Answer:15x^2-3x
Step-by-step explanation:
Find a cubic polynomial with integer coefficients that has $\sqrt[3]{2} + \sqrt[3]{4}$ as a root.
Find the powers [tex]a=\sqrt{2}+\sqrt{3}[/tex]
$a^{2}=5+2 \sqrt{6}$
$a^{3}=11 \sqrt{2}+9 \sqrt{3}$
The cubic term gives us a clue, we can use a linear combination to eliminate the root 3 term $a^{3}-9 a=2 \sqrt{2}$ Square $\left(a^{3}-9 a\right)^{2}=8$ which gives one solution. Expand we have $a^{6}-18 a^{4}-81 a^{2}=8$ Hence the polynomial $x^{6}-18 x^{4}-81 x^{2}-8$ will have a as a solution.
Note this is not the simplest solution as $x^{6}-18 x^{4}-81 x^{2}-8=\left(x^{2}-8\right)\left(x^{4}-10 x^{2}+1\right)$
so fits with the other answers.
Answer:
[tex]y^3 -6y-6[/tex]
Foram prescritos 500mg de dipirona para uma criança com febre.Na unidade tem disponivel ampola de 1g/2ml.Quantos g vão ser administrados no paciente
De acordo com a disponibilidade da unidade, há apenas a seguinte dosagem: 1g/2mL - ou seja, uma grama de dipirona a cada 2mL
O enunciado está meio mal formulado, pois é dito que foram prescritos 500mg de dipirona e é essa quantidade de farmaco que a criança tem que tomar. Deseja-se saber quantos mL deverao ser administrados.
Fazendo a classica regra de 3, podemos chegar no volume desejado:
(atentar que 500mg = 0,5g)
g mL
1 --------- 2
0,5 --------- X
1 . X = 0,5 . 2
X = 1mLFindℒ{f(t)}by first using a trigonometric identity. (Write your answer as a function of s.)f(t) = 12 cost −π6
Answer:
[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]
Step-by-step explanation:
Given that:
[tex]f(t) = 12 cos (t- \dfrac{\pi}{6})[/tex]
recall that:
cos (A-B) = cos AcosB + sin A sin B
∴
[tex]f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}][/tex]
[tex]f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}][/tex]
[tex]f(t) = 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t)[/tex]
[tex]L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ][/tex]
[tex]L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ][/tex]
[tex]L(f(t)) = 6 \sqrt{3} \dfrac{S}{S^2 + 1^2}+ 6 \dfrac{1}{S^2 +1^2}[/tex]
[tex]L(f(t)) = \dfrac{6 \sqrt{3} +6 }{S^2+1}[/tex]
[tex]L(f(t)) = \dfrac{6( \sqrt{3} \ S +1 }{S^2+1}[/tex]
[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]
Transform the given parametric equations into rectangular form. Then identify the conic.
Answer:
Solution : Option B
Step-by-Step Explanation:
We have the following system of equations at hand here.
{ x = 5 cot(t), y = - 3csc(t) + 4 }
Now instead of isolating the t from either equation, let's isolate cot(t) and csc(t) --- Step #1,
x = 5 cot(t) ⇒ x - 5 = cot(t),
y = - 3csc(t) + 4 ⇒ y - 4 = - 3csc(t) ⇒ y - 4 / - 3 = csc(t)
Now let's square these two equations. We know that csc²θ - cot²θ = 1, so let's subtract the equations as well. --- Step #2
( y - 4 / - 3 )² = (csc(t))²
- ( x - 5 / 1 )² = (cot(t))²
___________________
(y - 4)² / 9 - x² / 25 = 1
And as we are subtracting the two expressions, this is an example of a hyperbola. Therefore your solution is option b.
a
A solid metal cone of base radius a cm and height 2a cm is melted and solid
spheres of radius are made without wastage. How many such spheres can be
made?
volume of a cone
.
.
.
volume of sphere
.
.
number of spheres that can be made......
.
.
hence a hemisphere can be formed
Time
(minutes)
Water
(gallons)
1
16.50
1.5
24.75
2
33
find the constant of proportionality for the second and third row
Answer:
16.50
Step-by-step explanation:
Constant of proportionality = no of gallons of water per 1 minute.
In the first row, we have 16.50 gallons of water per 1 minute.
In the 2nd row, we have 24.75 gallons of water in 1.5 minutes. In 1 minute, we will have 24.75 ÷ 1.5 = 16.50 gallons
In the 3rd row, we have 33 gallons in 2 minutes. In 1 minute, we will have 33 ÷ 2 = 16.50 gallons.
We can see that there seems to be the same constant of proportionality for the 2nd and 3rd row, which is 16.50.
Thus, a relationship between gallons of water (w) and time (t), considering the constant, 16.50, can be written as: [tex] w = 16.50t [/tex]
This means the constant of proportionality, 16.50, is same for all rows.
What is the most precise name for quadrilateral ABCD with vertices A(–5,2), B(–3, 5),C(4, 5),and D(2, 2)?
Answer: ABCD is a parallelogram.
Step-by-step explanation:
First we plot these point on a graph as given in attachment.
From the attachment we can observe that AD || BC || x-axis .
also, AB ||CD, that will make ABCD a parallelogram , but to confirm we check the property of parallelogram "diagonals bisect each other" , i.e . "Mid point of both diagonals are equal".
Mid point of AC= [tex](\dfrac{-5+4}{2},\dfrac{2+5}{2})=(\dfrac{-1}{2},\dfrac{7}{2})[/tex]
Mid point of BD= [tex](\dfrac{-3+2}{2},\dfrac{5+2}{2})=(\dfrac{-1}{2},\dfrac{7}{2})[/tex]
Thus, Mid point of AC=Mid point of BD
i.e. diagonals bisect each other.
That means ABCD is a parallelogram.
Answer: ABCD is a parallelogram.
Step-by-step explanation:
First, we plot these points on a graph as given in the attachment. From the attachment, we can observe that AD || BC || x-axis. Also, AB ||CD, which will make ABCD a parallelogram, but to confirm, we check the parallelogram property "diagonals bisect each other," i.e., "Midpoint of both diagonals is equal."
The midpoint of AC=. The midpoint of BD=. Thus, the Midpoint of AC=Mid point of BD diagonals bisects each other. That means ABCD is a parallelogram.
Find usubscript10 in the sequence -23, -18, -13, -8, -3, ...
Step-by-step explanation:
utilise the formula a+(n-1)d
a is the first number while d is common difference
Answer:
22
Step-by-step explanation:
Using the formular, Un = a + (n - 1)d
Where n = 10; a = -23; d = 5
U10 = -23 + (9)* 5
U10 = -23 + 45 = 22
The cost of a daily rental car is as follows: The initial fee is $39.99 for the car, and it costs $0.20 per mile. If Julie's final bill was $100.00 before taxes, how many miles did she drive?
Answer:
300.05 miles
Step-by-step explanation:
initial fee= $39.99
final bill = $ 100
cost =$ 0.20 per mile
remaining amount = $ 60.01
solution,
she drive = remaining amount / cost
=60.01/0.20
=300.05 miles
Answer:
500 miles
Step-by-step explanation:
Let us use cross multiplication to find the unknown amount.
Given:
1) Cost for 1 mile=$0.20
2)Cost for x miles=$100
Solution:
No of miles Cost
1) 1 $0.20
2)x $100
By cross multiplying,
100 x 1= 0.20x
x=100/0.20
x=500 miles
Thank you!
please help me in these question ????
A school bag contains 12 pens of which 5 are red and the other are black. 4 pens are selected from the bag.
(a) How many different samples of size 4 pens are possible?
(b) How many samples have 3 red pens and 1 black pen?
(c) How many samples of size 4 contain at least two red pens?
(d) How many samples of size 4 contain
If the average yield of cucumber acre is 800 kg, with a variance 1600 kg, and that the amount of the cucumber follows the normal distribution.
1- What percentage of a cucumber give the crop amount between and 834 kg?
2- What the probability of cucumber give the crop exceed 900 kg ?
Answer:
Step-by-step explanation:
A school bag contains 12 pens of which 5 are red and the other are black. 4 pens are selected from the bag.
(a) How many different samples of size 4 pens are possible?
12C4=12!/(4!*8!)=495
(b) How many samples have 3 red pens and 1 black pen?
5C3*7C1
5C3=5!/(3!*2!)=10
7C1=7!/(1!*6!)=7
=>5C3*7C1=10*7=70
(c) How many samples of size 4 contain at least two red pens?
(5C2*7C2)+(5C3*7C1)+(5C4*7C0)
5C2=5!/(2!*3!)=10
7C2=7!/(2!*5!)=21
5C3=5!/(3!*2!)=10
7C1=7!/(1!*6!)=7
5C4=5!/(4!*1!)=5
7C0=7!/(0!*7!)=1
=>(5C2*7C2)+(5C3*7C1)+(5C4*7C0)=285
(d) How many samples of size 4 contain at most one black pen?
(7C1*5C3)+(7C0*5C4)
7C1=7!/(1!*6!)=7
7C0=7!/(0!*7!)=1
5C3=5!/(3!*2!)=10
5C4=5!/(4!*1!)=5
=>(7C1*5C3)+(7C0*5C4)=(7*10)+(1*5)=75
The cost, C, in United States Dollars ($), of cleaning up x percent of an oil spill along the Gulf Coast of the United States increases tremendously as x approaches 100. One equation for determining the cost (in millions $) is:
Complete Question
On the uploaded image is a similar question that will explain the given question
Answer:
The value of k is [tex]k = 214285.7[/tex]
The percentage of the oil that will be cleaned is [tex]x = 80.77\%[/tex]
Step-by-step explanation:
From the question we are told that
The cost of cleaning up the spillage is [tex]C = \frac{ k x }{100 - x }[/tex] [tex]x \le x \le 100[/tex]
The cost of cleaning x = 70% of the oil is [tex]C = \$500,000[/tex]
Now at [tex]C = \$500,000[/tex] we have
[tex]\$ 500000 = \frac{ k * 70 }{100 - 70 }[/tex]
[tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]
[tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]
[tex]k = 214285.7[/tex]
Now When [tex]C = \$900,000[/tex]
[tex]x = 80.77\%[/tex]
Salaries of 42 college graduates who took a statistics course in college have a mean, , of . Assuming a standard deviation, , of $, construct a % confidence interval for estimating the population mean .
Answer:
The 99% confidence interval for estimating the population mean μ is ($60,112.60, $68087.40).
Step-by-step explanation:
The complete question is:
Salaries of 42 college graduates who took a statistics course in college have a mean, [tex]\bar x[/tex] of, $64, 100. Assuming a standard deviation, σ of $10,016 construct a 99% confidence interval for estimating the population mean μ.
Solution:
The (1 - α)% confidence interval for estimating the population mean μ is:
[tex]CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]
The critical value of z for 99% confidence interval is:
[tex]z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.57[/tex]
Compute the 99% confidence interval for estimating the population mean μ as follows:
[tex]CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]
[tex]=64100\pm 2.58\times\frac{10016}{\sqrt{42}}\\\\=64100+3987.3961\\\\=(60112.6039, 68087.3961)\\\\\approx (60112.60, 68087.40)[/tex]
Thus, the 99% confidence interval for estimating the population mean μ is ($60,112.60, $68087.40).
Kenji earned the test scores below in English class.
79, 91, 93, 85, 86, and 88
What are the mean and median of his test scores?
Answer:
mean=87
median=87
Step-by-step explanation:
mean=sum of test score/number of subject
mean=79+91+93+85+86+88/6
mean=522/6
mean=87
Literal meaning of median is medium.
To find the number which lies in the medium, we must rearrange the number in ascending.
79, 91, 93, 85, 86, 88
79, 85, 86, 88, 91, 93
86+88/2=87
Hope this helps ;) ❤❤❤
Let me know if there is an error in my answer.
5x+4(-x-2)=-5x+2(x-1)+12
Answer:
x=9/2
Step-by-step explanation:
Let's solve your equation step-by-step.
5x+4(−x−2)=−5x+2(x−1)+12
Step 1: Simplify both sides of the equation.
5x+4(−x−2)=−5x+2(x−1)+12
5x+(4)(−x)+(4)(−2)=−5x+(2)(x)+(2)(−1)+12 (Distribute)
5x+−4x+−8=−5x+2x+−2+12
(5x+−4x)+(−8)=(−5x+2x)+(−2+12) (Combine Like Terms)
x+−8=−3x+10
x−8=−3x+10
Step 2: Add 3x to both sides.
x−8+3x=−3x+10+3x
4x−8=10
Step 3: Add 8 to both sides.
4x−8+8=10+8
4x=18
Step 4: Divide both sides by 4.
4x/4=18/4
x=9/2
602/100 into a decimal describe plz
Answer:
6.02
six point zero two
Step-by-step explanation:
Answer:
602 / 100= 6,02
Step-by-step explanation:
602 to divide 100 = 6,02
A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of ounces and a standard deviation of ounce. You randomly select cans and carefully measure the contents. The sample mean of the cans is ounces. Does the machine need to be reset? Explain your reasoning. ▼ Yes No , it is ▼ very unlikely likely that you would have randomly sampled cans with a mean equal to ounces, because it ▼ lies does not lie within the range of a usual event, namely within ▼ 1 standard deviation 2 standard deviations 3 standard deviations of the mean of the sample means.
Complete question is;
A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 128 ounces and a standard deviation of 0.20 ounce. You randomly select 35 cans and carefully measure the contents. The sample mean of the cans is 127.9 ounces. Does the machine need to be? reset? Explain your reasoning.
(yes/no)?, it is (very unlikely/ likely) that you would have randomly sampled 35 cans with a mean equal to 127.9 ?ounces, because it (lies/ does not lie) within the range of a usual? event, namely within (1 standard deviation, 2 standard deviations 3 standard deviations) of the mean of the sample means.
Answer:
Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.
Step-by-step explanation:
We are given;
Mean: μ = 128
Standard deviation; σ = 0.2
n = 35
Now, formula for standard error of mean is given as;
se = σ/√n
se = 0.2/√35
se = 0.0338
Normally, the range of values should be within 2 standard deviations of mean. In this case, normal range of values will be;
μ ± 2se = 128 ± 0.0338
This gives; 127.9662, 128.0338
So, Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.
The quotient of 3 and the cube
of y+2
Answer:
[tex]\dfrac{3}{(y+2)^3}[/tex]
Step-by-step explanation:
Maybe you want this written using math symbols. It will be ...
[tex]\boxed{\dfrac{3}{(y+2)^3}}[/tex]
Is {(4,2),(4,-2),(9,3),(9,-3)} a function
Answer:
no
Step-by-step explanation:
If any x-value is repeated, the relation is not a function. Both x=4 and x=9 are repeated values, so this relation is not a function.
Relating a Polynomial Identity to Pythagorean Triples
In this activity you'll relate polynomial identities with Pythagorean triples. Answer the following questions
based on this triangle with side lengths x^2 – 1, 2x, and x^2 + 1.
Answer:
Step-by-step explanation:
Hello, please consider the following.
For x > 1, we can apply Pythagoras theorem as below.
[tex]\text{Let's estimate this sum of two squares.} \\\\(2x)^2+(x^2-1)^2=4x^2+x^4-2x^2+1=x^4+2x^2+1\\\\\text{Let's estimate this square, now.} \\\\(x^2+1)^2=x^4+2x^2+1\\\\\text{The two expressions are equal, meaning.} \\\\(2x)^2+(x^2-1)^2=(x^2+1)^2\\\\\text{Using Pythagoras' theorem, we can say that this is a right triangle.}[/tex]
Thank you
Which of the following represents "next integer after the integer n"? n + 1 n 2n
Answer:
n + 1
Step-by-step explanation:
Starting with the integer 'n,' we represent the "next integer" by n + 1.
one third multiplied by the sum of a and b
Answer:
1/3(a+b)
hope it helps :>
The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100, and the standard deviation is 2. You wish to test H0: μ = 100 versus H1: μ ≠ 100 with a sample of n = 9 specimens.
A. If the acceptance region is defined as 98.5 le x- 101.5, find the type I error probability alpha.
B. Find beta for the case where the true mean heat evolved is 103.
C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?
Answer:
A.the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]
B. β = 0.0122
C. β = 0.0000
Step-by-step explanation:
Given that:
Mean = 100
standard deviation = 2
sample size = 9
The null and the alternative hypothesis can be computed as follows:
[tex]\mathtt{H_o: \mu = 100}[/tex]
[tex]\mathtt{H_1: \mu \neq 100}[/tex]
A. If the acceptance region is defined as [tex]98.5 < \overline x > 101.5[/tex] , find the type I error probability [tex]\alpha[/tex] .
Assuming the critical region lies within [tex]\overline x < 98.5[/tex] or [tex]\overline x > 101.5[/tex], for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is [tex]\mu = 100[/tex]
∴
[tex]\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}[/tex]
[tex]\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}[/tex]
when [tex]\mu = 100[/tex]
[tex]\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }[/tex]
[tex]\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }[/tex]
[tex]\mathtt{\alpha = P ( Z <-2.25 ) + P(Z > 2.25) }[/tex]
[tex]\mathtt{\alpha = P ( Z <-2.25 ) +( 1- P(Z < 2.25) })[/tex]
From the standard normal distribution tables
[tex]\mathtt{\alpha = 0.0122+( 1- 0.9878) })[/tex]
[tex]\mathtt{\alpha = 0.0122+( 0.0122) })[/tex]
[tex]\mathbf{\alpha = 0.0244 }[/tex]
Thus, the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]
B. Find beta for the case where the true mean heat evolved is 103.
The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis [tex]\mathtt{H_o}[/tex]
Thus;
β = P( type II error) - P( fail to reject [tex]\mathtt{H_o}[/tex] )
∴
[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]
Given that [tex]\mu = 103[/tex]
[tex]\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }[/tex]
[tex]\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }[/tex]
[tex]\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }[/tex]
[tex]\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}[/tex]
From standard normal distribution table
β = 0.0122 - 0.0000
β = 0.0122
C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?
[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]
Given that [tex]\mu = 105[/tex]
[tex]\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }[/tex]
[tex]\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }[/tex]
[tex]\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }[/tex]
[tex]\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}[/tex]
From standard normal distribution table
β = 0.0000 - 0.0000
β = 0.0000
The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.
The mean salary of federal government employees on the General Schedule is $59,593. The average salary of 30 state employees who do similar work is $58,800 with \sigmaσσ= $1500. At the 0.01 level of significance, can it be concluded that state employees earn on average less than federal employees? What is the critical value? Round your answer to the nearest hundredths.
Answer:
Yes it can be concluded that state employees earn on average less than federal employees
The critical value is [tex]Z_{\alpha } = - 2.33[/tex]
Step-by-step explanation:
From the question we are told that
The population mean is [tex]\mu = \$ 59593[/tex]
The sample size is n = 30
The sample mean is [tex]\= x = \$ 58800[/tex]
The standard deviation is [tex]\sigma = \$ 1500[/tex]
The significance level is [tex]\alpha = 0.01[/tex]
The null hypothesis is [tex]H_o : \mu = \$ 59593[/tex]
The alternative hypothesis is [tex]H_a : \mu < \$ 59593[/tex]
The critical value of [tex]\alpha[/tex] from the normal distribution table is [tex]Z_{\alpha } = - 2.33[/tex]
Generally the test statistics is mathematically evaluated as
[tex]t = \frac{\= x - \mu}{ \frac{ \sigma }{ \sqrt{n} } }[/tex]
=> [tex]t = \frac{ 58800 - 59593 }{ \frac{ 1500 }{ \sqrt{30} } }[/tex]
=> [tex]t = -2.896[/tex]
The p-value is obtained from the z-table
[tex]p-value = P(t < -2.896) = 0.0018898[/tex]
Since [tex]p-value < \alpha[/tex] , we reject the null hypothesis, hence it can be concluded that state employees earn on average less than federal employees
The following shape is based only on squares, semicircles, and quarter circles. Find the area of the shaded part.
Answer:
this? hope it helps ........
Answer:
The answer is area=32pi-64 and the perimeter is 8pi
Step-by-step explanation:
A mass of 5 kg stretches a spring 10 cm. The mass is acted on by an external force of 10sin( t ) N(newtons) and moves in a medium that imparts a viscous force of 2 N
when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate the initial value problem describing the motion of the mass.
A)Find the solution of the initial value problem in the above problem.
B)Plot the graph of the steady state solution
C)If the given external force is replaced by a force of 2 cos(ωt) of frequency ω , find the value of ω for which the amplitude of the forced response is maximum.
Answer:
A) C1 = 0.00187 m = 0.187 cm, C2 = 0.0062 m = 0.62 cm
B) A sample of how the graph looks like is attached below ( periodic sine wave )
C) w = [tex]\sqrt[4]{3}[/tex] is when the amplitude of the forced response is maximum
Step-by-step explanation:
Given data :
mass = 5kg
length of spring = 10 cm = 0.1 m
f(t) = 10sin(t) N
viscous force = 2 N
speed of mass = 4 cm/s = 0.04 m/s
initial velocity = 3 cm/s = 0.03 m/s
Formulating initial value problem
y = viscous force / speed = 2 N / 0.04 m/s = 50 N sec/m
spring constant = mg/ Length of spring = (5 * 9.8) / 0.1 = 490 N/m
f(t) = 10sin(t/2) N
using the initial conditions of u(0) = 0 m and u"(0) = 0.03 m/s to express the equation of motion
the equation of motion = 5u" + 50u' + 490u = 10sin(t/2)
A) finding the solution of the initial value
attached below is the solution and
B) attached is a periodic sine wave replica of how the grapgh of the steady state solution looks like
C attached below
Claire has to go to the movie theater the movie starts at 4:15 pm it is a 25min walk to the theater from her home what time dose the have to leave the house to get there on time
Answer:
claire has to leave at 3:50 from her house.
Answer:
She needs to leave by 3:50 to get there on time.
Step-by-step explanation:
4:15 - 0:25 = 3:50.
For the following graph, state the polar coordinate with a positive r and positive q (in radians). Explain your steps as to how you determined the coordinate (in your own words). I'm looking for answers that involve π, not degrees for your angles. State the polar coordinate with (r, -q). Explain how you found the new angle. State the polar coordinate with (-r, q). Explain how you found the new angle. State the polar coordinate with (-r, -q). Explain how you found the new angle.
the graph has 12 segments so angle enclosed by each segment is [tex] {2\pi\over 12}=\frac{\pi}6[/tex]
anti-clockwise is taken as positive, so if you want positive q, you need to rotate 8 segments [tex] q=8\frac,{\pi}6=\frac{4\pi}3 [/tex] , and and 8 circles or units so r=8
and for a negative angle, you need to rotate clockwise
Which is 4 segments from the horizontal line. so [tex]q=-\frac{2\pi}3[/tex] and r will be same, 8 units.
[not sure about -r so I won't include it in answer]
Answer:
Points : ( 8, - 2π/3 ), ( - 8, π/3 ), ( - 8, - 5π/3 )
Step-by-step explanation:
For the first two cases, ( r, θ ) r would be > 0, where r is the directed distance from the pole, and theta is the directed angle from the positive x - axis.
So when r is positive, we can tell that this point is 8 units from the pole, so r is going to be 8 in either case,
( 8, 240° ) - because r is positive, theta would have to be an angle with which it's terminal side passes through this point. As you can see that would be 2 / 3rd of 90 degrees more than a 180 degree angle,or 60 + 180 = 240 degrees.
( 8, - 120° ) - now theta will be the negative side of 360 - 240, or in other words - 120
Now let's consider the second two cases, where r is < 0. Of course the point will still be 8 units from the pole. Again for r < 0 the point will lay on the ray pointing in the opposite direction of the terminal side of theta.
( - 8, 60° ) - theta will now be 2 / 3rd of 90 degrees, or 60 degrees, for - r. Respectively the remaining degrees will be negative, 360 - 60 = 300, - 300. Thus our second point for - r will be ( - 8, - 300° )
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So we have the points ( 8, 240° ), ( 8, - 120° ), ( - 8, 60° ), and ( - 8, - 300° ). However we only want 3 cases, so we have points ( 8, - 120° ), ( - 8, 60° ), and ( - 8, - 300° ). Let's convert the degrees into radians,
Points : ( 8, - 2π/3 ), ( - 8, π/3 ), ( - 8, - 5π/3 )
HELP ASAP ROCKY!!! will get branliest.
Answer:
work pictured and shown
Answer:
Last one
Step-by-step explanation:
● [ ( 3^2 × 5^0) / 4 ]^2
5^0 is 1 since any number that has a null power is equal to 1.
●[ (3^2 ×1 ) / 4 ]^2
● (9/4)^2
● 81 / 16
Suppose that a sample mean is .29 with a lower bound of a confidence interval of .24. What is the upper bound of the confidence interval?
Answer:
The upper bound of the confidence interval is 0.34
Step-by-step explanation:
Here in this question, we want to calculate the upper bound of the confidence interval.
We start by calculating the margin of error.
Mathematically, the margin of error = 0.29 -0.24 = 0.05
So to get the upper bound of the confidence interval, we simply add this margin of error to the mean
That would be 0.05 + 0.29 = 0.34
logx-log(x-l)^2=2log(x-1)
Answer:
x = 1.00995066776
x = 2.52925492433
Step-by-step explanation:
This sort of equation is best solved using a graphing calculator. For that purpose, I like to rewrite the equation as a function whose zeros we're seeking. Here, that becomes ...
[tex]f(x)=\log{(x)}-\log{(x-1)}^2-2\log{(x-1)}[/tex]
The attached graph shows zeros at
x = 1.00995066776 and 2.52925492433
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Comment on the equation
Note that we have taken the middle term to be the square of the log, rather than the log of a square. For the latter interpretation, see mberisso's answer at https://brainly.com/question/17210068
Comment on the answer refinement
We have used Newton's method iteration to refine the solutions to this equation. The solution near 1.00995 requires the initial guess be very close for that method to work properly. Fortunately, the 1.01 value shown on the graph is sufficient for the purpose.
