, the conditional PDF of X given Y = y is given by:

fX|Y(x|y) = fY|X(y|x) fX(x) / fY(y)

where fX(x) is the PDF of X, which is given by fX(x) = (1-p)^(x-1) p for x = 1, 2, 3, ..., and fY|X(y|x) is the PDF of Y given X = x, which is given by fY|X(y|x) = 1 for 0 ≤ y ≤ p and 0 otherwise.

To find fY(y), we use the law of total probability:

fY(y) = ∑ fX(x) fY|X(y|x) for all x

Plugging in the values of fX(x) and fY|X(y|x), we get:

fY(y) = ∑ (1-p)^(x-1) p for 0 ≤ y ≤ p and 0 otherwise.

Since Y is uniform on [0,1], we have fY(y) = 1 for 0 ≤ y ≤ 1 and 0 otherwise. Therefore, the above sum simplifies to:

∑ (1-p)^(x-1) p = p / (1 - (1-p)) = 1

Now we can plug in the values of fY(y) and fX(x|y) into the formula for the conditional PDF of X given Y = y:

fX|Y(x|y) = fY|X(y|x) fX(x) / fY(y)

fX|Y(x|y) = (1/p) (1-p)^(x-1) p / 1 = (1-p)^(x-1)

Thus, the conditional PDF of X given Y = y is a geometric distribution with parameter 1-p.

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Answer:

Step-by-step explanation:

[tex]\frac{2^{-3}}{3^{-3} }[/tex]

=(-8)/(-27)

= 8/27

The simplified expression is 3a² + 3b² + 14a - 14b - 6ab -5.

We have,

3(a - b)² + 14(a - b)-5​

Simplifying the Expression as

Using Algebraic Identity

(a-b)² = a² -2ab + b²

So, 3 (a² -2ab + b²) + 14 (a-b) -5

= 3a² -6ab + 3b² + 14a - 14b -5

= 3a² + 3b² + 14a - 14b - 6ab -5

Thus, the simplified expression is 3a² + 3b² + 14a - 14b - 6ab -5.

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Step-by-step explanation:

they are similar, that means for our case here that they're is one central scaling factor for all sides between the 2 triangles.

by looking at the forms of both triangles, we see that

ET corresponds to TR.

FT corresponds to TS.

FE corresponds to RS.

for TE and TR we have the length information :

25 and 40

so, the scaling factor between these 2 corresponding sides can then be used for the other pairs of corresponding sides.

the scaling factor to go from the large to the small triangle is

25/40 = 5/8

therefore,

FT = TS × 5/8 = 22 × 5/8 = 11×5/4 = 55/4 = 13.75

FE = RS × 5/8 = 54 × 5/8 = 27×5/4 = 33.75

the perimeter of TFE is therefore

25 + 13.75 + 33.75 = 72.5

The reported value of [tex]nd 2 0 = 1.4110[/tex] for decane was obtained using light with a wavelength of [tex]589.3 nm[/tex].

The subscript "d" in the relative refractive index notation (nd) refers to the wavelength of light used to measure the refractive index. This notation is used to specify the particular wavelength of light used in a refractive index measurement.

When light passes through a medium, it is refracted or bent due to the change in speed of light as it passes from one medium to another. The amount of bending depends on the refractive index of the medium. The refractive index is a dimensionless quantity that describes how much the speed of light is reduced when it passes through a particular material. The refractive index of a material depends on the wavelength of light that is used to measure it.

Different wavelengths of light have different refractive indices when they pass through the same material. The refractive index of a material can be measured using different wavelengths of light, and the value obtained depends on the wavelength of light used. Therefore, it is essential to specify the wavelength of light used to measure the refractive index of a material.

In the case of decane, the subscript "d" in the relative refractive index notation (nd) stands for the "yellow doublet" line of sodium, which has a wavelength of 589.3 nanometers. Therefore, the reported value of [tex]nd 2 0 = 1.4110[/tex] for decane was obtained using light with a wavelength of [tex]589.3 nm[/tex].

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The total number of arrangements of the letters in the word "REPETITION" where the first occurrence of the letter E is before the first occurrence of the letter T is 120,960.. This can be answered by the concept of Combination.

To solve this problem, we can use the concept of permutations. The word "REPETITION" has a total of 10 letters.

Step 1: Calculate the total number of arrangements of all the letters without any restrictions.

The total number of arrangements of 10 letters without any restrictions can be calculated using the formula for permutations, which is n! (n factorial), where n is the number of items to be arranged. In this case, the total number of arrangements without any restrictions is 10! (10 factorial), which is equal to 3,628,800.

Step 2: Consider the restriction where the first occurrence of the letter E is before the first occurrence of the letter T.

In order to satisfy this restriction, we need to consider the positions of E and T in the arrangements. There are three possible cases:

Case 1: E is in the first position and T is in the second position.

In this case, we have fixed the positions of E and T, and the remaining 8 letters can be arranged in 8! (8 factorial) ways.

Case 2: E is in the first position and T is in a position other than the second.

In this case, we have fixed the position of E, and the position of T can be any of the remaining 8 positions. The remaining 8 letters can be arranged in 8! ways.

Case 3: E is not in the first position and T is in a position after E.

In this case, the position of T is fixed, and the position of E can be any of the positions before T. The remaining 8 letters can be arranged in 8! ways.

Step 3: Calculate the total number of arrangements that satisfy the restriction.

We add the total number of arrangements from each case calculated in Step 2:

8! + 8! + 8!

Step 4: Simplify the expression.

We can factor out 8! from the sum:

8! + 8! + 8! = 3 x 8!

Step 5: Calculate the final answer.

Substitute the value of 8! (8 factorial) into the expression:

3 x (8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) = 3 x 40,320 = 120,960

Therefore, the total number of arrangements of the letters in the word "REPETITION" where the first occurrence of the letter E is before the first occurrence of the letter T is 120,960.

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The sum of the tuple (1, 2, -2) and twice the tuple (-2, 3, 5) is (-3, 8, 8).

To calculate the sum of two tuples, we need to add the corresponding elements of the tuples. In this case, the tuples are (1, 2, -2) and twice (-2, 3, 5), which gives us (-4, 6, 10) when multiplied by 2. Then, we add the corresponding elements of both tuples: 1 + (-4) = -3 for the first element, 2 + 6 = 8 for the second element, and -2 + 10 = 8 for the third element.

Therefore, the sum of the two tuples is (-3, 8, 8).

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The given series is a geometric series with the formula: ∑ (1 + 6)^n / 7^n (from n=1 to infinity) In a geometric series,

The convergence or divergence is determined by the common ratio (r). In this case, the common ratio r is (1 + 6) / 7, which simplifies to 1.

Since the absolute value of the common ratio |r| is equal to 1, the series is inconclusive regarding convergence or divergence. Therefore, we need to use another test.

Notice that (1+6)/7 = 7/6 > 1. This means that the terms of the series do not approach zero as n approaches infinity. Therefore, the series sigma^[infinity]_n=1 (1+6)^n/7^n diverges by the divergence test. Therefore, the series does not have a sum.

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Let h(x) = x² - x. Then, for any f(x) = ax² + bx + c ∈ p2(r), we have h(0) = 0 and h′(1) = 1 - 1 = 0. Thus, g(h) = h(0)h′(1) = 0. This means that g is the zero functional on the subspace spanned by h.



In this question, we are given a linear functional g : p2(r) → r that is defined by g(f) = f(0)f′(1), where p2(r) is the space of polynomials of degree at most 2 with real coefficients. We need to find a polynomial h(x) ∈ p2(r) such that g(h) = 0 for any f(x) ∈ p2(r).

To find such an h(x), we can first consider the product f(0)f′(1) that appears in the definition of g. Since f(0) is the constant term of f(x) and f′(1) is the slope of the tangent to f(x) at x = 1, the product f(0)f′(1) measures the behavior of f(x) near x = 1.

Based on this observation, we can choose a polynomial h(x) that has a zero at x = 0 and a critical point at x = 1. One such polynomial is h(x) = x² - x, which has h(0) = 0 and h′(x) = 2x - 1, so h′(1) = 1.

Now, we can verify that g(h) = h(0)h′(1) = 0 for any f(x) ∈ p2(r). This is because, for any such f(x), we have f(0) = c and f′(1) = 2a + b, where f(x) = ax² + bx + c. Thus, g(f) = c(2a + b) = 2ac + bc, which is a linear function of a, b, and c.

Since g(h) = 0, we conclude that g is the zero functional on the subspace spanned by h(x). In other words, any polynomial that is a multiple of h(x) will be mapped to zero by g.

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Let h(x) = x² - x. Then, for any f(x) = ax² + bx + c ∈ p2(r), we have h(0) = 0 and h′(1) = 1 - 1 = 0. Thus, g(h) = h(0)h′(1) = 0. This means that g is the zero functional on the subspace spanned by h.



In this question, we are given a linear functional g : p2(r) → r that is defined by g(f) = f(0)f′(1), where p2(r) is the space of polynomials of degree at most 2 with real coefficients. We need to find a polynomial h(x) ∈ p2(r) such that g(h) = 0 for any f(x) ∈ p2(r).

To find such an h(x), we can first consider the product f(0)f′(1) that appears in the definition of g. Since f(0) is the constant term of f(x) and f′(1) is the slope of the tangent to f(x) at x = 1, the product f(0)f′(1) measures the behavior of f(x) near x = 1.

Based on this observation, we can choose a polynomial h(x) that has a zero at x = 0 and a critical point at x = 1. One such polynomial is h(x) = x² - x, which has h(0) = 0 and h′(x) = 2x - 1, so h′(1) = 1.

Now, we can verify that g(h) = h(0)h′(1) = 0 for any f(x) ∈ p2(r). This is because, for any such f(x), we have f(0) = c and f′(1) = 2a + b, where f(x) = ax² + bx + c. Thus, g(f) = c(2a + b) = 2ac + bc, which is a linear function of a, b, and c.

Since g(h) = 0, we conclude that g is the zero functional on the subspace spanned by h(x). In other words, any polynomial that is a multiple of h(x) will be mapped to zero by g.

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Answer: C) [tex]3x^3+ 6x^2 + 5x +10[/tex]

Step-by-step explanation:

Using the FOIL method since we're multiplying two binomials. We get...

F: Firsts

[tex]3x^2(x)[/tex]=[tex]3x^3[/tex]

O: Outside

[tex]3x^2(2) = 6x^2[/tex]

I: Insides

[tex]5(x) = 5x[/tex]

L: Lasts

[tex]5(2) = 10[/tex]

Put them together and we have [tex]3x^3+ 6x^2 + 5x +10[/tex]! These two are both equivalent since the form given in the question was factored form and this form is just the same thing expanded!

Answer:

C

Step-by-step explanation: