Answer:
It is not true
Step-by-step explanation:
Suppose your domain is the integer numbers. Define
P(x)="x is even"
Q(x)="x is odd"
So we have that the predicate [tex]\forall x(P(x) \vee Q(x))[/tex] is always true because the integers are always even or odd. But the predicate [tex]\forall x P(x) \vee \forall x Q(x)[/tex] means that all the integer numbers are even or all the integer numbers are odd, which is false. So we can't deduce [tex]\forall x P(x) \vee \forall x Q(x)[/tex] from [tex]\forall x(P(x) \vee Q(x))[/tex].
The Eco Pulse survey from the marketing communications firm Shelton Group asked individuals to indicate things they do that make them feel guilty (Los Angeles Times, August 15, 2012). Based on the survey results, there is a .39 probability that a randomly selected person will feel guilty about wasting food and a .27 probability that a randomly selected person will feel guilty about leaving lights on when not in a room. Moreover, there is a .12 probability that a randomly selected person will feel guilty for both of these reasons. a. What is the probability that a randomly selected person will feel guilty for either wasting food or leaving lights on when not in a room (to 2 decimals)? b. What is the probability that a randomly selected person will not feel guilty for either of these reasons (to 2 decimals)?
Answer:
a) There is a probability of 42% that the person will feel guilty for only one of those things.
b)There is a probability of 46% that a randomly selected person will not feel guilty for either of these reasons
Step-by-step explanation:
This probability problem can be solved by building a Venn like diagram for each probability.
I say that we have two sets:
-Set A, for those people that will feel guilty about wasting food.
-Set B, for those people that will feel guilty about leaving lights on when not in a room.
The most important information is that there is a .12 probability that a randomly selected person will feel guilty for both of these reasons. It means that [tex]P(A \cap B) = .12.[/tex]
The problem also states that there is a .39 probability that a randomly selected person will feel guilty about wasting food. It means that P(A) = 0.39. The probability of a person feeling guilty for only wasting food is PO(A) = .39-.12 = .27.
Also, there is a .27 probability that a randomly selected person will feel guilty about leaving lights on when not in a room. So, the probability of a person feeling guilty for only leaving the lights on is PO(B) = 0.27-0.12 = 0.15.
a) What is the probability that a randomly selected person will feel guilty for either wasting food or leaving lights on when not in a room?
This is the probability that the person feels guilt for only one of those things, so:
P = PO(A) + PO(B) = 0.27 + 0.15 = 0.42 = 42%
b) What is the probability that a randomly selected person will not feel guilty for either of these reasons
The sum of all the probabilities is always 1. In this problem, we have the following probabilies
- The person will not feel guilty for either of these reasons: P
- The person will feel guilty for only one of those things: PO(A) + PO(B) = 0.42
- The person will feel guilty for both reasons: PB = 0.12
So
`P + 0.42 + 0.12 = 1
P = 1-0.54
P = 0.46
There is a probability of 46% that a randomly selected person will not feel guilty for either of these reasons
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).
hugh and janet are dining out at a cafe that has on its menu 6 entrees, 5 salads and 10 desserts. They decide thy will each order a different entree, salad, and dessert, and share the portions. how many different meals are possible?
Answer: 300
Step-by-step explanation:
Given : Hugh and Janet are dining out at a cafe that has on its menu 6 entrees, 5 salads and 10 desserts.
They decide thy will each order a different entree, salad, and dessert, and share the portions.
Then, the number of different meals are possible :-
[tex]6\times5\times10=300[/tex]
Hence, the number of different meals are possible =300
1) Solve the word problem for the portion, rate, or base.
A quality control process finds 46.8 defects for every 7,800 units of production. What percent of the production is defective?
2) Solve the word problem for the portion, rate, or base.
A medical insurance policy requires Ana to pay the first $100 of her hospital expense. The insurance company will then pay 90% of the remaining expense. Ana is expecting a short surgical stay in the hospital, for which she estimates the total bill to be about $4,800.
How much (in $) of the total bill will Ana owe?
Answer:
1) 0.6% of the production is defective.
2) Ana will owe $570.
Step-by-step explanation:
Both questions here are percentage problems
Percentage problems can be explained as a rule of three problem
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. In this case, the rule of three is a cross multiplication.
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 case, the rule of three is a line multiplication.
Percentage problems have a direct relationship between the measures.
1) The problem states that a quality control process finds 46.8 defects for every 7,800 units of production. And asks what percent of the production is defective? We have to answers how many defects are there for 100 units of production. So:
46.8 defects - 7,800 units
x defects - 100 units
7,800x = 4680
[tex]x = \frac{4680}{7800}[/tex]
x = 0.6
0.6% of the production is defective.
2) The problem states that the medical insurance policy requires Ana to pay the first $100 of her hospital expense. The insurance company will then pay 90% of the remaining expense. The total bill is expected to be about $4,800.
Ana has to pay:
P = P1 + P2
-P1 :The first $100
-P2: 10% of the remaining expense. The remaining expense is $4,800-$100 = $4,700. Ana has to pay 10% of this. So
4700 - 100%
P2 - 10%
100P2 = 47000
[tex]P2 = \frac{47000}{100}[/tex]
P2 = $470
Ana will owe P = P1 + P2 = $100 + $470 = $570.
To find the percentage of defective units in the production, set up a ratio and calculate the percentage. Approximate 0.6% of the production is defective. To find how much Ana will owe of the total bill, subtract $100 from the total bill and calculate 10% of the remaining expense. Ana will owe $470 of the total bill.
Explanation:To find the percent of the production that is defective, we need to find the ratio of the number of defective units to the total number of units produced. In this case, we have 46.8 defects for every 7,800 units of production. So, we can set up the ratio as:
Defective Units / Total Units = 46.8 / 7,800
To find the percentage, we can multiply the ratio by 100:
Percentage of Defective Units = (46.8 / 7,800) * 100
Now, calculate the value of the ratio and simplify to find the percentage:
Percentage of Defective Units = 0.006 * 100 = 0.6%
Therefore, approximately 0.6% of the production is defective.
For the second question, to find how much of the total bill Ana will owe, we need to calculate the 10% of the remaining expense after she pays the first $100. First, subtract $100 from the total bill:
Remaining Expense = $4,800 - $100 = $4,700
Next, calculate 10% of the remaining expense:
Amount Ana will owe = 10% of $4,700 = ($4,700 * 10) / 100 = $470
Therefore, Ana will owe $470 of the total bill.
Al purchases a speedboat costing $24,500. State taxes are 5.5% and federal excise tax is 13%. What is the total purchase price? (Round your answer to the nearest cent if necessary)
Answer: The total purchase price is $ 29,032
Step-by-step explanation:
Hi, to solve this problem you have to solve the percentages of each taxes first.
So :
state taxes =$24,500 × 5.5% = $1,347.5federal tax = $24,500 × 13% = $3,185The next step is adding the taxes results to the speedboat cost.
so:
speedboat = $24,500
speedboat + taxes = $24,500 + $1,347.5 +$3,185 = $29,032
The total purchse price for the speedboat is $29,032.
If AA and BB are countable sets, then so is A∪BA∪B.
Answer with Step-by-step explanation:
We are given that A and B are two countable sets
We have to show that if A and B are countable then [tex]A\cup B[/tex] is countable.
Countable means finite set or countably infinite.
Case 1: If A and B are two finite sets
Suppose A={1} and B={2}
[tex]A\cup B[/tex]={1,2}=Finite=Countable
Hence, [tex]A\cup B[/tex] is countable.
Case 2: If A finite and B is countably infinite
Suppose, A={1,2,3}
B=N={1,2,3,...}
Then, [tex]A\cup B[/tex]={1,2,3,....}=N
Hence,[tex]A\cup B[/tex] is countable.
Case 3:If A is countably infinite and B is finite set.
Suppose , A=Z={..,-2,-1,0,1,2,....}
B={-2,-3}
[tex]A\cup B[/tex]=Z=Countable
Hence, [tex]A\cup B[/tex] countable.
Case 4:If A and B are both countably infinite sets.
Suppose A=N and B=Z
Then,[tex]A\cup B[/tex]=[tex]N\cup Z[/tex]=Z
Hence,[tex]A\cup B[/tex] is countable.
Therefore, if A and B are countable sets, then [tex]A\cup B[/tex] is also countable.
Answer:
To remedy confusions like yours and to avoid the needless case analyses, I prefer to define X to be countable if there is a surjection from N to X.
This definition is equivalent to a few of the many definitions of countability, so we are not losing any generality.
It is a matter of convention whether we allow finite sets to be countable or not (though, amusingly, finite sets are the only ones whose elements you could ever finish counting).
So, if A and B be countable, let f:N→A and g:N→B be surjections. Then the two sequences (f(n):n⩾1)=(f(1),f(2),f(3),…) and (g(n):n⩾1)=(g(1),g(2),g(3),…) eventually cover all of A and B, respectively; we can interleave them to create a sequence that will surely cover A∪B:
(h(n):n⩾1):=(f(1),g(1),f(2),g(2),f(3),g(3),…).
An explicit formula for h is h(n)=f((n+1)/2) if n is odd, and h(n)=g(n/2) if n is even.
Hope it helps uh mate...✌
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]
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
Medication Z is prescribed for a 50 kg woman and is to be given in enteric-coated pills of 125 mg per pill This medication is to be administered every 6 hours, but the total daily amount cannot exceed 8 mg.per kg of body weight per day. a) How many pills will this patient need? b) If the first dose is administered at 6 AM, when is the last dose given?
Answer:
a) The patient is going to need 3.2 pills. So four doses.
b) The last dose is at midnight, that i consider 12 PM
Step-by-step explanation:
This problem can be solved by a rule of three in which the measures are directly related, meaning that we have a cross multiplication.
a) How many pills will this patient need?
The first step is finding how many mg she is going to need.
She weighs 50kg, and the total daily amount cannot exceed 8 mg per kg of body. So:
1kg - 8 mg
50 kg - x mg
[tex]x = 50*8[/tex]
[tex]x = 400[/tex]mg.
Each dose has 125 mg, so she will need:
1 dose - 125 mg
x doses - 400mg
[tex]125x = 400[/tex]
[tex]x = \frac{400}{125}[/tex]
[tex]x = 3.2[/tex]
The patient is going to need 3.2 pills. So four doses.
b) If the first dose is administered at 6 AM, when is the last dose given?
There are four doses, administered every 6 hours.
6AM: First dose
12AM: Second dose
6PM: Third dose
12PM: Fourth Dose
The last dose is at midnight, that i consider 12 PM
Final answer:
To calculate the number of pills, we first determine the total daily allowable dosage based on the patient's weight and divide that by the dosage per pill, rounding up to provide whole pills. Then we schedule the medication every 6 hours starting from 6 AM, with the last dose at 12 AM.
Explanation:
The question involves calculating the appropriate dosage of medication for a patient based on weight and the maximum allowable daily dosage.
The medication prescribed is available as enteric-coated pills of 125mg each, and the patient requires a dosage that does not exceed 8mg per kg of their body weight per day.
The patient weighs 50 kg. To find the total daily dosage allowed, we multiply the patient's weight by the maximum mg per kg: 50 kg × 8 mg/kg = 400 mg.
Since the patient can have 400 mg of medication per day and the medication comes in 125 mg pills, we divide the total daily dosage by the dosage per pill: 400 mg ÷ 125 mg/pill = 3.2 pills.
Since we cannot give a patient a fraction of a pill, the patient will need 4 pills per day.
The medication must be administered every 6 hours. If the first dose is at 6 AM, the subsequent doses will be at 12 PM, 6 PM, and 12 AM, with the last dose given precisely 18 hours after the first dose.
Question 4 options:
Given the following definitions:
U = {a, b, c, d, e, f, g}
A = {a, c, e, g}
B = {a, b, c, d}
Find A'
Answer in roster form, with a single space after each comma.
Answer:
{ b, d, f }
Step-by-step explanation:
In the roster form we write the elements of a set by separating commas and enclose them within {} bracket.
We have give,
[tex]U = \{a, b, c, d, e, f, g\}[/tex],
[tex]A = \{a, c, e, g\}[/tex],
[tex]B = \{a, b, c, d\}[/tex],
[tex]\because A' = U - A[/tex]
[tex]\implies A' = \{a, b, c, d, e, f, g\} - \{a, c, e, g\}[/tex]
= { b, d, f }
Final answer:
A' is the complement of set A, consisting of elements in the universal set U that are not in A. Given U = {a, b, c, d, e, f, g} and A = {a, c, e, g}, A' is found to be {b, d, f}.
Explanation:
The complement of a set A, denoted by A', consists of all the elements in the universal set U that are not in A. Given the universal set U = {a, b, c, d, e, f, g} and set A = {a, c, e, g}, to find A' we need to identify all elements in U that are not in A.
To find A', we compare each element in U with the elements in A:
If the element is in A, we do not include it in A'.
If the element is not in A, we include it in A'.
After comparing, we find that A' consists of the elements {b, d, f} since these are the elements in U that are not present in set A.
Therefore, the roster form of A' with a single space after each comma is:
{b, d, f}
Learn more about Set Complement here:
https://brainly.com/question/30913259
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Solve. X1 – 3x2 + 4x3 = -4 3xı – 7x2 + 7x3 = -8 –4x1 + 6x2 – x3 = 7
Answer:
There is no solution for this system
Step-by-step explanation:
I am going to solve this system by the Gauss-Jordan elimination method.
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}1&-3&4|-4\\3&-7&7|-8\\-4&6&-1|7\end{array}\right][/tex]
We start reducing the first row. So:
[tex]L2 = L2 - 3L1[/tex]
[tex]L3 = L3 + 4L1[/tex]
Now the matrix is:
[tex]\left[\begin{array}{ccc}1&-3&4|-4\\0&2&-5|4\\0&-6&15|-9\end{array}\right][/tex]
We divide the second line by 2:
[tex]L2 = \frac{L2}{2}[/tex]
And we have the following matrix:
[tex]\left[\begin{array}{ccc}1&-3&4|-4\\0&1&\frac{-5}{2}|2\\0&-6&15|-9\end{array}\right][/tex]
Now we do:
[tex]L3 = L3 + 6L2[/tex]
So we have
[tex]\left[\begin{array}{ccc}1&-3&4|-4\\0&1&\frac{-5}{2}|2\\0&0&0|3\end{array}\right][/tex]
This reduced matrix means that we have:
[tex]0x_{3} = 3[/tex]
Which is not possible
There is no solution for this system
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]
Chose parameters h and k such that the system has a) a unique solution, b) many solutions, and c) no solution. X1 + 3x2 = 4 2x1 + kx2 = h
Answer:
a) The system has a unique solution for [tex]k\neq 6[/tex] and any value of [tex]h[/tex], and we say the system is consisted
b) The system has infinite solutions for [tex]k=6[/tex] and [tex]h=8[/tex]
c) The system has no solution for [tex]k=6[/tex] and [tex]h\neq 8[/tex]
Step-by-step explanation:
Since we need to base the solutions of the system on one of the independent terms ([tex]h[/tex]), the determinant method is not suitable and therefore we use the Gauss elimination method.
The first step is to write our system in the augmented matrix form:
[tex]\left[\begin{array}{cc|c}1&3&4\\2&k&h\end{array}\right][/tex]
The we can use the transformation [tex]r_0\rightarrow r_0 -2r_1[/tex], obtaining:
[tex]\left[\begin{array}{cc|c}1&3&4\\0&k-6&h-8\end{array}\right][/tex].
Now we can start the analysis:
If [tex]k\neq 6[/tex] then, the system has a unique solution for any value of [tex]k[/tex], meaning that the last row will transform back to the equation as:[tex](k-6)x_2=h-8\\x_2=h-8/(k-6)[/tex]
from where we can see that only in the case of [tex]k=6[/tex] the value of [tex]x_2[/tex] can not be determined.
if [tex]k=6[/tex] and [tex]h=8[/tex] the system has infinite solutions: this is very simple to see by substituting these values in the equation resulting from the last row:[tex](k-6)x_2=h-8\\0=0[/tex] which means that the second equation is a linear combination of the first one. Therefore, we can solve the first equation to get [tex]x_1[/tex] as a function of [tex]x_2[/tex] o viceversa. Thus, [tex]x_2[/tex] ([tex]x_1[/tex]) is called a parameter since there are no constraints on what values they can take on.
if [tex]k=6[/tex] and [tex]h\neq 8[/tex] the system has no solution. Again by substituting in the equation resulting from the last row:
[tex](k-6)x_2=h-8\\0=h-8[/tex] which is false for all values of [tex]h\neq 8[/tex] and since we have something that is not possible [tex](0\neq h-8,\ \forall \ h\neq 8)[/tex] the system has no solution
Is the following relation a function?
Yes
NO
Answer: The relation is a function.
Step-by-step explanation: In this situation, we are given a relation in the form of a graph and we are asked if it represents a function. In this situation, we would you something called the vertical line test. In other words, if we can draw a vertical line that passes through more than one point on the graph, then the relation is not a function. Notice that in this problem, it's impossible to draw a vertical line that passes through more than one point on the graph so the relation is a function.
Therefore, this relation must be a function.
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
What is meant by the "complexity of an algorithm"?
Answer: Algorithm complexity or the complexity of an algorithm is known as a measure under which one evaluates degree of count of operations, that are specifically performed by an algorithm which is taken in consideration as a function of size of the data. In rudimentary terms, it is referred to as a rough approximation of number of stages required in order to enforce an algorithm.
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
Ravi takes classes at both Westside Community College and Pinewood Community College. At Westside, class fees are $98 per credit hour, and at Pinewood, class fees are S115 per credit hour. Ravi is taking a combined total of 18 credit hours at the two schools. Suppose that he is taking w credit hours at Westside. Write an expression for the combined total dollar amount he paid for his class fees. total paid (in dollars) - 0 x 5 ?
Answer:
98w +115(18-w)
Step-by-step explanation:
The fees per credit hour at Westside are $98.
The fees per credit hour at Pinewood are $115.
The total amount of hours is 18, this means the hours at Westside plus the hours at Pinewood add up to 18.
The problem asks us to name "w" the credit hours at Westside, therefore, the credit hours at Pinewood would be "18 - w"
Combining this information, we have that the combined total amount he paid would be:
(Fee per hour at Westside)(hours at Westside) + (Fee per hour at Pinewood)(hours at Pinewood) = Total
⇒98w + 115(18 - w) = Total
Final answer:
The expression for the combined total cost of Ravi's classes is 98w + 115(18 - w), where w represents the number of credit hours taken at Westside College, and (18 - w) represents the number of credit hours taken at Pinewood College.
Explanation:
To calculate the combined total cost of Ravi's classes at Westside Community College and Pinewood Community College, we can create an expression based on the number of credit hours he is taking at each school. If he is taking w credit hours at Westside, where the cost is $98 per credit hour, the total cost at Westside would be 98w. Since Ravi is taking a total of 18 credit hours at both schools combined, he would be taking (18 - w) credit hours at Pinewood, where the cost is $115 per credit hour. The total cost at Pinewood would therefore be 115(18 - w).
The expression for the combined total amount paid for Ravi's classes would be the sum of the costs for each school:
Total Paid = 98w + 115(18 - w)
pls tell me the
domain
range
and if its a function or not thx
Answer:
both domain and range are {-4, -3, -2, -1, 0, 1, 2, 3, 4}
it IS a function
Step-by-step explanation:
The domain is the list of x-values of the plotted points. It is all the integers from -4 to +4, inclusive.
The range is the list of y-values of the plotted points. It is all the integers from -4 to +4, inclusive. (domain and range are the same for this function)
No x-value has more than one y-value associated with it, so this relation IS A FUNCTION.
The provider orders digoxin pediatric elixir 60 mcg, PO, BID. The bottle of elixir contains 0.05 mg per 1 ml. Calculate the dosage in ml. (round to nearest tenth)
Answer:
The dosage has 1.2ml
Step-by-step explanation:
This problem can be solved as a rule of three problem.
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.
Unit conversion problems, like this one, is an example of a direct relationship between measures.
First step: The first step is the conversion from mcg to mg.
The bottle has 60 mcg. How much is this in ml? 1 mcg has 0.001mg. So:
1 mcg - 0.001mg
60mcg - x mg
x = 60*0.001
x = 0.06mg
Final step: The bottle of elixir contains 0.05 mg per 1 ml. Calculate the dosage in ml.
The dosage has 0.06 mg, so:
0.05mg - 1 ml
0.06mg - xml
0.05x = 0.06
[tex]x = \frac{0.06}{0.05}[/tex]
x = 1.2ml
The dosage has 1.2ml
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
When 23 mL of water for injection is added to a drug-lyophilized powder, the resulting concentration is 200,000 units per mL. What is the volume of the dry powder if the amount of drug in the vial was 5,000,000 units? .
Answer:
2 mL
Step-by-step explanation:
Given:
Volume of water for injection = 23 mL
Resulting concentration = 200,000 units per mL
Amount of drug in the vial = 5,000,000 units
Now,
Let the final volume of the solution be 'x' mL
Now, concentration = [tex]\frac{\textup{units of the powder}}{\textup{Total volume of the soltuion}}[/tex]
thus,
200,000 = [tex]\frac{\textup{5,000,000}}{\textup{x}}[/tex]
or
x = 25 mL
also,
Total volume 'x' = volume of water + volume of powder
or
25 mL = 23 + volume of powder
or
Volume of powder = 2 mL
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.
Find the equation of the line going through the points (2,-1) and (5,2) 3x 2y
Answer:
The equation of the line is:
[tex]y = x - 3[/tex]
Step-by-step explanation:
The general equation of a straight line is given by:
[tex]y = ax + b[/tex]
Being given two points, we can replace x and y, solve the system and find the values for a and b.
Solution:
The line goes through the point [tex](2,-1)[/tex]. It means that when [tex]x = 2, y = -1[/tex]. Replacing in the equation:
[tex]y = ax + b[/tex]
[tex]-1 = 2a + b[/tex]
[tex]2a + b = -1[/tex]
The line also goes through the point [tex](5,2)[/tex]. It means that when [tex]x = 5 y = 2[/tex]. Replacing in the equation:
[tex]y = ax + b[/tex]
[tex]2 = 5a + b[/tex]
[tex]5a + b = 2[/tex]
Now we have to solve the following system of equations:
[tex]1) 2a + b = -1[/tex]
[tex]2) 5a + b = 2[/tex]
From 1), we have:
[tex]b = -1 - 2a[/tex]
Replacing in 2)
[tex]5a - 1 - 2a = 2[/tex]
[tex]3a = 3[/tex]
[tex]a = \frac{3}{3}[/tex]
[tex]a = 1[/tex]
[tex]b = -1 - 2a = -1 - 2 = -3[/tex]
The equation of the line is:
[tex]y = x - 3[/tex]
Prove: If n is a positiveinteger and n2 is
divisible by 3, then n is divisible by3.
Answer:
If [tex]n^2[/tex] is divisible by 3, the n is also divisible by 3.
Step-by-step explanation:
We will prove this with the help of contrapositive that is we prove that if n is not divisible by 3, then, [tex]n^2[/tex] is not divisible by 3.
Let n not be divisible by 3. Then [tex]\frac{n}{3}[/tex] can be written in the form of fraction [tex]\frac{x}{y}[/tex], where x and y are co-prime to each other or in other words the fraction is in lowest form.
Now, squaring
[tex]\frac{n^2}{9} = \frac{x^2}{y^2}[/tex]
Thus,
[tex]n^2 = \frac{9x^2}{y^2}[/tex]
[tex]\frac{n^2}{3} = \frac{3x^2}{y^2}[/tex]
It can be clearly seen that the fraction [tex]\frac{3x^2}{y^2}[/tex] is in lowest form.
Hence, [tex]n^2[/tex] is not divisible by 3.
Thus, by contrapositivity if [tex]n^2[/tex] is divisible by 3, the n is also divisible by 3.
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.
"The diagram above represents a square garden. If each side of the garden is increased in length by 50%, by what percent is the area of the garden increased?" Whats the explanation for this problem, I cant seem to figure it out?
Answer:
Step-by-step explanation:
Let's say that each side of the square is of length [tex]s[/tex]. The area of this square would then be:
[tex]A = s^{2}[/tex]
If we increase the length of each side by 50%, then the length becomes [tex]1.5s[/tex], which will result in the area being:
[tex]A = (1.5s)(1.5s)[/tex]
[tex]A = 2.25s^{2}[/tex]
This means the area has increased by [tex]1.25[/tex] the original amount, or 125%.
If triangle MNP is congruent to triangle PNM, classifytriangle
MNP by its sides.
Answer:
MNP is a Congruent Triangle according to Side-Side-Side Congruence.
Step-by-step explanation:
Whenever we talk about Euclidean Geometry and Congruence of Triangles. We are taking into account that in any given plane, three given points, in this case, M, N, and P is and its segments between two points make up a Triangle.
In this case MNP and an identical one and PNM
To be called congruent, it's necessary to have the same length each side and when it comes to angles, congruent angles have the same measure.
A postulate, cannot be proven since it's self-evident. And there's one that fits for this case which says
"Every SSS (Side-Side-Side) correspondence is a congruence"
So this is why we can classify MNP as Side-Side-Side congruence since its segments are the same size MN, NP, and MP for both of them. The order of the letters does not matter.
Need Help Fast!!!!!!!!!!!!!!!!!!
In what proportion should 10% ethanol be mixed with 65% ethanol to obtain 50% ethanol?
Answer:
3 : 8
Step-by-step explanation:
Let x quantity of 10% ethanol is mixed with y quantity of 65% ethanol to obtain 50% ethanol mixture,
Thus, the total quantity of resultant mixture = x + y
Also, ethanol in 10% ethanol mixture + ethanol in 65% ethanol mixture = ethanol in resultant mixture,
⇒ 10% of x + 65% of y = 50% of (x+y)
[tex]\implies \frac{10x}{100}+\frac{65y}{100}=\frac{50(x+y)}{100}[/tex]
⇒ 10x + 65y = 50(x+y)
⇒ 10x + 65y = 50x+50y
⇒ 10x - 50x = 50y - 65y
⇒ -40x = -15y
[tex]\implies \frac{x}{y}=\frac{15}{40}=\frac{3}{8}[/tex]