Answer:
F(x)=[tex]\frac{3^x}{ln(3)}[/tex]+7cosh(x)+C
Step-by-step explanation:
The function is f(x)=3ˣ+7sinh(x), so we can define it as f(x)=g(x)+h(x) where g(x)=3ˣ and h(x)=7sinh(x).
Now we have to find the most general antiderivative of the function this means that we have to calculate [tex]\int\ {f(x)} \, dx[/tex] wich is the same as [tex]\int\ {(g+h)(x)} \, dx[/tex]
The sum rule in integration states that the integral of a sum of two functions is equal to the sum of their integrals. Then,
[tex]\int\ {(g+h)(x)} \, dx[/tex] = [tex]\int\ {g(x)} \, dx + \int\ {h(x)} \, dx[/tex]
1- [tex]\int\ {g(x)} \, dx =[/tex][tex]\int\ {3^x} \, dx = \frac{3^x}{ln(3)}+C[/tex] this is because of the rule for integration of exponencial functions, this rule is:
[tex]\int\ {a^x} \, dx =\frac{a^x}{ln(x)}[/tex], in this case a=3
2-[tex]\int\ {h(x)} \, dx =[/tex][tex]\int\ {7sinh(x)} \, dx =7\int\ {sinh(x)} \, dx =7cosh(x)+C[/tex] , number seven is a constant (it doesn´t depend of "x") so it "gets out" of the integral.
The result then is:
F(x)= [tex]\int\ {(h+g)(x)} \, dx=\int\ {h(x)} \, dx +\int\ {g(x)} \, dx[/tex]
[tex]\int\ {3^x} \, dx +\int\ {7sinh(x)} \, dx = \frac{3^x}{ln(3)} +7cosh(x) + C[/tex]
The letter C is added because the integrations is undefined.
The population of a mining city grows at a rate proportional to that population, in two years the population has doubled and a year later there were 10,000 inhabitants.
What was the initial population?
Answer:
The initial population was approximatedly 3535 inhabitants.
Step-by-step explanation:
The population of the city can be given by the following differential equation.
[tex]\frac{dP}{dt} = Pr[/tex],
In which r is the rate of growth of the population.
We can solve this diffential equation by the variable separation method.
[tex]\frac{dP}{dt} = Pr[/tex]
[tex]\frac{dP}{P} = r dt[/tex]
Integrating both sides:
[tex]ln P = rt + c[/tex]
Since ln and the exponential are inverse operations, to write P in function of t, we apply ln to both sides.
[tex]e^{ln P} = e^{rt + C}[/tex]
[tex]P(t) = Ce^{rt}[/tex]
C is the initial population, so:
[tex]P(t) = P(0)e^{rt}[/tex]
Now, we apply the problem's statements to first find the growth rate and then the initial population.
The problem states that:
In two years the population has doubled:
[tex]P(2) = 2P(0)[/tex]
[tex]P(t) = P(0)e^{rt}[/tex]
[tex]2P(0) = P(0)e^{2r}[/tex]
[tex]2 = e^{2r}[/tex]
To isolate r, we apply ln both sides
[tex]e^{2r} = 2[/tex]
[tex]ln e^{2r} = ln 2[/tex]
[tex]2r = 0.69[/tex]
[tex]r = \frac{0.69}{2}[/tex]
[tex]r = 0.3466[/tex]
So
[tex]P(t) = P(0)e^{0.3466t}[/tex]
In two years the population has doubled and a year later there were 10,000 inhabitants.
[tex]P(3) = 10,000[/tex]
[tex]P(t) = P(0)e^{0.3466t}[/tex]
[tex]10,000= P(0)e^{0.3466*3}[/tex]
[tex]P(0) = \frac{10,000}{e^{1.04}}[/tex]
[tex]P(0) = 3534.55[/tex]
The initial population was approximatedly 3535 inhabitants.
Disks of polycarbonate plastic from a supplier are analyzed for scratch and shock resistance. The results from 100 disks are summarized as follows:
shock resistance
scratch resistance high low
high 70 9
low 16 5
Let A denote the event that a disk has high shock resistance, and let B denote the event that a disk has high scratch resistance. If a disk is selected at random, determine the following probabilities. Input your answers in the fractional form (do not simplify).
P(A)=86/100
P(B)=79/100
P(A')=7/50
P(A U B)=95/100
P(A' U B)= ???
Answer:
0.84
Step-by-step explanation:
Given that Disks of polycarbonate plastic from a supplier are analyzed for scratch and shock resistance. The results from 100 disks are summarized as follows:
P(A) = 0.86, P(B) = 0.79, P(A') = 0.14, P(AUB) = 0.95
We are to find out P(A'UB)
We have
[tex]P(AUB) =P(A)+P(B)-P(A\bigcap B)\\0.95=0.86+0.79-P(A\bigcap B)\\P(A\bigcap B)=0.70[/tex]
[tex]P(A'UB) = P(A')+P(B)-P(A' \bigcap B)\\= 1-P(A) +P(B)-[P(B)-P(A \bigcap B)]\\= 1-0.86+0.79-P(B)+[tex]P(A'UB)=0.14+0.79-0.79+0.70\\=0.84[/tex]P(A \bigcap B)[/tex]
Given that [tex]z=x+iy[/tex], find the value of x and of y such that [tex]z+4iz^*=-3+18i[/tex] where z* is the complex conjugate of z.
Answer:
[tex]z= 5-2i[/tex]
Step-by-step explanation:
Start replacing!
[tex](x+iy) + 4i (x-iy) = -3+18i \\ x+iy +4ix -4i^2 y = -3 + 18i \\\\x + i (4x+y) +4y = -3 + 18i\\x+4y + (4x+y) i = -3 + 18i[/tex]
Now two complex numbers are equal if both the real and imaginary parts are equal, which gives you the system of equation [tex]\left \{x+4y=-3} \atop {4x+y=18} \right.[/tex]
Pick any method to solve it and you'll get [tex] x=5, y= -2 [/tex]
PLEASE HELP I AM ON A TIME LIMIT
Answer:
(a) even: J, K, M, O; odd: L, N
(b) L and O are connected to J
(c) N is of degree 3
Step-by-step explanation:
Count the edge ends that intersect each vertex. You get ...
J-2, K-2, L-3, M-2, N-3, O-4
These numbers are the degree of the vertex.
a) Vertices with even degree are J, K, M, O, since these have degrees of 2, 2, 2, and 4, respectively--all even numbers.
Vertices with odd degree are L and N, since these both have degree 3, an odd number.
__
b) The vertices that are adjacent to J are the ones at the other ends of the edges that intersect vertex J. One of those two edges connects to vertex L; the other to vertex O. J is adjacent to L and O.
__
c) When we counted edges in part (a), we found vertex N to be of degree 3.
The recommended dose for a brand nicotine patch is one 21-mg dose per day for 6 weeks, followed by 14 mg per day for 2 weeks, and then 7 mg per day for 2 more weeks. What total quantity, in grams, would a patient receive during this course of treatment?
Answer:
1.176 grams
Step-by-step explanation:
Given:
Recommended dose
21 mg per day for 6 weeks
Now,
1 week = 7 days
Thus,
number of days in 6 weeks = 6 × 7 = 42 days
Therefore, the total dose = dose per days × number of days
= 21 × 42 = 882 mg
further,
14 mg per day for 2 weeks
Now,
1 week = 7 days
Thus,
number of days in 2 weeks = 2 × 7 = 14 days
Therefore, the total dose = dose per days × number of days
= 14 × 14 = 196 mg
further,
7 mg per day for 2 weeks
Now,
1 week = 7 days
Thus,
number of days in 6 weeks = 2 × 7 = 14 days
Therefore, the total dose = dose per days × number of days
= 7 × 14 = 98 mg
Hence, the total dose = 882 + 196 + 98 = 1176 mg
also,
1 g = 1000 mg
thus,
1176 mg = 1.176 grams
total quantity received during this course is 1.176 grams
The eye of a hurricane passes over Grand Bahama Island in a direction 60.0° north of west with a speed of 43.5 km/h. Three hours later, the course of the hurricane suddenly shifts due north, and its speed slows to 23.5 km/h. How far from Grand Bahama is the hurricane 4.95 h after it passes over the island?
Answer:
D = 170.6Km
Step-by-step explanation:
First of all, we set the reference (origin) at Grand Bahama.
Nw, from the first displacement of 3h we calculate the distance:
D1 = V1*t = 43.5 * 3 = 130.5 Km
The coordinates of this new location is given by:
r1 = ( -D1*cos(60°); D1*sin(60°)) = (-62.5; 158.835) Km
For the second displacement, the duration was of 1.95 hours (4.95 -3), so the distance traveled was:
D2 = V2*t = 23.5 * 1.95 = 45.825 Km
The coordinates of this new location is given by:
r2 = r1 + ( 0; D2) = (-62.25; 158.835) Km
Now we just need to calculate the magnitude of that vector to know the distance to Grand Bahama:
[tex]Dt = \sqrt{D_{2x}^{2}+D_{2y}^{2}}=170.6 Km[/tex]
Yusuf has between 10 and 20 toffees. If he counts his toffees in 4s, he has 3 left over. If he counts his toffees in 5s, he has 4 left over. How many toffees has Yusuf got?
Answer:
Yusuf got 19 toffees
Step-by-step explanation:
The problem tells us that if Yusuf counts his toffees in 4's, he has 3 left. Let's call the number of toffees "t", then we know that [tex]t[/tex]≡[tex]3(mod4)[/tex]
We also know that [tex]t[/tex]≡[tex]4(mod 5)[/tex]
Between 10 and 20, there are only 2 numbers t≡[tex]4(mod 5)[/tex], these numbers are 14 and 19.
However, 14≡[tex]2(mod 4)[/tex] so it doesn't add up.
On the other hand, 19 ≡3[tex](mod4)[/tex] and therefore, this is the answer.
Answer:
19 toffees.
Step-by-step explanation:
It is given that Yusuf has between 10 and 20 toffees.
Let n be the number of toffees.
If he counts his toffees in 4s, he has 3 left over.
[tex]n\equiv 3(mod4)[/tex] ... (1)
So, the possible values of n between 10 and 20 are 11, 15 and 19.
If he counts his toffees in 5s, he has 4 left over.
[tex]n\equiv 4(mod5)[/tex] ... (2)
So, the possible values of n between 10 and 20 are 14 and 19.
From (1) and (2) the possible value of n between 10 and 20 is 19.
Therefore, Yusuf has 19 toffees.
A solution initially contains 200 bacteria. 1. Assuming the number y increases at a rate proportional to the number present, write down a differential equation connecting y and the time t. 2. If the rate of increase of the number is initially 100 per hour, how many bacteria are there after 2 hours? Solution:
Answer:
1.[tex]\frac{dy}{dt}=ky[/tex]
2.543.6
Step-by-step explanation:
We are given that
y(0)=200
Let y be the number of bacteria at any time
[tex]\frac{dy}{dt}[/tex]=Number of bacteria per unit time
[tex]\frac{dy}{dt}\proportional y[/tex]
[tex]\frac{dy}{dt}=ky[/tex]
Where k=Proportionality constant
2.[tex]\frac{dy}{y}=kdt[/tex],y'(0)=100
Integrating on both sides then, we get
[tex]lny=kt+C[/tex]
We have y(0)=200
Substitute the values then , we get
[tex]ln 200=k(0)+C[/tex]
[tex]C=ln 200[/tex]
Substitute the value of C then we get
[tex]ln y=kt+ln 200[/tex]
[tex]ln y-ln200=kt[/tex]
[tex]ln\frac{y}{200}=kt[/tex]
[tex]\frac{y}{200}=e^{kt}[/tex]
[tex]y=200e^{kt}[/tex]
Differentiate w.r.t
[tex]y'=200ke^{kt}[/tex]
Substitute the given condition then, we get
[tex]100=200ke^{0}=200 \;because \;e^0=1[/tex]
[tex]k=\frac{100}{200}=\frac{1}{2}[/tex]
[tex]y=200e^{\frac{t}{2}}[/tex]
Substitute t=2
Then, we get [tex]y=200e^{\frac{2}{2}}=200e[/tex]
[tex]y=200(2.718)=543.6=543.6[/tex]
e=2.718
Hence, the number of bacteria after 2 hours=543.6
6r+7=13+7r
steps too pls
Answer:
r = -6
Step-by-step explanation:
We presume you want to find the value of r that satisfies the equation.
Subtract 6r+13 from both sides:
(6r+7) -(6r+13) = (13 +7r) -(6r+13)
-6 = r . . . . . simplify
_____
More detailed explanation
When we look at the equation we see the only variable is r, and that terms containing it appear on both sides of the equal sign. There is only one "r" term on each side, so we don't need to do any consolidation. We observe that the term with the smallest coefficient is 6r and that it is on the left side.
When we subtract 6r from the equation we will have the remaining "r" term on the right, but we will also have a constant there that we don't want. So, we can subtract that constant as well. That is why we choose to subtract 6r+13 from the equation. Doing so leaves the constants on the left and the "r" terms on the right.
As it happens, the difference between the "r" terms is plain "r", so we're done after we finish the subtraction.
__
When considering the "r" terms, we choose to subtract the term with the smallest coefficient so that the result has a positive coefficient for "r". This helps reduce mistakes in later steps, if there are any later steps.
___
Alternate "steps"
For a linear equation like this one, you can subtract one side from both sides. This might look like ...
0 = 6+r . . . . . after subtracting 6r+7 (left side) from both sides
Then you can divide by the coefficient of r (which does nothing to this equation), and subtract the resulting constant (on the side with the variable). Here, that would give ...
-6 = r
These three steps will work to solve any linear equation. Simplification steps may be required depending on the complexity. Again, it might be helpful, though is not essential, to subtract the side with the smallest coefficient of the variable.
___
Final note
The rules of equality say you can do anything you like to an equation, as long as you do the same thing to both sides. We can say "subtract the constant" because we are assured that you know it must be subtracted from both sides of the equation. Beware of any instruction that tells you to do something to one side of an equation and something different to the other side.
Let P(x) be the statement"x= x2", If the domain consists of the integers, what are these truth values? (a) P(0) (b) P(1) (c) P(2) (d) P(-1) (e)
Answer: i guess the problem is with P(x) => "x = [tex]x^{2}[/tex]", then P(x) is true if that equality is true, and is false if the equality is false.
so lets see case for case.
a) x = 0, and [tex]0^{2}[/tex] = 0. So p(0) is true.
b) x = 1 and [tex]1^{2}[/tex] = 1, so P(1) is true.
c) x = 2, and [tex]2^{2}[/tex] = 4, and 2 ≠ 4, then P(2) is false.
d) x= -1 and [tex]1^{2}[/tex] = 1, and 1 ≠ -1, so P(-1) is false.
The truth value of P(0) and P(1) is true while the truth value of P(2) and P(-1) is false
The statement is given as:
[tex]x = x^2[/tex]
For P(0), we have:
[tex]0 = 0^2[/tex]
[tex]0 = 0[/tex] --- this is true
For P(1), we have:
[tex]1 = 1^2[/tex]
[tex]1 = 1[/tex] --this is true
For P(2), we have:
[tex]2 = 2^2[/tex]
[tex]2= 4[/tex] -- this is false
For P(-1), we have:
[tex](-1) = (-1)^2[/tex]
[tex](-1) = 1[/tex] --- this is false
Hence, the truth value of P(0) and P(1) is true while the truth value of P(2) and P(-1) is false
Read more about truth values at:
https://brainly.com/question/10678994
The Titanic was 883 feet long and 175 feet tall.
A. If a model is built to have a scale ratio of 1in : 36ft , how long will the model be?
B. If a model is built to have a scale ratio of 1 in : 22 ft , how tall will the model be?
C. If a model is built to have a ratio of 1: 30 , how tall will the model be?
Answer:
Actual length = 883 feet
Actual height = 175 feet
A) If a model is built to have a scale ratio of 1 in : 36ft , how long will the model be?
36 feet = 1 inch
Actual length = 883 feet
So, 883 feet = [tex]\frac{883}{36} inch[/tex]
883 feet = [tex]24.527 inch[/tex]
So, If a model is built to have a scale ratio of 1in : 36ft ,the model will be 24.527 inch long.
B) If a model is built to have a scale ratio of 1 in : 22 ft , how tall will the model be?
22 feet = 1 inch
Actual height = 175 feet
So, 175 feet = [tex]\frac{175}{22} inch[/tex]
883 feet = [tex]7.9545 inch[/tex]
So, If a model is built to have a scale ratio of 1 in : 22 ft , the model will be 7.9545 inch tall.
C) If a model is built to have a ratio of 1: 30 , how tall will the model be?
Let the height be x
Actual length = 883 feet
Ratio of 1: 30
So, [tex]\frac{1}{30}=\frac{883}{x}[/tex]
[tex]x=\frac{883}{30}[/tex]
[tex]x=29.433[/tex]
So, If a model is built to have a ratio of 1: 30 ,the model will be 29.433 inches tall .
u + 3b - 2a + 2 solve for a
Answer:
a=2
Step-by-step explanation:
u=3b-2(2) or u=3b-4
Final answer:
To solve the equation for a, rearrange the terms and isolate 'a' by dividing by the coefficient.
Explanation:
Step 1: Start with the equation u + 3b - 2a + 2 = 0.
Step 2: Rearrange the terms to isolate 'a', which gives -2a = -u - 3b - 2.
Step 3: Divide by -2 to solve for 'a', resulting in a = (u + 3b + 2) / 2.
A particular brand of dishwasher soap is sold in three sizes: 25 oz, 45 oz, and 60 oz. Twenty percent of all purchasers select a 25-oz box, 50% select a 45-oz box, and the remaining 30% choose a 60-oz box. Let X1 and X2 denote the package sizes selected by two independently selected purchasers. (a) Determine the sampling distribution of X. x 25 35 45 42.5 52.5 60 p(x) Calculate E(X). E(X) = oz Compare E(X) to μ. E(X) > μ E(X) < μ E(X) = μ
Answer:
(a) Sampling distribution
P(25) = 0,04
P(35) = 0.1 + 0.1 = 0,2
P(42,5) = 0.06 + 0.06 = 0,12
P(45) = 0,25
P(52,5) = 0.15 + 0.15 = 0,3
P(60) = 0,09
(b) E(X) = 45.5 oz
(c) E(X) = μ
Step-by-step explanation:
The variable we want to compute is
[tex]X=(X1+X2)/2[/tex]
For this we need to know all the possible combinations of X1 and X2 and the probability associated with them.
(a) Sampling distribution
Calculating all the 9 combinations (3 repeated, so we end up with 6 unique combinations):
P(25) = P(X1=25) * P(X2=25) = p25*p25 = 0.2 * 0.2 = 0,04
P(35) = p25*p45+p45*p25 = 0.2*0.5 + 0.5*0.2 = 0.1 + 0.1 = 0,2
P(42,5) = p25*p60 + p60*p25 = 0.2*0.3 + 0.3*0.2 = 0.06 + 0.06 = 0,12
P(45) = p45*p45 = 0.5 * 0.5 = 0,25
P(52,5) = p45*p60 + p60*p45 = 0.5*0.3 + 0.3*0.5 = 0.15 + 0.15 = 0,3
P(60) = p60*p60 = 0.3*0.3 = 0,09
(b) Using the sample distribution, E(X) can be expressed as:
[tex]E(X)=\sum_{i=1}^{6}P_{i}*X_{i}\\E(X)=0.04*25+0.2*35+0.12*42.5+0.3*52.5+0.09*60 = 45.5[/tex]
The value of E(X) is 45.5 oz.
(c) The value of μ can be calculated as
[tex]\mu=\sum_{i=1}^{3}P_{i}*X_{i}\\\mu=0.2*25+0.5*45+0.3*60=45.5[/tex]
We can conclude that E(X)=μ
We could have arrived to this conclusion by applying
[tex]E(X)=E((X1+X2)/2)=E(X1)/2+E(X2)/2\\\\\mu = E(X1)=E(X2)\\\\E(X)=\mu /2+ \mu /2 = \mu[/tex]
A patient is to receive Taxol 100 mg/m2. The patient weighs 120 lbs and is 5'8" tall. What dose of Taxol in milligrams should the patient receive? Round answers to the nearest tenth and DO NOT include units
Answer:
164.6 mg
Step-by-step explanation:
Given:
Weight of the patient= 120 lbs
Height of patient = 5'8" = 5 × 12 + 8 = 68 inches
Dose of Taxol to be administered= 100 mg/ m²
Now,
the surface area of the body of patient = [tex]\textup{(Weight in kg)}^{0.425}\times\textup{(Height in cms)}^{0.725}\times0.007184[/tex]
Also,
weight of patient in kg = 120 × 0.454 = 54.48 kg
Height of patient in cm = 68" × 2.54 = 172.72 cm
therefore,
Body surface area = [tex]\textup{(54.48)}^{0.425}\times\textup{(172.72)}^{0.725}\times0.007184[/tex]
or
= 5.47 × 41.89 × 0.007184
or
= 1.646 m²
Hence,
Dose of Taxol to be received by the patient
= 100 mg/m² × surface area of the patient
= 100 × 1.646
= 164.6 mg
Jack has a collection of 10 pairs of gloves in his wardrobe. Before a business trip, he has to pack his luggage, and he selects 8 gloves, without looking at them. We assume that any set of 8 gloves is equally likely to be chosen. Find the probability that these 8 gloves do not include any matching pair of gloves, that is, that there are no two (left and right) gloves, coming from the same pair.
Answer:
[tex]\frac{6\cdot 8 \cdot 10 \cdot 12}{13 \cdot 15 \cdot 17\cdot 19}\approx 0.091 [/tex]
Step-by-step explanation:
To compute the probability of not including any matching pairs, we can compute the number of ways in which he could pick 8 gloves with no matching pairs, and divide it by the total number of ways in which he could pick 8 gloves.
The process of choosing 8 gloves with no matching pair can be seen as follows:
He first picks any random gloves out of the 20, in this 1st step he has then 20 available choices. Then he needs to pick some other glove, BUT it cannot be the other glove from the pair he already picked one from. So at this step there aren't 19 choices, but 18 available choices. Now he has 2 gloves from different pairs, and needs to pick another glove. There are 18 gloves left, BUT he cannot pick the remaining glove from any of the 2 pairs he already has chosen one from. Therefore he only has 16 choices left. The process continues like this, until he chooses 8 gloves in total. Hence the total number of ways to choose 8 gloves with no matching pair is
[tex] 20 \cdot 18 \cdot 16 \cdot 14 \cdot 12 \cdot 10 \cdot 8 \cdot 6[/tex]
Now, the total numer of ways in which he could pick any 8 gloves out of 20 can be seen as follows: At the start he has 20 available gloves, he chooses any of them, so having 20 available choices on that first step. He then needs to choose any other glove, so he has 19 choices. He then picks any other glove, so he has 18 choices. And so on, until he has chosen the 8 gloves. Hence the total number of ways to choose 8 gloves is
[tex] 20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 [/tex]
Therefore the probability of choosing 8 gloves with no matching pair is
[tex]\frac{20 \cdot 18 \cdot 16 \cdot 14 \cdot 12 \cdot10 \cdot 8 \cdot 6}{20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13}=\frac{6\cdot 8 \cdot 10 \cdot 12}{13 \cdot 15 \cdot 17\cdot 19}\approx 0.091 [/tex]
Evaluate 1^3 + 2^3 +3^3 +.......+ n^3
Notice that
[tex](n+1)^4-n^4=4n^3+6n^2+4n+1[/tex]
so that
[tex]\displaystyle\sum_{i=1}^n((n+1)^4-n^4)=\sum_{i=1}^n(4i^3+6i^2+4i+1)[/tex]
We have
[tex]\displaystyle\sum_{i=1}^n((i+1)^4-i^4)=(2^4-1^4)+(3^4-2^4)+(4^4-3^4)+\cdots+((n+1)^4-n^4)[/tex]
[tex]\implies\displaystyle\sum_{i=1}^n((i+1)^4-i^4)=(n+1)^4-1[/tex]
so that
[tex]\displaystyle(n+1)^4-1=\sum_{i=1}^n(4i^3+6i^2+4i+1)[/tex]
You might already know that
[tex]\displaystyle\sum_{i=1}^n1=n[/tex]
[tex]\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2[/tex]
[tex]\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6[/tex]
so from these formulas we get
[tex]\displaystyle(n+1)^4-1=4\sum_{i=1}^ni^3+n(n+1)(2n+1)+2n(n+1)+n[/tex]
[tex]\implies\displaystyle\sum_{i=1}^ni^3=\frac{(n+1)^4-1-n(n+1)(2n+1)-2n(n+1)-n}4[/tex]
[tex]\implies\boxed{\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4}[/tex]
If you don't know the formulas mentioned above:
The first one should be obvious; if you add [tex]n[/tex] copies of 1 together, you end up with [tex]n[/tex].The second one is easily derived: If [tex]S=1+2+3+\cdots+n[/tex], then [tex]S=n+(n-1)+(n-2)+\cdots+1[/tex], so that [tex]2S=n(n+1)[/tex] or [tex]S=\dfrac{n(n+1)}2[/tex].The third can be derived using a similar strategy to the one used here. Consider the expression [tex](n+1)^3-n^3=3n^2+3n+1[/tex], and so on. Graph the system of equations. Then determine whether the system has no solution, one solution, or infinitely many solutions. If the system has one solution, name it.
y = -x + 5
y = x - 3
a) one solution; (1, 4)
b) infinitely many
c) no solution
d) one solution; (4, 1)
Answer:
d) one solution; (4, 1)
Step-by-step explanation:
It often works well to follow problem directions. A graph is attached, showing the one solution to be (4, 1).
_____
You know there will be one solution because the lines have different slopes. Each is in the form ...
y = mx + b
where m is the slope and b is the y-intercept.
The first line has slope -1 and y-intercept +5; the second line has slope 1 and y-intercept -3. The slope is the number of units of "rise" for each unit of "run", so it can be convenient to graph these by starting at the y-intercept and plotting points with those rise and run from the point you know.
Write an expression with (-1) as its base that will produce a positive product, and explain why your answer is valid
Here are some possible answers
In a fraternity with 32 members, 18 take mathematics, 5 take both mathematics and literature, and 8 take neither mathematics nor literature. How many take literature but not mathematics?
Answer:
1 member took literature but not mathematics.
Step-by-step explanation:
We can draw a Venn diagram for the given question.
In a fraternity total number of members = 32
Number of members who took mathematics M = 18
Number of members who took both mathematics and literature (M∩L) = 5
And number of members who took neither mathematics nor literature = 8
Therefore, number of members who took literature but not mathematics
= 32 - [(18 + 5) + 8]
= 32 - [23 + 8]
= 32 - 31
= 1
Therefore, 1 member took literature but not mathematics.
Answer:
1 member took literature but not mathematics.
Step-by-step explanation:
The image of (6, 9) under a dilation is (4, 6).
The scale factor is
0 -2
Answer:
The scale factor is 2/3.
Step-by-step explanation:
When the dilation is about the origin, the ratio of image coordinates to original coordinates is the scale factor:
4/5 = 6/9 = 2/3
The scale factor is 2/3.
Find equations of the line that is parallel to the z-axis and passes through the midpoint between the two points (0, −4, 9) and (−8, 5, 1). (Enter your answers as a comma-separated list of equations.) x=−8, y=5+9t, z=1−8t
The equations of the line are:
x = -4,
y = 0.5,
z = 5 + t
A line that is parallel to the z-axis lies in the xy-plane. Since it's parallel to the z-axis, its direction in the xy-plane is determined by the coefficients of x and y in its direction vector.
Let's first find the midpoint between the two given points:
Midpoint = [(x₁ + x₂) / 2, (y₁ + y₂) / 2, (z₁ + z₂) / 2]
Midpoint = [(-8 + 0) / 2, (5 - 4) / 2, (1 + 9) / 2]
Midpoint = [-4, 0.5, 5]
So, the midpoint is (-4, 0.5, 5).
Now, let's create a line that passes through the midpoint and is parallel to the z-axis. The equation of such a line in vector form is:
r(t) = Midpoint + t Direction
Where r(t) is the position vector of a point on the line, t is a scalar parameter, Midpoint is the midpoint we calculated, and Direction is the direction vector.
Since the line is parallel to the z-axis, its direction vector is (0, 0, 1). Thus, the equation of the line is:
r(t) = (-4, 0.5, 5) + t (0, 0, 1)
r(t) = (-4, 0.5, 5 + t)
In component form, the equations of the line are:
x = -4
y = 0.5
z = 5 + t
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The midpoint of the two points is (-4, 0.5, 5). The line parallel to the z-axis passing through this point has the equations: x=-4, y=0.5, z=t.
Explanation:Firstly, we need to find out the midpoint between the two points (0, -4, 9) and (-8, 5, 1).
The formula to calculate the midpoint of two points in three dimensions is ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2).
Applying this formula to our points, we get the midpoint as ((0-8)/2, (-4+5)/2, (9+1)/2) which is (-4, 0.5, 5).
Lines parallel to the z-axis in three-dimensional space have equations of the form x=a, y=b, z=t, where 'a' and 'b' are constants representing any particular point through which the line passes, and 't' represents a variable that can take any real value.
Since the line we want passes through the point (-4, 0.5, 5), our equations for the desired line become: x=-4, y=0.5, z=t.
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Preliminary data analyses indicate that it is reasonable to use a t-test to carry out the specified hypothesis test. Perform the t-test. Be sure to state the hypotheses and the P-Value. State your conclusion in a sentence. A test of sobriety involves measuring a subject's motor skills. The mean score for men who are sober is known to be 35.0. A researcher would like to perform a hypothesis test to determine whether the mean score for sober women differs from 35.0. Twenty randomly selected sober women take the test and produce a mean score of 41.0 with a standard deviation of 3.7. Perform the hypothesis test at the 0.01 level of significance.
By considering the given information, we have
Null hypothesis : [tex]H_0: \mu=35.0[/tex]
Alternative hypothesis : [tex]H_1: \mu\neq35.0[/tex]
Since the alternative hypothesis is two-tailed , so the test is a two-tailed test.
Given : Sample size : n= 20, since sample size is less than 30 so the test applied is a t-test.
[tex]\overline{x}=41.0[/tex] ; [tex]\sigma= 3.7[/tex]
Test statistic : [tex]t=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
i.e. [tex]t=\dfrac{41.0-35.0}{\dfrac{3.7}{\sqrt{20}}}=7.252112359\approx7.25[/tex]
Degree of freedom : n-1 = 20-1=19
Significance level = 0.01
For two tailed, Significance level [tex]=\dfrac{0.01}{2}=0.005[/tex]
By using the t-distribution table, the critical value of t =[tex]t_{19, 0.005}=2.861[/tex]
Since , the observed t-value (7.25) is greater than the critical value (2.861) .
So we reject the null hypothesis, it means we have enough evidence to support the alternative hypothesis.
We conclude that there is some significance difference between the mean score for sober women and 35.0.
A division of the Gibson Corporation manufactures bicycle pumps. Each pump sells for $10, and the variable cost of producing each unit is 60% of the selling price. The monthly fixed costs incurred by the division are $50,000. What is the break-even point for the division? (Round your answers to the nearest whole number.)
(x,y)=
Answer:
12,500 units
Step-by-step explanation:
Given:
Selling cost of the pump = $10
Variable cost of producing each unit = 60% of selling cost
or
Variable cost of producing each unit = 0.60 × $10 = $6
Monthly fixed cost = $50,000
Now,
The profit per unit = Selling cost - Variable cost
or
The profit per unit = $10 - $6 = $4
Therefore,
The breakeven point = [tex]\frac{\textup{Fixed cost}}{\textup{Profit per unit}}[/tex]
or
the breakeven point = [tex]\frac{\textup{50,000}}{\textup{4}}[/tex]
or
the breakeven point = 12,500 units
Evaluate C_n.xP^xQn-x For the given n=7, x=2, p=1/2
Answer:
The value of given expression is [tex]\frac{21}{128}[/tex].
Step-by-step explanation:
Given information: n=7, x=2, p=1/2
[tex]q=1-p=1-\frac{1}{2}=\frac{1}{2}[/tex]
The given expression is
[tex]C(n,x)p^xq^{n-x}[/tex]
It can be written as
[tex]^nC_xp^xq^{n-x}[/tex]
Substitute n=7, x=2, p=1/2 and q=1/2 in the above formula.
[tex]^7C_2(\frac{1}{2})^2(\frac{1}{2})^{7-2}[/tex]
[tex]\frac{7!}{2!(7-2)!}(\frac{1}{2})^2(\frac{1}{2})^{5}[/tex]
[tex]\frac{7!}{2!5!}(\frac{1}{2})^{2+5}[/tex]
[tex]\frac{7\times 6\times 5!}{2\times 5!}(\frac{1}{2})^{2+5}[/tex]
[tex]21(\frac{1}{2})^{7}[/tex]
[tex]\frac{21}{128}[/tex]
Therefore the value of given expression is [tex]\frac{21}{128}[/tex].
What is the range of the relation?
1. {3, 9, 12}
2. {−6, −1, −0.5}
3. {−6, 3, 9}
4. {−1, −0.5, 9, 12}
Answer:
the range of the relation is 3
Find the optimal solution for the following problem
Minimize C = 13x + 3y
subject to 12x + 14y ≥ 21
15x + 20y ≥ 37
and x ≥ 0, y ≥ 0.
1. What is the optimal value of x?
2. What is the optimal value of y?
3.What is the minimum value of the objective function?
Answer:
Minimize C =[tex]13x + 3y[/tex]
[tex]12x + 14y \geq 21[/tex]
[tex]15x + 20y \geq 37[/tex]
and x ≥ 0, y ≥ 0.
Plot the the lines on the graph and find the feasible region
[tex]12x + 14y \geq 21[/tex] -- Blue
[tex]15x + 20y \geq 37[/tex] --- Green
So, the boundary points of feasible region are (-3.267,4.3) , (0,1.85) and (2.467,0)
Substitute the value in Minimize C
Minimize C =[tex]13x + 3y[/tex]
At (-3.267,4.3)
Minimize C =[tex]13(-3.267) + 3(4.3)[/tex]
Minimize C =[tex]-29.571[/tex]
At (0,1.85)
Minimize C =[tex]13(0) + 3(1.85)[/tex]
Minimize C =[tex]5.55[/tex]
At (2.467,0)
Minimize C =[tex]13(2.467) + 3(0)[/tex]
Minimize C =[tex]32.071[/tex]
So, the optimal value of x is -3.267
So, the optimal value of y is 4.3
So, the minimum value of the objective function is -29.571
How many ways can six of the letters of the word ALGORITHM be selected 8. How many ways can the letters of the word ALGORITHM be arranged in a be seated together in the row? and written in a row? row if the letters GOR must remain together (in this order)?
The number of letters in word "ALGORITHM" = 9
The number of combinations to select r things from n things is given by :-
[tex]C(n,r)=\dfrac{n!}{r!(n-r)!}[/tex]
Now, the number of combinations to select 6 letters from 9 letters will be :-
[tex]C(9,8)=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!\times3!}=84[/tex]
Thus , the number of ways can six of the letters of the word ALGORITHM=84
The number of ways to arrange n things in a row :[tex]n![/tex]
So, the number of ways can the letters of the word ALGORITHM be arranged in a be seated together in the row :-
[tex]9!=362880[/tex]
If GOR comes together, then we consider it as one letter, then the total number of letters will be = 1+6=7
Number of ways to arrange 9 letters if "GOR" comes together :-
[tex]7!=5040[/tex]
Thus, the number of ways to arrange 9 letters if "GOR" comes together=5040
solve the lenear system by using the inverse of the coefficient matrix:
x + 2z = -1
2x - y = 2
3y + 4z = 1
Answer:
The solution of this system is x=9/4, y=5/2, and z=-13/8
Step-by-step explanation:
1. Writing the equations in matrix form
The system of linear equations given can be written in matrix form as
[tex]\left[\begin{array}{ccc}1&0&2\\2&-1&0\\0&3&4\end{array}\right]\left[\begin{array}{c}x&y&z\end{array}\right] = \left[\begin{array}{c}-1&2&1\end{array}\right][/tex]
Writing
A = [tex]\left[\begin{array}{ccc}1&0&2\\2&-1&0\\0&3&4\end{array}\right][/tex]
X = [tex]\left[\begin{array}{c}x&y&z\end{array}\right][/tex]
B = [tex]\left[\begin{array}{c}-1&2&1\end{array}\right][/tex]
we have
AX=B
This is the matrix form of the simultaneous equations.
2. Solving the simultaneous equations
Given
AX=B
we can multiply both sides by the inverse of A
[tex]A^{-1}AX=A^{-1}B[/tex]
We know that [tex]A^{-1}A=I[/tex], the identity matrix, so
[tex]X=A^{-1}B[/tex]
All we need to do is calculate the inverse of the matrix of coefficients, and finally perform matrix multiplication.
3. Calculate the inverse of the matrix of coefficients
A = [tex]\left[\begin{array}{ccc}1&0&2\\2&-1&0\\0&3&4\end{array}\right][/tex]
To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.
[tex]\left[\begin{array}{ccc|ccc}1&0&2&1&0&0\\2&-1&0&0&1&0\\0&3&4&0&0&1\end{array}\right][/tex]
Make zeros in column 1 except the entry at row 1, column 1. Subtract row 1 multiplied by 2 from row 2[tex]\left[\begin{array}{ccc|ccc}1&0&2&1&0&0\\0&-1&-4&-2&1&0\\0&3&4&0&0&1\end{array}\right][/tex]
Make zeros in column 2 except the entry at row 2, column 2. Add row 2 multiplied by 3 to row 3[tex]\left[\begin{array}{ccc|ccc}1&0&2&1&0&0\\0&-1&-4&-2&1&0\\0&0&-8&-6&3&1\end{array}\right][/tex]
Multiply row 2 by −1[tex]\left[\begin{array}{ccc|ccc}1&0&2&1&0&0\\0&1&4&2&-1&0\\0&0&-8&-6&3&1\end{array}\right][/tex]
Make zeros in column 3 except the entry at row 3, column 3. Divide row 3 by −8[tex]\left[\begin{array}{ccc|ccc}1&0&2&1&0&0\\0&1&4&2&-1&0\\0&0&1&3/4&-3/8&-1/8\end{array}\right][/tex]
Subtract row 3 multiplied by 2 from row 1[tex]\left[\begin{array}{ccc|ccc}1&0&0&-1/2&3/4&1/4\\0&1&4&2&-1&0\\0&0&1&3/4&-3/8&-1/8\end{array}\right][/tex]
Subtract row 3 multiplied by 4 from row 2[tex]\left[\begin{array}{ccc|ccc}1&0&0&-1/2&3/4&1/4\\0&1&0&-1&1/2&1/2\\0&0&1&3/4&-3/8&-1/8\end{array}\right][/tex]
As can be seen, we have obtained the identity matrix to the left. So, we are done.
[tex]A^{-1} = \left[\begin{array}{ccc}-1/2&3/4&1/4\\-1&1/2&1/2\\3/4&-3/8&-1/8\end{array}\right][/tex]
4. Find the solution [tex]X=A^{-1}B[/tex]
[tex]X= \left[\begin{array}{ccc}-1/2&3/4&1/4\\-1&1/2&1/2\\3/4&-3/8&-1/8\end{array}\right]\cdot \left[\begin{array}{c}-1&2&1\end{array}\right] = \left[\begin{array}{c}9/4&5/2&-13/8\end{array}\right][/tex]
Paco bought 3 CDs that cost d dollars each and a pack of gum for C cents. Write an expression for the total cost of his purchase. 3d + c 3C + d 3d/c
Answer:
The correct option is A) [tex]3d+C[/tex].
Step-by-step explanation:
Consider the provided information.
Paco bought 3 CDs that cost d dollars each.
Let d is the cost of each CDs.
The cost of 3 CDs will be 3 times of d.
This can be written as:
[tex]3d[/tex]
He bought a pack of gum for C cent. Thus, we can say that the cost of pack of gum is C.
Now add the cost of gum in above expression.
[tex]3d+C[/tex]
Hence, the required expression is [tex]3d+C[/tex].
Thus, the correct option is A) [tex]3d+C[/tex].
You have an order for Vasopressin 18 units/hr IV infusion. You have available vasopressin 200 units in 5000 mL D5W. Please calculate the drip rate in mL/hr.
Answer:
450 mL/hr
Step-by-step explanation:
Given:
Order for vasopressin = 18 units/hour
Available vasopressin = 200 units in 5000 mL
Now,
Volume of vasopressin per unit = [tex]\frac{\textup{Volume of vasopressin}}{\textup{Number of units}}[/tex]
or
Volume of vasopressin per unit = [tex]\frac{\textup{5000}}{\textup{200}}[/tex]
or
Volume of vasopressin per unit = 25 mL/unit
Thus,
Drip rate in mL/hr
= volume of vasopressin per unit × order for vassopressin
or
Drip rate in mL/hr = 25 × 18 = 450 mL/hr
The drip rate for an order of Vasopressin 18 units/hr, given a solution concentration of 200 units in 5000 mL, is calculated to be 450 mL/hr.
Explanation:To find the drip rate in mL/hr, we start by determining the concentration of the vasopressin solution. It is 200 units in 5000 mL D5W, so the concentration is 0.04 units/mL (200 units/5000 mL).
Next, we know the doctor prescribed 18 units/hr of vasopressin. To find out how many mL this corresponds to, we divide the order of 18 units/hr by the concentration in units/mL, which gives us 450 mL/hr (18 units/hr / 0.04 units/mL).
Therefore, the drip rate for the Vasopressin order is 450 mL/hr.
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