pls help real quick its two questions

The given expressions can be written in the complex form a + bi as follows,
1. log(1) = 0 + 0i
2. log(-1) = 0 + πi
3. log(i) = 0 + (1/2)πi
4. ii = e^(-π/2) + 0i

The given expressions can be written in the form a + bi, where a and b are real numbers, and i is the imaginary unit.

1. log(1)
Since log(1) = 0, we can write it as 0 + 0i.

2. log(-1)
Using the complex logarithm, log(-1) can be expressed as πi, so it is 0 + πi.

3. log(i)
The complex logarithm of i is (1/2)πi, so we can write it as 0 + (1/2)πi.

4. ii
To find ii, we first need to express i in exponential form. i = \(e^{\frac{i\pi }{2} }\), so:
ii = \((e^{\frac{i\pi }{2} })^{i}\)
Using the power rule for exponentials (a^(mn) = (a^m)^n):
ii = \(e^{\frac{-\pi }{2} }\)
This can be written as \(e^{\frac{-\pi }{2} }\) + 0i.

So, the given expressions can be written in the form a + bi as follows:
1. log(1) = 0 + 0i
2. log(-1) = 0 + πi
3. log(i) = 0 + (1/2)πi
4. ii = \(e^{\frac{-\pi }{2} }\) + 0i

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Answer:

6x²-26x+31

Step-by-step explanation:

(x²-3x+5)(2x-7)​

=2x³-7x²-6x²+21x+10x-35

=2x³-13x²+31x-35

Now to differentiate:

dx=6x²-26x+31

Tara needs to earn n+20 score in the 5th equally weighted test that is 20 number higher than n.

Average number is calculated by adding all the values and then divided the sum by the total number of values. It is also said the middle number of a set.

Given, Tara gets average score "n" point on 4 equally weighted test.

After the 5th test the average score become "n+4"

let Tara achieve X score in the 5th test.

Now, calculating average score of 5 equally weighted test.

Scored "n" for the first 4 test so (n x 4 + X)/ 5 = "n+4"

                                        4n+X = 5n+20

                                      X = n+20

Hence, in the 5th test she earned "n+20" score which is 20 score higher than n.

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Answer:

46 own neither a cat or dog

Step-by-step explanation:

Add up the dog owners and the cat owners

65+17 = 82

Subtract those who subtract both, so they are not counted twice

82-4 = 78

The total is 124, so subtract the dog and cat owners from this number

124-78

46

The property of each given rational number operation are respectively; Commutative Property; Closure Property; Associative Property; Closure Property; Distributive Property

The main properties of rational numbers are:

a) The property used here is commutative property which says that;

a + b = b + a.

This tallies with the operation used on the rational numbers.

b) The property used here is called Closure Property. This is because for two rational numbers say a and b, the results of addition, subtraction and multiplication operations gives a rational number.

c) The property used here is called associative property because it states that; a * (b * c) = (a * b) * c.

d) The property used here is called Closure Property. This is because for two rational numbers say a and b, the results of addition, subtraction and multiplication operations gives a rational number.

e)  The property used here is called distributive property because it states that; a * (b * c) = (ab * ac)

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For a related-samples t test, the source of variability attributed to individual differences that remains after difference scores are computed is typically referred to as residual variance.

This residual variance can be thought of as the unexplained variability that is still present in the data even after accounting for the systematic differences between the two related groups. In order to accurately interpret the results of a related-samples t test, it is important to consider not only the mean difference between the two groups, but also the amount of residual variance that remains in the data. The source of variability attributed to individual differences that remains after difference scores are computed is the "within-subjects variability."

Lastly, we can address the within-subjects variability. Although the difference scores help to reduce the impact of individual differences, there will still be some within-subjects variability. This remaining variability is due to factors such as random error, measurement error, or individual response variability that are not completely accounted for by the difference scores. It is important to consider this variability when interpreting the results of a related-samples t-test.

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A) The biomass and glucose concentrations entering the second reactor are 9.29 g/L and 9.51 g/L.  B) The product and substrate concentrations in the effluent of the second reactor are 0.28 g/L and 8.66 g/L. C) The product and substrate concentrations leaving the second reactor are 8.36 g/L and 8.42 g/L.

A) Determine the biomass and glucose concentrations entering the second reactor.

The biomass and glucose concentrations entering the second reactor can be determined using the following equations:

\(X_e\) = \(X_i\) * exp(-\(H_m\) * t)

\(S_e\) = \(S_i\) - \(Y_x\)/s * \(X_i\) * exp(-\(H_m\) * t)

where:

\(X_e\) is the biomass concentration entering the second reactor (g/L)

\(X_i\) is the biomass concentration entering the first reactor (g/L)

\(S_e\) is the glucose concentration entering the second reactor (g/L)

\(S_i\) is the glucose concentration entering the first reactor (g/L)

\(H_m\) is the specific growth rate (h-1)

t is the residence time in the first reactor (h)

\(Y_x\)/s is the yield coefficient of biomass (g cells/g glucose)

In this case, we have:

\(X_i\) = 10 g/L

\(S_i\) = 10 g/L

\(H_m\) = 0.3 h-1

t = 40 / 600 = 0.067 h

Therefore, the biomass and glucose concentrations entering the second reactor are:

\(X_e\) = 10 * exp(-0.3 * 0.067) = 9.29 g/L

\(S_e\) = 10 - 0.4 * 10 * exp(-0.3 * 0.067) = 9.51 g/L

B) Assume there is negligible growth in the second stage and the specific rate of product formation is qp = 0.03 g P/g cell h, and Yps = 0.8 g P/g S. Determine the product and substrate concentrations in the effluent of the second reactor.

The product and substrate concentrations in the effluent of the second reactor can be determined using the following equations:

\(P_e\) = \(q_p\) * \(X_e\)

\(S_e\) = \(S_i\) - \(Y_p\)s * \(P_e\)

where:

\(P_e\) is the product concentration in the effluent of the second reactor (g/L)

\(q_p\) is the specific rate of product formation (g P/g cell h)

\(Y_{ps\) is the yield coefficient of product (g P/g S)

In this case, we have:

\(q_p\) = 0.03 g P/g cell h

\(Y_{ps\) = 0.8 g P/g S

\(X_e\) = 9.29 g/L

Therefore, the product and substrate concentrations in the effluent of the second reactor are:

\(P_e\) = 0.03 * 9.29 = 0.28 g/L

\(S_e\) = 9.51 - 0.8 * 0.28 = 8.66 g/L

C) If 10% of the cells are removed from the effluent of the second reactor and recycled into the feed of the first reactor; calculate the product and substrate concentrations leaving the second reactor.

The product and substrate concentrations leaving the second reactor can be determined using the following equations:

\(X_r\) = \(X_e\) - 0.1 * \(X_e\)

\(S_r\) = \(S_e\) - \(Y_{ps\) * \(P_e\) + 0.1 * \(Y_x\)/s * \(S_i\)

where:

\(X_r\) is the biomass concentration leaving the second reactor (g/L)

\(S_r\) is the substrate concentration leaving the second reactor (g/L)

In this case, we have:

0.1 * \(X_e\) = 0.1 * 9.29 = 0.929 g/L

0.1 * \(Y_x\)/s * \(S_i\) = 0.1 * 0.4 * 10 = 0.4 g/L

Therefore, the product and substrate concentrations leaving the second reactor are:

\(X_r\) = 9.29 - 0.929 = 8.36 g/L

\(S_r\) = 8.66 - 0.8 * 0.28 + 0.4 = 8.42 g/L

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Answer:

I am pretty sure the answer is  64

Step-by-step explanation

Volume=lxwxh

4x4x4=64

Answer: 13. 30

Step-by-step explanation:

13. Area = 18x * 10y = 180xy

Length = 6th

With ?

180xy = 6xy x width

180 x/6xy

Width = 30

14. 3^(9-5)= 3^4 = 81 the truck weighs 81 times as the driver

3^5

No, these two events cannot be disjoint. Disjoint events, also known as mutually exclusive events, are events that cannot occur simultaneously.

In other words, if events A and B are disjoint, it means that the occurrence of one event excludes the possibility of the other event occurring.

However, in this case, we are given that P(A) > 0 and P(B) > 0. This means that both events A and B have non-zero probabilities of occurring. Since both events have a positive probability, they cannot be disjoint because there is a possibility of their intersection, where both events occur simultaneously.

Disjoint events would have the property that P(A ∩ B) = 0, indicating that their intersection is an empty set. But given that P(A) > 0 and P(B) > 0, it implies that there is some non-zero probability for the events A and B to occur together, making them not disjoint.

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1) Considering the equation

3x -10 = -16 Add 10 to both sides

3x -10+10= -16 +10

Answer:

12 oak trees

Step-by-step explanation:

1/4 of 48 =12

Answer:

A 38.91

Step-by-step explanation:

distance formula = d= √((x2-x1)²+(y2-y1)²)

Answer:

60

Step-by-step explanation:

\(x \: + \: y \: + 30 \: + 30 \: = 180 \: \: \: \: \: \: \: \: \: (1)\\ \\ y + 30 = 90 \\ y = 90 - 30 \\ y = 60 \\ \\ (1) \: \: \: \: \: \: \: \: \: \: \: \: \: x + 60 + 30 + 30 = 180 \\ x = 180 - 120 \\ x = 60\)

Answer:

  25.2 gallons per week

Step-by-step explanation:

  \(\dfrac{12\text{ cups}}{5\text{ h}}\times\dfrac{24\text{ h}}{1\text{ da}}\times\dfrac{7\text{ da}}{1\text{ wk}}\times\dfrac{1\text{ gal}}{16\text{ cups}}=\dfrac{12\cdot24\cdot7\text{ gal}}{5\cdot16\text{ wk}}=\boxed{25.2\text{ gal/wk}}\)

Answer:

c

Step-by-step explanation:

Answer:

So first we have to figure out what $1200 per month for 12 months is since it was a year. 1200x12=14,400 a year which is outrageous for an apartment just saying.

Then we find out what 12% of the 1200 is and then add it to 1200.

To find the percentage of that we need to know how many times 12 fits into 100% 12 fits into 100 a total of 8.33333333333 times.

So now we divide 1200 by 8.33333333333 to see what 12% of it is.

1200 divided by 8.33333333333 is 144. So 1200+144=1344.

Now since one of the nine years has passed already we only have 8 remaining.

Now we see what 1344 a month for 12 months is 16,128 a year.

Now we multiply 16,128 by 8 to get the outcome of 8 more years. So 16,128x8=129,024

Thats it pls mark brainliest

Step-by-step explanation:

We should use a sample size of at least 154 to estimate the mean fuel efficiency for the new type of motorcycle within 0.5 mpg. The confidence interval in Part A is between 60.725 and 65.555 mpg.

To construct a 95% confidence interval for the mean fuel efficiency for this vehicle, we can use the formula:

Confidence interval = sample mean ± (critical value) * (standard deviation / sqrt(sample size))

First, let's calculate the sample mean and standard deviation for the given data:

Sample mean = (57.5 + 62.3 + 67.2 + 58.1 + 59.6 + 71.9 + 55.2 + 57.6 + 71.9 + 68.8 + 69.1 + 65.3 + 62.3 + 64.7 + 59.9 + 62.1 + 63.5 + 62.9 + 66.7 + 61.3) / 20

= 63.14

Sample standard deviation = √(((57.5 - 63.14)² + (62.3 - 63.14)² + (67.2 - 63.14)² + ... + (61.3 - 63.14)²) / 19)

= 4.97

Next, we need to find the critical value for a 95% confidence level. Since the sample size is small (20), we will use a t-distribution. The degrees of freedom for this calculation is 20 - 1 = 19. Consulting a t-table or using statistical software, we find the critical value to be approximately 2.093.

Plugging the values into the formula, we get:

Confidence interval = 63.14 ± (2.093) * (4.97 / √(20)) ≈ 63.14 ± 2.415

Interpreting the confidence interval, we can say with 95% confidence that the true mean fuel efficiency for this vehicle falls between 60.725 and 65.555 mpg.

To determine if the new motorcycle has greater fuel efficiency than the previous model, we can perform a hypothesis test. The null hypothesis (H0) is that there is no difference in fuel efficiency between the two models, and the alternative hypothesis (Ha) is that the new model has greater fuel efficiency.

We can use a t-test to compare the means of two independent samples. The test statistic is calculated as:
t = (sample mean 1 - sample mean 2) / sqrt((sample variance 1 / sample size 1) + (sample variance 2 / sample size 2))

For this case, the sample mean for the new model is 65.9 mpg, and the sample standard deviation is 3.2 mpg. The sample mean and standard deviation for the previous model were calculated in Part A. We can use the same sample size of 20 for both models.

Calculating the test statistic:

t = (65.9 - 63.14) / √((4.97² / 20) + (3.2² / 20))

t ≈ 1.376

Looking up the critical value for a t-distribution with 20 - 1 = 19 degrees of freedom at a 95% confidence level, we find it to be approximately 2.093.

Since the test statistic (1.376) is less than the critical value (2.093), we do not have enough evidence to convincingly claim that the new model has greater fuel efficiency than the previous model.

To construct a 98% confidence interval for the mean fuel efficiency of the older model, we can use the same formula as in Part A. The only difference is the critical value. For a 98% confidence level, the critical value for a t-distribution with 20 - 1 = 19 degrees of freedom is approximately 2.861.

Plugging the values into the formula:

Confidence interval = 63.14 ± (2.861) * (4.97 / √(20))

≈ 63.14 ± 3.303

Interpreting the confidence interval, we can say with 98% confidence that the true mean fuel efficiency for the older model falls between 59.837 and 66.443 mpg.

The change in confidence level from 95% to 98% does not change the conclusion in Part B. The hypothesis test is based on comparing the test statistic to the critical value, and the confidence interval is based on the sample mean and standard deviation. The change in confidence level does not alter these values or the comparison between them.

To estimate the mean fuel efficiency for the new type of motorcycle within 0.5 mpg, we need to calculate the sample size required. The formula for the sample size is:

Sample size = (Z-score * standard deviation / margin of error)²

Plugging in the values:

Z-score = critical value for the desired confidence level (let's assume 95%)

Standard deviation = 3.2 mpg (from the given data)

Margin of error = 0.5 mpg

Using a Z-score of approximately 1.96 for a 95% confidence level:

Sample size = (1.96 * 3.2 / 0.5)² ≈ 153.86

We should use a sample size of at least 154 to estimate the mean fuel efficiency for the new type of motorcycle within 0.5 mpg.

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Answer:

See below for proof.

\(S_{500}=125250\)

Step-by-step explanation:

Given arithmetic series:

Therefore:

\(S_n=1+2+3+...+(n-2)+(n-1)+n\)

\(S_n=1+(1+1)+(1+2)+...+(1+n-3)+(1+n-2)+(1+n-1)\)

\(S_n=1+(1+1)+(1+2(1))+...+(1+(n-3)(1))+(1+(n-2)(1))+(1+(n-1)(1))\)

Let:

Therefore:

\(S_n=a+(a+d)+(a+2d)+...+(a+(n-3)d)+(a+(n-2)d)+(a+(n-1)d)\)

Reverse the order:

\(S_n=(a+(n-1)d)+(a+(n-2)d)+(a+(n-3)d)+...+(a+2d)+(a+d)+a\)

Add the two expressions for Sₙ:

\(2S_n=(2a+(n-1)d)+(2a+(n-1)d)+(2a+(n-1)d)+...+(2a+(n-1)d)\)

Therefore, the term (2a + (n – 1)d) has been repeated n times:

\(2S_n=n(2a+(n-1)d)\)

Divide both sides by 2:

\(S_n=\dfrac{1}{2}n(2a+(n-1)d)\)

\(S_n=\dfrac{1}{2}n(a+a+(n-1)d)\)

Replace  a with  a₁  (first term) and a + (n – 1)d  with  aₙ (last term):

\(S_n=\dfrac{1}{2}n(a_1+a_n)\)

To find the sum of the series 1 + 2 + 3 + ... + 500, substitute the following values into the formula:

Therefore:

\(\implies S_{500}=\dfrac{1}{2}(500)(1+500)\)

\(\implies S_{500}=250(501)\)

\(\implies S_{500}=125250\)

The volume of oil exiting the pipe is approximately 100 cu in/hr, 2,400 cu in/day, and 1.39 cu ft/day when converting cu in/day to cubic feet per day.



To calculate the volume of oil exiting the pipe every hour, you would need to know the flow rate of the oil in cubic inches per hour. Let's assume the flow rate is 100 cubic inches per hour.To find the volume of oil exiting the pipe every day, you would multiply the flow rate by the number of hours in a day. There are 24 hours in a day, so the volume of oil exiting the pipe every day would be 100 cubic inches per hour multiplied by 24 hours, which equals 2,400 cubic inches per day.

To convert the volume from cubic inches per day to cubic feet per day, you would need to divide the volume in cubic inches by the number of cubic inches in a cubic foot. There are 1,728 cubic inches in a cubic foot. So, dividing 2,400 cubic inches per day by 1,728 cubic inches per cubic foot, we get approximately 1.39 cubic feet per day.

Therefore, the volume of oil exiting the pipe is approximately 100 cubic inches per hour, 2,400 cubic inches per day, and 1.39 cubic feet per day.

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Answer: The missing side length of the quadrilateral is 13x - 16.

Step-by-step explanation:

We are given that the perimeter (P) of the quadrilateral is 31x - 12.

We are also given 3 of the 4 side lengths of the quadrilateral.

We are asked to find the missing side length of the quadrilateral.

To find the missing side length we should:

Step 1: Add the given sides:

7x + 9 + 13x - 15 + 10 - 2x

After combining like terms you get,

18x + 4

Step 2: Subtract -> Perimeter - total side lengths

(31x - 12) - (18x + 4)

distribute the negative:

31x - 12 - 18x - 4

Combine liker terms:

13x - 16

The missing side length of the quadrilateral is 13x - 16.

Answer:

3

Step-by-step explanation:

m= 18

m/6 --> 18/6 = 3

Set up as a system of equations,

Let x represent the number of 4 credit courses and y represent the 3 credit courses.

The total number of courses the student took was 15. So we can say\(x+y=15\)

The total credits earned was 54. So we can say, \(4x+3y=54\)

Now we have the following system of equations, solve for the value of x.

\(\left \{ {{x+y=15} \atop {4x+3y=54}} \right.\)

Take the top equation and solving for y. We get,

=> \(x+y=15\)

=>\(y=15-x\)

Take the value of y above and plug it into the second equation.

=> \(4x+3y=54\)

=> \(4x+3(15-x)=54\)

=> \(4x+45-3x=54\)

=> \(x+45=54\)

=> \(x=54-45\)

=> \(x=9\)

Thus, the student took 9, 4 credit classes.

Answer:

4%

Answer:

72 bucks

Step-by-step explanation:

pints to quarts 6/2=3

quarts to money 3*24=72

Answer:

ur a god

Step-by-step explanation:

extraextraextra

Answer: thanks for the points

Step-by-step explanation:

We need radius

Let that be r

Answer:

12 feet

Step-by-step explanation:

It asks how many feet away from the sprinkler can water be spread. Therefore, in this scenario, we require the radius.

Area of a circle :

Area (circle) = πr²

Solving :

⇒ 452.16 = 3.14 × r²

⇒ r² = 452.16 ÷ 3.14

⇒ r = √144

⇒ r = 12 feet