An airplane left airport A flying on a course of 72 degrees.
After flying 495 km, the plane was just able to pick up
communication signals from airport B which is due east of airport
A. If airport B's communication signals can be picked up within a
radius of 300 km from airport B, for how many kilometers can the
airplane fly and still be in contact with airport B? Give you
answer to one decimal place.
Answers
Distance for which Aeroplane can be in contact with Airport B is = 396.34 km
Step-by-step explanation:In the question,
We have an Airport at point A and another at point B.
Now,
Airplane flying at the angle of 72° with vertical catches signals from point D.
Distance travelled by Airplane, AD = 495 km
Now, Let us say,
AB = x
So,
In triangle ABD, Using Cosine Rule, we get,
[tex]cos(90-72) =cos18= \frac{AB^{2}+AD^{2}-BD^{2}}{2.AD.AB}[/tex]
So,
On putting the values, we get,
[tex]cos18 = \frac{x^{2}+495^{2}-300^{2}}{2(495)(x)}\\0.951(990x)=x^{2}+245025-90000\\x^{2}-941.54x+155025=0\\[/tex]
Therefore, x is given by,
x = 212.696, 728.844
So,
The value of x can not be 212.696 as the length of LB (radius) itself is 300 km.
So,
x = 728.844 km
So,
AL = AB - BL
AL = x - 300
AL = 728.844 - 300
AL = 428.844 km
Now, in the circle from a property of secants we can say that,
AL x AM = AD x AC
So,
428.844 x (728.844 + 300) = 495 x AC
441213.576 = 495 x AC
AC = 891.34 km
So,
The value of CD is given by,
CD = AC - AD
CD = 891.34 - 495
CD = 396.34 km
Therefore, the distance for which the Aeroplane can still be in the contact with Airport B is 396.34 km.
Related Questions
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) ________
Answers
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.
Prove the following statement using a proof by contraposition. Yr EQ,s ER, if s is irrational, then r + 1 is irrational.
Answers
Answer:
I think that what you are trying to show is: If [tex]s[/tex] is irrational and [tex]r[/tex] is rational, then [tex]r+s[/tex] is rational. If so, a proof can be as follows:
Step-by-step explanation:
Suppose that [tex]r+s[/tex] is a rational number. Then [tex]r[/tex] and [tex]r+s[/tex] can be written as follows
[tex]r=\frac{p_{1}}{q_{1}}, \,p_{1}\in \mathbb{Z}, q_{1}\in \mathbb{Z}, q_{1}\neq 0[/tex]
[tex]r+s=\frac{p_{2}}{q_{2}}, \,p_{2}\in \mathbb{Z}, q_{2}\in \mathbb{Z}, q_{2}\neq 0[/tex]
Hence we have that
[tex]r+s=\frac{p_{1}}{q_{1}}+s=\frac{p_{2}}{q_{2}}[/tex]
Then
[tex]s=\frac{p_{2}}{q_{2}}-\frac{p_{1}}{q_{1}}=\frac{p_{2}q_{1}-p_{1}q_{2}}{q_{1}q_{2}}\in \mathbb{Q}[/tex]
This is a contradiction because we assumed that [tex]s[/tex] is an irrational number.
Then [tex]r+s[/tex] must be an irrational number.
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.
Answers
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
**Question 1\.\** The data were gathered by the following procedure, reported in the study. "Between January and June 1998, parents of children aged 2-16 years [...] that were seen as outpatients in a university pediatric ophthalmology clinic completed a questionnaire on the child’s light exposure both at present and before the age of 2 years." Was this study observational, or was it a controlled experiment? Explain.
Answers
Answer:
OBSERVATIONAL
Step-by-step explanation:
Let`s see the two kind of experiments mentioned:
Observational experiments: are those where individuals are observed or certain outcomes are measured. There is no intervention by the scientist, for example: no treatment at all is givenControlled experiments: Are those experiments that are directly manipulaed by the sciencist in order to study the controlled variable's reaction or change.In this case, the patient's parent only answer a questionnairewith no intervention from the doctors into the pacients at all, so it's an observational experiment
827 divieded by 26 with a fraction remainder
Answers
Solve the system of linear equations using the Gauss-Jordan elimination method. 2x1 − x2 + 3x3 = −10 x1 − 2x2 + x3 = −3 x1 − 5x2 + 2x3 = −7 (x1, x2, x3) =
Answers
Answer:
The solution is: [tex](x_{1}, x_{2}, x_{3}) = (1,0,-4)[/tex]
Step-by-step explanation:
The Gauss-Jordan elimination method is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.
We have the following system:
[tex]2x_{1} - x_{2} + 3x_{3} = -10[/tex]
[tex]x_{1} - 2x_{2} + x_{3} = -3[/tex]
[tex]x_{1} - 5x_{2} + 2x_{3} = -7[/tex]
This system has the following augmented matrix:
[tex]\left[\begin{array}{ccc}2&-1&3|-10\\1&-2&1|-3\\1&-5&2| -7\end{array}\right][/tex]
To make the reductions easier, i am going to swap the first two lines. So
[tex]L1 <-> L2[/tex]
Now the matrix is:
[tex]\left[\begin{array}{ccc}1&-2&1|-3\\2&-1&3|-10\\1&-5&2| -7\end{array}\right][/tex]
Now we reduce the first row, doing the following operations
[tex]L2 = L2 - 2L1[/tex]
[tex]L3 = L3 - L1[/tex]
So, the matrix is:
[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&3&1|-4\\0&-3&1| -4\end{array}\right][/tex]
Now we divide L2 by 3
[tex]L2 = \frac{L2}{3}[/tex]
So we have
[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&1&\frac{1}{3}|\frac{-4}{3}\\0&-3&1| -4\end{array}\right][/tex]
Now we have:
[tex]L3 = 3L2 + L3[/tex]
So, now we have our row reduced matrix:
[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&1&\frac{1}{3}|\frac{-4}{3}\\0&0&2| -8\end{array}\right][/tex]
We start from the bottom line, where we have:
[tex]2x_{3} = -8[/tex]
[tex]x_{3} = \frac{-8}{2}[/tex]
[tex]x_{3} = -4[/tex]
At second line:
[tex]x_{2} + \frac{x_{3}}{3} = \frac{-4}{3}[/tex]
[tex]x_{2} - \frac{4}{3} = -\frac{4}{3}[/tex]
[tex]x_{2} = 0[/tex]
At the first line
[tex]x_{1} -2x_{2} + x_{3} = -3[/tex]
[tex]x_{1} - 4 = -3[/tex]
[tex]x_{1} = 1[/tex]
The solution is: [tex](x_{1}, x_{2}, x_{3}) = (1,0,-4)[/tex]
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?
Answers
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.
The percent increase in Americans in prison for drug related offenses from 1980 to 2015 was 1048%. In 1980 the number of Americans in prison for drug related offenses was 40,900. How many American's were in prison for drug related offenses in 2015?
Answers
Answer: There were 469,532 American prison for drug related offenses in 2015.
Step-by-step explanation:
Since we have given that
Number of Americans in prison for drug related offenses in 1980 = 40,900
Rate of increment in Americans in prison from 1980 to 2015 = 1048%
So, Number of Americans who were in prison for drug related offenses in 2015 is given by
[tex]\dfrac{100+1048}{100}\times 40900\\\\\dfrac{1148}{100}\times 40900\\\\=1148\times 409\\\\=469,532[/tex]
Hence, there were 469,532 American prison for drug related offenses in 2015.
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? .
Answers
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
Assume that MTA Sandwiches sells sandwiches for $2.85 each. The cost of each sandwich follows:
Materials $ 0.80
Labor 0.40
Variable overhead 0.40
Fixed overhead ($18,400 per month, 18,400 units per month) 1.00
Total cost per sandwich $ 2.60
One of MTA's regular customers asked the company to fill a special order of sandwiches at a selling price of $1.85 each for a fund-raising event sponsored by a social club at the local college. MTA has capacity to fill it without affecting total fixed costs for the month. MTA's general manager was concerned about selling the sandwiches below the cost of $2.60 per sandwich and has asked for your advice.
Required:
a. Prepare a schedule to show the impact on MTA's profits of providing 800 sandwiches in addition to the regular production and sales of 18,400 sandwiches per month. (Select option "higher" or "lower", keeping Status Quo as the base. Select "None" if there is no effect.)
b. Based solely on the data given, what is the lowest price per sandwich at which the special order can be filled without reducing MTA's profits? (Round your answer to 2 decimal places.)
Answers
Final answer:
Explaining the impact of selling additional sandwiches on MTA's profits.
Explanation:
To prepare the schedule showing the impact on MTA's profits:
Calculate the total income from selling 800 additional sandwiches: $2.85 x 800 = $2,280.
Calculate the total cost of producing 800 sandwiches: $5 x 800 = $4,000.
Subtract the cost from the income to find the profit impact: $2,280 - $4,000 = lower profit.
Explanation (150 words): Adding 800 sandwiches at a cost of $5 each while selling at $2.85 results in a loss, impacting MTA's profits negatively. This shift reduces the overall profitability due to the increased production costs outweighing the revenue generated from the additional sandwich sales.
Is the following relation a function?
Yes
NO
Answers
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.
Find the equation of the line going through the points (2,-1) and (5,2) 3x 2y
Answers
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]
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?
Answers
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
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 ?
Answers
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)
Is 4320 perfect, abundant, or deficient? also perfect numbers? Explain why all positive multiples of 6 greater than 6 are aburndant numbers.
Answers
Final answer:
The number 4320 is an abundant number because the sum of its proper divisors exceeds 4320. All positive multiples of 6 greater than 6 are abundant because their smallest divisor, 6, is the sum of its proper divisors, and they have additional divisors that increase the sum.
Explanation:
To determine if the number 4320 is perfect, abundant, or deficient, we must compare the sum of its proper divisors (excluding the number itself) with the number. A perfect number is equal to the sum of its divisors. A number is abundant if the sum of its divisors is greater than the number, and it is deficient if the sum is less.
For 4320, the divisors are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 27, 30, 32, 36, 40, 45, 48, 54, 60, 64, 72, 80, 90, 96, 108, 120, 135, 144, 160, 180, 192, 216, 240, 270, 288, 320, 360, 432, 480, 540, 576, 720, 864, 1080, and 1440. Their sum is greater than 4320, so it is an abundant number.
Regarding positive multiples of 6 greater than 6 being abundant, the smallest divisor of such a multiple is always 6, which is already the sum of its divisors (1, 2, and 3). Since there are additional divisors beyond 1, 2, and 3, the sum of divisors must exceed the number, making it abundant.
If triangle MNP is congruent to triangle PNM, classifytriangle
MNP by its sides.
Answers
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!!!!!!!!!!!!!!!!!!
Answers
Find a general solution of y" - 6y' +10y=0.
Answers
Answer:
[tex]y(x) = e^{3x} [Acos x+Bsin x][/tex]
Step-by-step explanation:
Given is a differential equation
[tex]y" - 6y' +10y=0.[/tex]
We have characteristic equation as
[tex]m^2-6m+10 =0[/tex]
The above quadratic cannot be factorised hence use formula
[tex]m=\frac{6+/-\sqrt{36-40} }{2} \\=3+i, 3-i[/tex]
Hence general solution would be
[tex]y(x) = e^{3x} [Acos x+Bsin x][/tex]
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? ________
Answers
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).
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)
Answers
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
Given: ABCD trapezoid, BK ⊥ AD , AB=DC AB=8, AK=4 Find: m∠A, m∠B
Answers
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]
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)
Answers
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.
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)
Answers
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]
what are the measures of angle A and angle B when angle A is
half as large as its complement, angle b?
Answers
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
A penny collection contains twelve 1967 pennies, seven 1868 pennies and eleven 1971 pennies. If you are to pick some pennies without looking at the dates, how many must you pick to be sure of getting at least five pennies from the same year. Show work.
Answers
Answer:
You must pick at least 13 pennies to be sure of getting at least five from the same year.
Step-by-step explanation:
You have:
12 1967 pennies
7 1868 pennies
11 1971 pennies.
how many must you pick to be sure of getting at least five pennies from the same year?
This value is the multiplication of the number of different pennies by the antecessor of the number of pennies you want, added by 1.
So
You have 3 differennt pennies
You want to get at least five from the same year.
[tex]3*4 + 1 = 13[/tex]
You must pick at least 13 pennies to be sure of getting at least five from the same year.
For example, if you pick 12 pennies, you can have four from each year.
Adding three values, 13 is the smallest number that you need at least one term of the addition being equal or bigger than 5.
To be sure of getting at least five pennies from the same year, you need to pick 35 pennies in total.
Explanation:To ensure that you get at least five pennies from the same year, you need to consider the worst-case scenario. In this case, the worst-case scenario is where you pick pennies from each of the three different years first before getting five from the same year. So, you need to pick the maximum number of pennies from each year first before reaching the desired goal. The maximum number of pennies you need to pick is:
12 + 7 + 11 + 5 = 35 pennies
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Solve the given differential equation by undetermined coefficients.
y''' − 3y'' + 3y' − y = ex − x + 21
Answers
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
In what proportion should 10% ethanol be mixed with 65% ethanol to obtain 50% ethanol?
Answers
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]
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?
Answers
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
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.
Answers
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}
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Prove that if n is a perfect square, then n+1 can never be a perfect square
Answers
Answer:
Proved
Step-by-step explanation:
To prove that if n is a perfect square, then n+1 can never be a perfect square
Let n be a perfect square
[tex]n=x^2[/tex]
Let [tex]n+1 = y^2[/tex]
Subtract to get
[tex]1 = y^2-x^2 =(y+x)(y-x)[/tex]
Solution is y+x=y-x=1
This gives x=0
So only 0 and 1 are consecutive integers which are perfect squares
No other integer satisfies y+x=y-x=1