Answer:
Step-by-step explanation:
Given that
[tex]r(t) = (6cost, 6sint, t), 0\leq t\leq 2\pi\\r'(t) = (-6sint, 6cost, 1),\\||r'(t)||=\sqrt{(-6sint)^2 +(6cost)^2+1} =\sqrt{37}[/tex]
Hence arc length = [tex]\int\limits^a_b {||r'(t)||} \, dt[/tex]
Here a = 0 b = 2pi and r'(t) = sqrt 37
Hence integrate to get
[tex]\int\limits^{2\pi} _0 {\sqrt{37} } \, dt\\ =\sqrt{37} (t)\\=2\pi\sqrt{37}[/tex]
**Question 1\.\** The data were gathered by the following procedure, reported in the study. "Between January and June 1998, parents of children aged 2-16 years [...] that were seen as outpatients in a university pediatric ophthalmology clinic completed a questionnaire on the child’s light exposure both at present and before the age of 2 years." Was this study observational, or was it a controlled experiment? Explain.
Answer:
OBSERVATIONAL
Step-by-step explanation:
Let`s see the two kind of experiments mentioned:
Observational experiments: are those where individuals are observed or certain outcomes are measured. There is no intervention by the scientist, for example: no treatment at all is givenControlled experiments: Are those experiments that are directly manipulaed by the sciencist in order to study the controlled variable's reaction or change.In this case, the patient's parent only answer a questionnairewith no intervention from the doctors into the pacients at all, so it's an observational experiment
A closed rectangular container with a square base is to have a volume of 686 in3. The material for the top and bottom of the container will cost $4 per in2, and the material for the sides will cost $2 per in2. Find the dimensions of the container of least cost.
Answer:
Length = Width = 7 inches
Height = 14 inches
Step-by-step explanation:
Let's call
x = length and width of the bottom and top
y = heigth
The bottom and top are squares, so their total area is
[tex]A_{bt}=x^2+x^2=2x^2[/tex]
the area of one side is xy. As we have 4 sides, the total area of the sides is
[tex]A_s=4xy[/tex]
The cost for the bottom and top would be
[tex]C_{bt}=\$4.2x^2=\$8x^2[/tex]
The cost for the sides would be
[tex]C_s=\$2.4xy=\$8xy[/tex]
The total cost to made the box is
[tex]C_t=\$(8x^2+8xy)[/tex]
But the volume of the box is
[tex]V=x^2y=686in^3[/tex]
isolating
[tex]y=\frac{686}{x^2}[/tex]
Replacing this expression in the formula of the total cost
[tex]C_t(x)=8x^2+8x(\frac{686}{x^2})=8x^2+\frac{5488}{x}[/tex]
The minimum of the cost should be attained at the point x were the derivative of the cost is zero.
Taking the derivative
[tex]C'_t(x)=16x-\frac{5488}{x^2}[/tex]
If C' = 0 then
[tex]16x=\frac{5488}{x^2}\Rightarrow x^3=\frac{5488}{16}\Rightarrow x^3=343\Rightarrow x=\sqrt[3]{343}=7[/tex]
We have to check that this is actually a minimum.
To check this out, we take the second derivative
[tex]C''_t(x)=16+\frac{10976}{x^3}[/tex]
Evaluating this expression in x=7, we get C''>0, so x it is a minimum.
Now that we have x=7, we replace it on the equation of the volume to get
[tex]y=\frac{686}{49}=14in[/tex]
The dimensions of the most economical box are
Length = Width = 7 inches
Height = 14 inches
To find the dimensions of the container of least cost, assume the length of the base is x and the width of the base is also x. The height of the container is 686/(x^2). The dimensions of the container of least cost are approximately 10.33 inches by 10.33 inches by 6.68 inches.
Explanation:To find the dimensions of the container of least cost, we need to minimize the cost function. Let's assume the length of the base is x, then the width of the base will also be x since it's a square. The height of the container will be 686/(x^2) since the volume is given as 686 in³. The total cost, C, is given by C = 4(2x^2) + 2(4x(686/(x^2))). Simplifying this expression gives C = 8x^2 + 5488/x. To find the minimum cost, we can take the derivative of C with respect to x and set it equal to zero:
dC/dx = 16x - 5488/x^2 = 0
Solving this equation gives x = √(343/2) ≈ 10.33. Therefore, the dimensions of the container of least cost are approximately 10.33 inches by 10.33 inches by 6.68 inches.
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Find an equation of the circle that satisfies the given conditions. (Give your answer in terms of x and y.)
Center
(4, −5)
and passes through
(7, 4)
Answer:
[tex] (x-4)^2+(y+5)^2=90[/tex]
Step-by-step explanation:
The equation of a circle of radius r, centered at the point (a,b) is
[tex](x-a)^2+(y-b)^2=r^2[/tex]
We already know the center is at [tex](4,-5)[/tex], we are just missing the radius. To find the radius, we can use the fact that the circle passes through the point (7,4), and so the radius is just the distance from the center to this point (see attached image). So we find the distance by using distance formula between the points (7,4) and (4,-5):
radius[tex]=\sqrt{(7-4)^2+(4-(-5))^2}=\sqrt{3^2+9^2}=\sqrt{90}[/tex]
And now that we know the radius, we can write the equation of the circle:
[tex] (x-4)^2+(y-(-5))^2=\sqrt{90}^2[/tex]
[tex] (x-4)^2+(y+5)^2=90[/tex]
The percent increase in Americans in prison for drug related offenses from 1980 to 2015 was 1048%. In 1980 the number of Americans in prison for drug related offenses was 40,900. How many American's were in prison for drug related offenses in 2015?
Answer: There were 469,532 American prison for drug related offenses in 2015.
Step-by-step explanation:
Since we have given that
Number of Americans in prison for drug related offenses in 1980 = 40,900
Rate of increment in Americans in prison from 1980 to 2015 = 1048%
So, Number of Americans who were in prison for drug related offenses in 2015 is given by
[tex]\dfrac{100+1048}{100}\times 40900\\\\\dfrac{1148}{100}\times 40900\\\\=1148\times 409\\\\=469,532[/tex]
Hence, there were 469,532 American prison for drug related offenses in 2015.
For what value(s) of k will the relation not be a function?
A = {(1.5k−4, 7), (−0.5k+8, 15)}
Answer:
K=6
Step-by-step explanation:
we know that
A function is: a relation from a set of inputs to a set of possible outputs where each input is related to exactly one output
so
In this problem
If the x-coordinate of the first point is equal to the x-coordinate of the second point, then the relation will not be a function
Remember that the x-coordinate is the input value and the y-coordinate is the output value
Equate the x-coordinates and solve for k
[tex]1.5k-4=-0.5k+8[/tex]
[tex]1.5k+0.5k=8+4[/tex]
[tex]2k=12[/tex]
[tex]k=6[/tex]
therefore
If the value of k is 6 the, the relation will not be a function
Final answer:
To determine when the relation A is not a function, set the first elements of the ordered pairs equal to each other. Solving for k, we find that k must equal 6 to make the relation not a function, as this value of k would result in the same input for both pairs, violating the definition of a function.
Explanation:
For a relation to be considered a function, each input value must be associated with exactly one output value. Looking at the provided relation A = {(1.5k-4, 7), (-0.5k+8, 15)}, we can see that there are two ordered pairs. For A to not be a function, the first element of both ordered pairs must be the same because this would mean that a single input is associated with two different outputs, which violates the definition of a function.
We set the first elements equal to each other to find the value of k that would make the relation not a function: 1.5k - 4 = -0.5k + 8. Solving this equation, we add 0.5k to both sides and add 4 to both sides to obtain 2k = 12, and then divide both sides by 2 to find that k = 6.
Therefore, when k equals 6, the relation A will not be a function because both ordered pairs will have the first element equal to 5, which corresponds to two different output values (7 and 15).
Cantwell Associates, a real estate devel- oper,is planning to
build a new apartment complex con- sisting ofone-bedroomunits and
two- and three-bedroom townhouses. A total of192 units is planned,
and the number of family units (two- andthree-bedroom town- houses)
will equal the number of one-bedroomunits. If the number of
one-bedroom units will be three times thenumber of three-bedroom
units, find how many units of each typewill be in the complex.
Answer:
number of 1 bedroom units are 96
number of 1 bedroom units are 64
number of 1 bedroom units are 32
Step-by-step explanation:
Let the number of 1 bedroom units be 'a'
number of 2 bedroom units be 'b'
and,
number of 3 bedroom units be 'c'
now,
according to the question
a + b + c = 192 ................. (1)
also,
b + c = a ..............(2)
and,
a = 3c ...................(3)
now,
substituting value of 'a' from 3 into 2, we get
b + c = 3c
or
b = 2c ...................(4)
also,
from 3 and 1
3c + b + c = 192
or
4c + b = 192 ................(5)
now from 4 and 5,
4c + 2c = 192
or
6c = 192
or
c = 32 units
now, substituting c in equation 4, we get
b = 2 × 32 = 64 units
and, substituting c in equation (3), we get
a = 3 × 32 = 96 units
Therefore,
number of 1 bedroom units are 96
number of 1 bedroom units are 64
number of 1 bedroom units are 32
Based on the imperial unit system, currently in use only in Liberia, Myanmar, and the United States, engineers use the acre-foot, as a volume unit. It is defined as 1 acre of land to a depth of 1 ft. Last spring, a severe thunderstorm dumped 2.5 in. of rain in 30 min on the town of Avon, which has an area of 101 km2. What volume of water, in acre-feet, fell on the town?
In this type of problems what we have to do is unit conversion. In order to do so we need all the equivalences which we will be mentioning during the explanation of the problem:
First of all the answer is asked to be in acre-feet and we can see the data we are getting from the rain is in [tex]in*Km^{2}[/tex] not even a volume unit.
To calculate the volume of poured rain we need to have both numbers in the same units, we will convert [tex]Km^{2}[/tex] to [tex]in^{2}[/tex] using the equivalence [tex]1 Km^{2}=1550001600in^{2}[/tex] like this:
[tex]101Km^{2}*\frac{1550001600in^{2} }{1Km^{2}}=156550161600in^{2}[/tex]
it is possible now to calculate the volume ([tex]Volume_{cuboid}=Area*Height[/tex]) like this:
[tex]Volume_{cuboid}=156550161600in^{2}*2.5in=391375404000in^{3}[/tex]
Now we just need to convert this volume to acre-feet and we will do so using the equivalence [tex]1acre-foot=751271680in^{3}[/tex] like this:
[tex]391375404000in^{3}*\frac{1acre-foot}{751271680in^{3}}=5199.50403658 acre-feet[/tex]
5199.50403658 acre-feet would be the answer to our problem
Given: ABCD trapezoid, BK ⊥ AD , AB=DC AB=8, AK=4 Find: m∠A, m∠B
Answer:
[tex]m\angle B=m\angle C=120^{\circ}[/tex]
[tex]m\angle A=m\angle D=60^{\circ}[/tex]
Step-by-step explanation:
Trapezoid ABCD is isosceles trapezoid, because AB = CD (given). In isosceles trapezoid, angles adjacent to the bases are congruent, then
[tex]\angle A\cong \angle D;[/tex][tex]\angle B\cong \angle C.[/tex]Since BK ⊥ AD, the triangle ABK is right triangle. In this triangle, AB = 8, AK = 4. Note that the hypotenuse AB is twice the leg AK:
[tex]AB=2AK.[/tex]
If in the right triangle the hypotenuse is twice the leg, then the angle opposite to this leg is 30°, so,
[tex]m\angle ABK=30^{\circ}[/tex]
Since BK ⊥ AD, then BK ⊥ BC and
[tex]m\angle KBC=90^{\circ}[/tex]
Thus,
[tex]m\angle B=30^{\circ}+90^{\circ}=120^{\circ}\\ \\m\angle B=m\angle C=120^{\circ}[/tex]
Now,
[tex]m\angle A=m\angle D=180^{\circ}-120^{\circ}=60^{\circ}[/tex]
A penny collection contains twelve 1967 pennies, seven 1868 pennies and eleven 1971 pennies. If you are to pick some pennies without looking at the dates, how many must you pick to be sure of getting at least five pennies from the same year. Show work.
Answer:
You must pick at least 13 pennies to be sure of getting at least five from the same year.
Step-by-step explanation:
You have:
12 1967 pennies
7 1868 pennies
11 1971 pennies.
how many must you pick to be sure of getting at least five pennies from the same year?
This value is the multiplication of the number of different pennies by the antecessor of the number of pennies you want, added by 1.
So
You have 3 differennt pennies
You want to get at least five from the same year.
[tex]3*4 + 1 = 13[/tex]
You must pick at least 13 pennies to be sure of getting at least five from the same year.
For example, if you pick 12 pennies, you can have four from each year.
Adding three values, 13 is the smallest number that you need at least one term of the addition being equal or bigger than 5.
To be sure of getting at least five pennies from the same year, you need to pick 35 pennies in total.
Explanation:To ensure that you get at least five pennies from the same year, you need to consider the worst-case scenario. In this case, the worst-case scenario is where you pick pennies from each of the three different years first before getting five from the same year. So, you need to pick the maximum number of pennies from each year first before reaching the desired goal. The maximum number of pennies you need to pick is:
12 + 7 + 11 + 5 = 35 pennies
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Assume that MTA Sandwiches sells sandwiches for $2.85 each. The cost of each sandwich follows:
Materials $ 0.80
Labor 0.40
Variable overhead 0.40
Fixed overhead ($18,400 per month, 18,400 units per month) 1.00
Total cost per sandwich $ 2.60
One of MTA's regular customers asked the company to fill a special order of sandwiches at a selling price of $1.85 each for a fund-raising event sponsored by a social club at the local college. MTA has capacity to fill it without affecting total fixed costs for the month. MTA's general manager was concerned about selling the sandwiches below the cost of $2.60 per sandwich and has asked for your advice.
Required:
a. Prepare a schedule to show the impact on MTA's profits of providing 800 sandwiches in addition to the regular production and sales of 18,400 sandwiches per month. (Select option "higher" or "lower", keeping Status Quo as the base. Select "None" if there is no effect.)
b. Based solely on the data given, what is the lowest price per sandwich at which the special order can be filled without reducing MTA's profits? (Round your answer to 2 decimal places.)
Final answer:
Explaining the impact of selling additional sandwiches on MTA's profits.
Explanation:
To prepare the schedule showing the impact on MTA's profits:
Calculate the total income from selling 800 additional sandwiches: $2.85 x 800 = $2,280.
Calculate the total cost of producing 800 sandwiches: $5 x 800 = $4,000.
Subtract the cost from the income to find the profit impact: $2,280 - $4,000 = lower profit.
Explanation (150 words): Adding 800 sandwiches at a cost of $5 each while selling at $2.85 results in a loss, impacting MTA's profits negatively. This shift reduces the overall profitability due to the increased production costs outweighing the revenue generated from the additional sandwich sales.
25% of all who enters a race do not complete. 30 haveentered.
what is the probability that exactly 5 are unable tocomplete the
race?
Answer:
The probability that exactly 5 are unable to complete the race is 0.1047
Step-by-step explanation:
We are given that 25% of all who enters a race do not complete.
30 have entered.
what is the probability that exactly 5 are unable to complete the race?
So, We will use binomial
Formula : [tex]P(X=r) =^nC_r p^r q^{n-r}[/tex]
p is the probability of success i.e. 25% = 0.25
q is the probability of failure = 1- p = 1-0.25 = 0.75
We are supposed to find the probability that exactly 5 are unable to complete the race
n = 30
r = 5
[tex]P(X=5) =^{30}C_5 (0.25)^5 (0.75)^{30-5}[/tex]
[tex]P(X=5) =\frac{30!}{5!(30-5)!} \times(0.25)^5 (0.75)^{30-5}[/tex]
[tex]P(X=5) =0.1047[/tex]
Hence the probability that exactly 5 are unable to complete the race is 0.1047
Consider the following table.
x 1 3 5 7 9
y 9 7 7 3 2
(a) Find the equation of the least-squares line for the data.
Answer:
y = -0.9x+10.1
Step-by-step explanation:
The equation of the line is:
[tex]y=mx+b[/tex]
You have been asked to stimate m and b. To do so, first find the product between each pair of x and y and the value of x squared:
[tex]\left[\begin{array}{cccc}x&y&x*y&x^2\\1&9&9&1\\3&7&21&9\\5&7&35&25\\7&3&21&49\\9&2&18&81\end{array}\right][/tex]
Then calculate the total sum of all columns:
[tex]\left[\begin{array}{cccc}x&y&x*y&x^2\\1&9&9&1\\3&7&21&9\\5&7&35&25\\7&3&21&49\\9&2&18&81\\\bold{25}&\bold{28}&\bold{104}&\bold{165}\end{array}\right][/tex]
m can be calculated following the next equation:
[tex]m=\frac{\frac{\sum{xy}-\sum{y}}{n}}{\sum{x^2}-\frac{(\sum{x})^2}{n}}[/tex]
where n is the number of (x, y) couples (5 in our case).
Replacing the values calculated previously:
[tex]m=\frac{104-\frac{25*28}{5} }{165-\frac{25^2}{5} }=\frac{104-\frac{700}{5} }{165-\frac{625}{5} } = \frac{104-140}{165-125 } = \frac{-36}{40} = -0.9[/tex]
For b:
[tex]b=\bar{y}- m\bar{x}=\frac{\sum{y}}{n}-m\frac{\sum{x}}{n}=\frac{28}{5}-(-0.9)\frac{25}{5}= \frac{28}{5}+\frac{22.5}{5}=\frac{50.5}{5}=10.1[/tex]
In the figure attached you can see the points given and the stimated line.
what are the measures of angle A and angle B when angle A is
half as large as its complement, angle b?
Answer:
Angle A = 60º and Angle B = 30º
Step-by-step explanation:
We know that A and B are complementary angles, therefore ∠A + ∠B =90. (+)
On the other hand, ∠A is half as large as ∠B; this can be written algebraically as [tex]A=\frac{B}{2}[/tex]. (*)
If we substitute (*) in (+) we get:
[tex]\frac{B}{2}+B = 90\\\\ \frac{B+2B=180}{2} \\\\ B+2B=180\\\\ 3B=180\\\\ B=\frac{180}{3}\\\\ B=60[/tex]
And now we substitute the value of B in (+) and we get:
∠A+60 = 90
∠A = 90-60
∠A = 30
30 units Humulin R insulin in 300 mL of normal saline (NS) to infuse for 12 hours. (Round to the nearest tenth if applicable) a. How many units per hour will be infused? ________ b. How many milliliters per hour will be infused? ________
Answer:
Part 1.
In 12 hours 30 units of Humulin R insulin in 300 mL is to be infused.
So, per hour = [tex]\frac{30}{12}= 2.5[/tex] units are to be infused.
Part 2.
Now 30 units Humulin R insulin in 300 mL of normal saline (NS).
So, 2.5 units will be in : [tex]\frac{300\times2.5}{30}= 25[/tex] mL
Hence. 2.5 units Humulin R insulin in 25 mL of normal saline (NS).
2-23 Ace Machine Works estimates that the probability its lathe tool is properly adjusted is 0.8. When the lathe is properly adjusted, there is a 0.9 probability that the parts produced pass inspection. If the lathe is out of adjustment, however, the probability of a good part being produced is only 0.2. A part randomly chosen is inspected and found to be acceptable. At this point, what is the posterior probability that the lathe tool is properly adjusted?
Answer:
The posterior probability that the lathe tool is properly adjusted is 94.7%
Step-by-step explanation:
This can be formulated as the following problem:
What is the probability of B happening, knowing that A has happened.
It can be calculated by the following formula
[tex]P = \frac{P(B).P(A/B)}{P(A)}[/tex]
Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.
In your problem we have that:
-A is the probability that the part chosen is found to be acceptable.
The problem states that the probability its lathe tool is properly adjusted is 0.8. When it happens, there is a 0.9 probability that the parts produced pass inspection. There is also a 0.2 probability of the lathe is out of adjustment, when it happens the probability of a good part being produced is only 0.2.
So, P(A) = P1 + P2 = 0.8*0.9 + 0.2*0.2 = 0.72 + 0.04 = 0.76
Where P1 is the probability of a good part being produced when lathe tool is properly adjusted and P2 is the probability of a good part being produced when lathe tool is not properly adjusted.
- P(B) is the the probability its lathe tool is properly adjusted. The problem states that P(B) = 0.8
P(A/B) is the probability of A happening given that B has happened. We have that A is the probability that the part chosen is found to be acceptable and B is the probability its lathe tool is properly adjusted. The problem states that when the lathe is properly adjusted, there is a 0.9 probability that the parts produced pass inspection. So P(A/B) = 0.9
So, probability of B happening, knowing that A has happened, where B is the lathe tool is properly adjusted and A is that the part randomly chosen is inspected and found to be acceptable is:
[tex]P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.8*0.9}{0.76} = \frac{0.72}{0.76} = 0.947 = 94.7%[/tex]
The posterior probability that the lathe tool is properly adjusted is 94.7%
Using Bayes' Theorem, the posterior probability that the lathe tool is properly adjusted given a part passes inspection is approximately 0.947 (94.7%). We determined this by combining the prior probabilities and the conditional probabilities to find the total probability of a part passing inspection.
To find the posterior probability, we can use Bayes' Theorem. Let's define the events:
A : Lathe tool is properly adjustedA' : Lathe tool is out of adjustmentB : Part passes inspectionPrior Probabilities
The probability that the lathe tool is properly adjusted is 0.8. Therefore, P(A) = 0.8 and P(A') = 0.2.
Conditional Probabilities
When the lathe is properly adjusted, the probability that a part passes inspection is P(B|A) = 0.9.When the lathe is out of adjustment, the probability that a part passes inspection is P(B|A') = 0.2.Applying Bayes' Theorem:
Bayes' Theorem states:
[tex]\[P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}\][/tex]First, we need to find P(B), the total probability that a part passes inspection:
[tex]\[P(B) = P(B|A) \cdot P(A) + P(B|A') \cdot P(A')\][/tex]Substituting values:
[tex]\[P(B) = 0.9 \cdot 0.8 + 0.2 \cdot 0.2 = 0.72 + 0.04 = 0.76\][/tex]Now, using Bayes' Theorem, we can calculate P(A|B):
[tex]\[P(A|B) = \frac{0.9 \cdot 0.8}{0.76} = \frac{0.72}{0.76} \approx 0.947\][/tex]Thus, the posterior probability that the lathe tool is properly adjusted given that a part is found to be acceptable is approximately 0.947 (94.7%).
Reduce 1.256 g to micrograms, to milligrams, and to kilograms.
The conversion of 1.256 grams is as follows: 1,256,000 micrograms, 1256 milligrams, and 0.001256 kilograms.
Explanation:To convert 1.256 grams (g) to different units of mass, you use the following conversion factors:
Micrograms (μg): 1 g = 1 × 10⁶ μg. So, 1.256 g = 1.256 × 10⁶ μg = 1,256,000 μg. Milligrams (mg): 1 g = 1000 mg. Thus, 1.256 g = 1.256 × 10³ mg = 1256 mg. Kilograms (kg): 1 g = 1 × 10⁻³ kg. Hence, 1.256 g = 1.256 × 10⁻³ kg = 0.001256 kg. Learn more about Mass Conversion here:
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1.256 grams is equal to 1,256,000 micrograms, 1,256 milligrams, and 0.001256 kilograms. These conversions use basic multiplication and division of the metric system.
To convert 1.256 grams to different units, you should be familiar with the respective conversion factors:
1 gram (g) = 1,000 milligrams (mg)1 gram (g) = 1,000,000 micrograms (µg)1 gram (g) = 0.001 kilograms (kg)Steps to Convert:
Micrograms: Multiply the number of grams by 1,000,000:Therefore, the conversions are:
1.256 grams = 1,256,000 micrograms1.256 grams = 1,256 milligrams1.256 grams = 0.001256 kilogramsProve the following statement using a proof by contraposition. Yr EQ,s ER, if s is irrational, then r + 1 is irrational.
Answer:
I think that what you are trying to show is: If [tex]s[/tex] is irrational and [tex]r[/tex] is rational, then [tex]r+s[/tex] is rational. If so, a proof can be as follows:
Step-by-step explanation:
Suppose that [tex]r+s[/tex] is a rational number. Then [tex]r[/tex] and [tex]r+s[/tex] can be written as follows
[tex]r=\frac{p_{1}}{q_{1}}, \,p_{1}\in \mathbb{Z}, q_{1}\in \mathbb{Z}, q_{1}\neq 0[/tex]
[tex]r+s=\frac{p_{2}}{q_{2}}, \,p_{2}\in \mathbb{Z}, q_{2}\in \mathbb{Z}, q_{2}\neq 0[/tex]
Hence we have that
[tex]r+s=\frac{p_{1}}{q_{1}}+s=\frac{p_{2}}{q_{2}}[/tex]
Then
[tex]s=\frac{p_{2}}{q_{2}}-\frac{p_{1}}{q_{1}}=\frac{p_{2}q_{1}-p_{1}q_{2}}{q_{1}q_{2}}\in \mathbb{Q}[/tex]
This is a contradiction because we assumed that [tex]s[/tex] is an irrational number.
Then [tex]r+s[/tex] must be an irrational number.
Is 4320 perfect, abundant, or deficient? also perfect numbers? Explain why all positive multiples of 6 greater than 6 are aburndant numbers.
Final answer:
The number 4320 is an abundant number because the sum of its proper divisors exceeds 4320. All positive multiples of 6 greater than 6 are abundant because their smallest divisor, 6, is the sum of its proper divisors, and they have additional divisors that increase the sum.
Explanation:
To determine if the number 4320 is perfect, abundant, or deficient, we must compare the sum of its proper divisors (excluding the number itself) with the number. A perfect number is equal to the sum of its divisors. A number is abundant if the sum of its divisors is greater than the number, and it is deficient if the sum is less.
For 4320, the divisors are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 27, 30, 32, 36, 40, 45, 48, 54, 60, 64, 72, 80, 90, 96, 108, 120, 135, 144, 160, 180, 192, 216, 240, 270, 288, 320, 360, 432, 480, 540, 576, 720, 864, 1080, and 1440. Their sum is greater than 4320, so it is an abundant number.
Regarding positive multiples of 6 greater than 6 being abundant, the smallest divisor of such a multiple is always 6, which is already the sum of its divisors (1, 2, and 3). Since there are additional divisors beyond 1, 2, and 3, the sum of divisors must exceed the number, making it abundant.
Are the rational numbers closed under addition?
Answer:
Yes, rational numbers are closed under addition.
Step-by-step explanation:
Rational numbers are number that can be expressed in the form of fraction [tex]\frac{x}{y}[/tex], where x and y are integers and y ≠ 0.
Now, the closure property of addition of rational number states that if we add two rational number, then the sum of these two rational number will also be a rational number.
Let a and b be two rational number, then,
a+b = c, where c is the sum of and b
c is also a rational number.
Thus, rational numbers are closed under addition.
This can be explained with the help of a example.
[tex]\frac{1}{7} + \frac{2}{7}[/tex] = [tex]\frac{2}{7}[/tex]
It is clear that [tex]\frac{1}{7}, \frac{2}{7}, \frac{3}{7}[/tex] are rational numers.
Thus, rational numbers are closed under addition.
Prove that if n is a perfect square, then n+1 can never be a perfect square
Answer:
Proved
Step-by-step explanation:
To prove that if n is a perfect square, then n+1 can never be a perfect square
Let n be a perfect square
[tex]n=x^2[/tex]
Let [tex]n+1 = y^2[/tex]
Subtract to get
[tex]1 = y^2-x^2 =(y+x)(y-x)[/tex]
Solution is y+x=y-x=1
This gives x=0
So only 0 and 1 are consecutive integers which are perfect squares
No other integer satisfies y+x=y-x=1
A pile of newspapers in Ms McGrath's art class was 17 3/4 inches high .Each Consecutive week,for the next 5 weeks,the height of the pile of newspapers increased by 8 7/12inches,What was the height ,in inches,of the pile after 3 weeks?
Final answer:
The height of the pile after 3 weeks is 43 181/192 inches.
Explanation:
To find the height of the pile after 3 weeks, we need to add the increase in height for each week. The initial height of the pile is 17 3/4 inches. For each week, the height increases by 8 7/12 inches.
So after 1 week, the height is (17 3/4 + 8 7/12) inches.
After 2 weeks, the height is [(17 3/4 + 8 7/12) + 8 7/12] inches.
And after 3 weeks, the height is [((17 3/4 + 8 7/12) + 8 7/12) + 8 7/12] inches.
Let's calculate:
After 1 week: 17 3/4 + 8 7/12 = 26 43/48 inchesAfter 2 weeks: 26 43/48 + 8 7/12 = 34 89/96 inchesAfter 3 weeks: 34 89/96 + 8 7/12 = 43 181/192 inchesSo, after 3 weeks, the height of the pile of newspapers is 43 181/192 inches.
Solve the given differential equation by undetermined coefficients.
y''' − 3y'' + 3y' − y = ex − x + 21
Answer:
Y = [tex]e^{t}[/tex] + [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex] + t - 18
Step-by-step explanation:
y''' − 3y'' + 3y' − y = ex − x + 21
Homogeneous solution:
First we propose a solution:
Yh = [tex]e^{r*t}[/tex]
Y'h = [tex]r*e^{r*t}[/tex]
Y''h = [tex]r^{2}*e^{r*t}[/tex]
Y'''h = [tex]r^{3}*e^{r*t}[/tex]
Now we solve the following equation:
Y'''h - 3*Y''h + 3*Y'h - Yh = 0
[tex]r^{3}*e^{r*t}[/tex] - 3*[tex]r^{2}*e^{r*t}[/tex] + 3*[tex]r*e^{r*t}[/tex] - [tex]e^{r*t}[/tex] = 0
[tex]r^{3} - 3r^{2} + 3r - 1 = 0[/tex]
To solve the equation we must propose a solution to the polynomial :
r = 1
To find the other r we divide the polynomial by (r-1) as you can see
attached:
solving the equation:
(r-1)([tex]r^{2} - 2r + 1[/tex]) = 0
[tex]r^{2} - 2r + 1[/tex] = 0
r = 1
So we have 3 solution [tex]r_{1} = r_{2} =r_{3}[/tex] = 1
replacing in the main solution
Yh = [tex]e^{t}[/tex] + [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex]
The t and [tex]t^{2}[/tex] is used because we must have 3 solution linearly independent
Particular solution:
We must propose a Yp solution:
Yp = [tex]c_{1} (t^{3} + t^{2} + t + c_{4} )e^{t} + c_{2} t + c_{3}[/tex]
Y'p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + c_{1}( 3t^{2} + 2t + 1 )e^{t} + c_{2}[/tex]
Y''p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + c_{1}(6t + 2)e^{t}[/tex]
Y'''p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + 6c_{1}e^{t}[/tex]
Y'''p - 3*Y''p + 3*Y'p - Yp = [tex]e^{t} - t + 21[/tex]
[tex]6c_{1}e^{t} - 18c_{1} te^{t} - 6c_{1} e^{t} + 6c_{1} te^{t} + 9c_{1} t^{2} e^{t} + 3c_{1}e^{t} + 3c_{2} - c_{2} t - c_{3}[/tex] = [tex]e^{t} - t + 21[/tex]
equalizing coefficients of the same function:
- 12c_{1} = 0
9c_{1} = 0
3c_{1} = 0
c_{1} = 0
3c_{2} - c_{3} = 21 => c_{5} = [tex]\frac{1}{3}[/tex]
-c_{2} = -1
c_{2} = 1
c_{3} = -18
Then we have:
Y = [tex]e^{t}[/tex] + [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex] + t - 18
Some conference-goers saunter over to the Healthy Snack Box Machine, where they each choose one of five kinds of fruit, one of three herbal teas, and one of six flavors of wrap sandwich to get packed in a box. How many possible snack boxes are there?
Answer: There are 90 snack boxes.
Step-by-step explanation:
Given : The number of kinds of fruits = 5
The number of kinds of herbal teas = 3
The number of kinds of flavors of wrap sandwich = 6
Then by using the fundamental principal of counting, the number of possible snack are there will be :_
[tex]5\times3\times6= 90[/tex]
Therefore, the number of possible snacks = 90
Adalimumab (Humira), a recombinant human monoclonal antibody, is available in a prefilled syringe containing 40 mg/0.8 mL. Calculate the concentration of drug on a mg/mL basis.
Answer:
The concentration of the drug is 50mg/mL.
Step-by-step explanation:
This problem can be solved by a rule of three.
In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.
When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too.
When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease.
In this problem, as the number of mg increases, so does the number of mL. It means the we have a direct rule of three.
Tha problem states that the antibody is available in a prefilled syringe containing 40 mg/0.8 mL. Calculate the concentration of drug on a mg/mL basis.
The problem wants to know how many mg are there in a mL of the drug.
40mg - 0.8mL
x mg - 1 mL
0.8x = 40
[tex]x = \frac{40}{0.8}[/tex]
x = 50mg.
The concentration of the drug is 50mg/mL.
2. Perform the calculations 11540+5972 in BCD arithmetic.
Answer:
0001 0110 1110 1011 0010
Step-by-step explanation:
First decimal number = 11540
second decimal number = 5972
BCD of 0 to 9 decimal number
Decimal number BCD
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
8 1000
9 1001
Hence, we can write above two numbers in BCD form as
11540 => 0001 0001 0101 0100 0000
5972 => 0101 1001 0111 0010
So, sum of 11540 and 5972 in BCD form can be given by
0001 0001 0101 0100 0000
+ 01 01 1001 01 11 001 0
0001 01 10 11 10 10 11 001 0
So, the sum of 11540 and 5972 in BCD form is
0001 0110 1110 1011 0010
Drug Ordered: Drug E, 1 mg/kg, SubQ, q12h for treatment of DVT. Drug Available: Drug E, 40 mg/0.4 mL syringe. Patient weight: 167 lb. a. How many kilograms does the patient weigh? (Round to the nearest tenth) ________ b. How many milligrams should the patient receive per day? (Round to the nearest tenth) ________ c. How many milliliters should the patient receive per dose? (Round to the nearest hundredth) ________
Answer:
a) The patient weighs 75.15kg = 75.1kg, rounded to the nearest tenth.
b) The patient should 75.1mg a day of the drug.
c) The patient should receive 0.37mL per dose, rounded to the nearest hundreth.
Step-by-step explanation:
These problems can be solved by direct rule of three, in which we have cross multiplication.
a. How many kilograms does the patient weigh?
The problem states that patient weighs 167lb. Each lb has 0.45kg. So:
1 lb - 0.45kg
167 lb - xkg
[tex]x = 167*0.45[/tex]
[tex]x = 75.15[/tex]kg
The patient weighs 75.15kg = 75.1kg, rounded to the nearest tenth.
b. How many milligrams should the patient receive per day?
The drug has 1mg/kg. The patient weighs 75.1kg. So
1 mg - 1 kg
x mg - 75.1kg
[tex]x = 75.1[/tex]mg
The patient should 75.1mg a day of the drug.
c. How many milliliters should the patient receive per dose?
The drug is SubQ, q12h. This means that the drug is administered twice a day, so there are 2 doses. 75.1mg of the drug are administered a day. so:
2 doses - 75.1mg
1 dose - xmg
[tex]2x = 75.1[/tex]
[tex]x = \frac{75.1}{2}[/tex]
[tex]x = 37.5[/tex] SubQ, q12h
For each dose, the patient should receive 37.5mg. Each 40mg of the drug has 0.4mL. So:
40mg - 0.4ml
37.5mg - xmL
[tex]40x = 0.4*37.4[/tex]
[tex]x = \frac{0.4*37.4}{40}[/tex]
[tex]x = 0.374mL[/tex]
The patient should receive 0.37mL per dose, rounded to the nearest hundreth.
827 divieded by 26 with a fraction remainder
Solve the system of linear equations using the Gauss-Jordan elimination method. 2x1 − x2 + 3x3 = −10 x1 − 2x2 + x3 = −3 x1 − 5x2 + 2x3 = −7 (x1, x2, x3) =
Answer:
The solution is: [tex](x_{1}, x_{2}, x_{3}) = (1,0,-4)[/tex]
Step-by-step explanation:
The Gauss-Jordan elimination method is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.
We have the following system:
[tex]2x_{1} - x_{2} + 3x_{3} = -10[/tex]
[tex]x_{1} - 2x_{2} + x_{3} = -3[/tex]
[tex]x_{1} - 5x_{2} + 2x_{3} = -7[/tex]
This system has the following augmented matrix:
[tex]\left[\begin{array}{ccc}2&-1&3|-10\\1&-2&1|-3\\1&-5&2| -7\end{array}\right][/tex]
To make the reductions easier, i am going to swap the first two lines. So
[tex]L1 <-> L2[/tex]
Now the matrix is:
[tex]\left[\begin{array}{ccc}1&-2&1|-3\\2&-1&3|-10\\1&-5&2| -7\end{array}\right][/tex]
Now we reduce the first row, doing the following operations
[tex]L2 = L2 - 2L1[/tex]
[tex]L3 = L3 - L1[/tex]
So, the matrix is:
[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&3&1|-4\\0&-3&1| -4\end{array}\right][/tex]
Now we divide L2 by 3
[tex]L2 = \frac{L2}{3}[/tex]
So we have
[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&1&\frac{1}{3}|\frac{-4}{3}\\0&-3&1| -4\end{array}\right][/tex]
Now we have:
[tex]L3 = 3L2 + L3[/tex]
So, now we have our row reduced matrix:
[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&1&\frac{1}{3}|\frac{-4}{3}\\0&0&2| -8\end{array}\right][/tex]
We start from the bottom line, where we have:
[tex]2x_{3} = -8[/tex]
[tex]x_{3} = \frac{-8}{2}[/tex]
[tex]x_{3} = -4[/tex]
At second line:
[tex]x_{2} + \frac{x_{3}}{3} = \frac{-4}{3}[/tex]
[tex]x_{2} - \frac{4}{3} = -\frac{4}{3}[/tex]
[tex]x_{2} = 0[/tex]
At the first line
[tex]x_{1} -2x_{2} + x_{3} = -3[/tex]
[tex]x_{1} - 4 = -3[/tex]
[tex]x_{1} = 1[/tex]
The solution is: [tex](x_{1}, x_{2}, x_{3}) = (1,0,-4)[/tex]
Find a general solution of y" - 6y' +10y=0.
Answer:
[tex]y(x) = e^{3x} [Acos x+Bsin x][/tex]
Step-by-step explanation:
Given is a differential equation
[tex]y" - 6y' +10y=0.[/tex]
We have characteristic equation as
[tex]m^2-6m+10 =0[/tex]
The above quadratic cannot be factorised hence use formula
[tex]m=\frac{6+/-\sqrt{36-40} }{2} \\=3+i, 3-i[/tex]
Hence general solution would be
[tex]y(x) = e^{3x} [Acos x+Bsin x][/tex]
If venom A is four times as potent as venom B and venom C is 2.5 times as potent as venom B. How many times as potent is venom C compared to venom A?
Answer:
Venom C is 0.625 times as potent as venom A.
Step-by-step explanation:
We are given that venom A is four times as potent as venom B.
Venom C is 2.5 times as potent as Venom B.
We have to find that how many times as potent is venom C is compared to venom A.
Let Venom B =x
Then Venom A=[tex]4\times x=4x[/tex]
Venom C=[tex]2.5\times x=2.5 x[/tex]
[tex]\frac{Venom\;C}{venom\;A}=\frac{2.5x}{4x}=\frac{25}{40}=\frac{5}{8}=0.625[/tex]
Hence, venom C is 0.625 times as potent as venom A.
In relative terms, venom C is 0.625 times as potent as venom A, using venom B as a common comparison base.
Explanation:To solve this, we first need to establish a common comparison base. If venom A is four times as potent as venom B, let's assign a relative potency value to venom B, let's say 1. Therefore, the potency of venom A is 4. Similarly, if venom C is 2.5 times as potent as venom B, then the potency of venom C is 2.5.
To find out how much more potent venom C is compared to venom A, we divide the potency of venom C by the potency of venom A. Mathematically, it looks like this: Potency of C ÷ Potency of A = 2.5 ÷ 4 = 0.625. Therefore, venom C is 0.625 times as potent as venom A.
Learn more about Relative Potency here:https://brainly.com/question/31589467
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