Step-by-step explanation:
Part A:
The two factors in 9(7+2x) are 9 and 7+2x
Part B:
First term: 9
Second term: 7+2x
Part C:
9(7+2x)
Open bracket
63+18x
The coefficient is 18x
A. The two factors are 9 and (7+2x).
B. In first factor, only one term and in second factor , two terms i.e. 7 and 2x are present.
C. The coefficient of the variable term is 18.
Algebraic expression:Given expression is;
[tex]9(7+2x)[/tex]
In given expression, there are two factors fist is 9 and second one is [tex](7+2x)[/tex]
In first factor, only one term and in second factor , two terms i.e. 7 and 2x are present.
To find the coefficient of variable term, we have to to expand given expression.
[tex]9(7+2x)=63+18x[/tex]
The coefficient of the variable term is 18.
Learn more about the algebraic expression:
https://brainly.com/question/4344214
Draw a triangle ABC, where AB = 8 cm , BC = 6 cm and angle B=70^ and locate its circumcentre and draw the circumcircle.
Step-by-step explanation:
ΔABC, where AB = 8 cm, BC = 6 cm, B = 70° Construction: (i) Draw the ∆ABC with the given measurements. (ii) Construct the perpendicular bisector at any two sides (AB and BC) and let them meet at S which is the circumcircle. (iii) S as centre and SA = SB = SC as radius, draw the circumcircle to pass through A, B, and C. Circum radius = 4.3cm .draw-triangle-abc-where-cm-bc-and-70-and-locate-its-circumcentre-and-draw-the-circumcircle
3. The size of a red blood cell is 0.000007 m and the size of a plant
cell is 0.0000127 m. Compare these two.
Given:
Size of a red blood cell = 0.000007 m
Size of a plant cell = 0.0000127 m
To find:
The comparison of these two values.
Solution:
We have,
Size of a red blood cell = 0.000007 m
Size of a plant cell = 0.0000127 m
Clearly, [tex]0.0000127>0.000007[/tex]. Now, the difference between these two values is:
[tex]0.0000127-0.000007=0.0000057[/tex]
Therefore, the size of a plant cell is 0.0000057 m more than the size of a red blood cell.
HELP AGAIN
235 ≤-8(1+5x)+3
i need the steps as well
Answer:
x ≤ -6
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to RightEquality Properties
Multiplication Property of EqualityDivision Property of EqualityAddition Property of EqualitySubtraction Property of EqualityStep-by-step explanation:
Step 1: Define
Identify
235 ≤ -8(1 + 5x) + 3
Step 2: Solve for x
[Subtraction Property of Equality] Subtract 3 on both sides: 232 ≤ -8(1 + 5x)[Division Property of Equality] Divide -8 on both sides: -29 ≥ 1 + 5x[Subtraction Property of Equality] Subtract 1 on both sides: -30 ≥ 5x[Division Property of Equality] Divide 5 on both sides: -6 ≥ xRewrite: x ≤ -6Step-by-step explanation:
To solve for x, make sure you move everything else to the other side of the ≤ sign.
So,
[tex]235\leq -8(1+5x)+3\\232\leq -8-40x\\240\leq -40x\\-6\geq x[/tex]
* Remember that the sign changes anytime you divide by a negative number!
So your answer is:
[tex]x\leq -6[/tex], x is less than or equal to -6.
What is the correct line graph for y=3x+5?
Answer:
The equation y=−3x+5 is in slope intercept form, and represents a straight line in which -3 is the slope, and 5 is the y -intercept.
Can you please help me with this question
14. The Elizabeth Tower is 320 feet tall. At what time or times during your ride on the London Eye are you at the same height as the top of the tower? Show your work. (4 points: 2 points for finding the correct time(s), 2 points for work shown)
Answer:
Ok so on a clock there is 12 numbers where 12 is on top so at 12 am and 12 pm noon and midnight you will be at the top of the clock
Hope This Helps!!!
During the ride on the London Eye, you will be at the same height as the top of the Elizabeth Tower at approximately 21 minutes and 43.16 seconds after the start of the ride.
To determine the time(s) during the ride on the London Eye when you are at the same height as the top of the Elizabeth Tower (commonly known as Big Ben), we need to consider the height of the London Eye and its rotational motion.
Given that the Elizabeth Tower is 320 feet tall, we need to find the position of the London Eye when its height aligns with the top of the tower.
The London Eye has a height of 443 feet, and it completes one full rotation in approximately 30 minutes (or 1800 seconds). This means that it moves at a constant angular velocity of 360 degrees per 1800 seconds.
To find the time(s) when the heights align, we can set up a proportion:
(Height of the Elizabeth Tower) / (Height of the London Eye) = (Angle covered by the London Eye) / 360 degrees
Substituting the given values:
320 / 443 = (Time to align) / 1800
Simplifying the equation:
(Time to align) = (320 / 443) * 1800
Calculating the value:
(Time to align) ≈ 1303.16 seconds
Converting the time to minutes and seconds:
(Time to align) ≈ 21 minutes and 43.16 seconds
Therefore, during the ride on the London Eye, you will be at the same height as the top of the Elizabeth Tower at approximately 21 minutes and 43.16 seconds after the start of the ride.
To know more about London Eye. here
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I’m stuck please help .
Answer:
me too
Step-by-step explanation:
me too
14. In this picture, three straight lines intersect at a point. Form an equation in x and solve for x.
Answer:
6x = 180
x = 30
Step-by-step explanation:
Solve the equation for x.
2/3x-1/9x+5=20
Answer:
x = 27
Step-by-step explanation:
I'm assuming the equation looks like this:
[tex]\frac{2}{3}x-\frac{1}{9}x+5=20[/tex]
Here's how to solve for x:
[tex]\frac{2}{3}x-\frac{1}{9}x+5=20[/tex]
(subtract 5 from both sides)
[tex]\frac{2}{3}x-\frac{1}{9}x=15[/tex]
(Find the GCF of 3 and 9, which is 3. Multiply 2/3 by 3/3. You get 6/9)
[tex]\frac{6}{9}x-\frac{1}{9}x=15[/tex]
(add like terms)
[tex]\frac{5}{9}x=15[/tex]
(multiply 9/5 to both sides, which is the same as dividing both sides by 5/9)
x = 27
Hope it helps (●'◡'●)
How would I solve the question below? In what order would I solve it?
4 ⋅ 3 + 2 ⋅ 9 − 40
Step-by-step explanation:
You would multiply 4 and 3, and 2 and 9 separately, then add them, then subtract 40. Remember PEMDAS.
(4*3) + (2*9) - 40
12 + 18 - 40
-10
Hope that helps
Can the range of a function be written like this {6,7,8,10} instead of like this [tex]6\leq x\leq 10[/tex]?
Answer:
No unless x is being used to define only elements of an integer set.
Step-by-step explanation:
No, not in general unless x is defined as a integer or a subset of the integers like the naturals, whole numbers....
Usually 6<=x<=10 means all real numbers between 6 and 10, inclusive. This means example that 6.6 or 2pi are in this set with infinitely other numbers that I can't name.
{6,7,8,9,10} just means the set containing the numbers 6,7,8,9,10 and that's only those 5 numbers.
Find dy/dx of the function y = √x sec*-1 (√x)
Hi there!
[tex]\large\boxed{\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \frac{1}{2|\sqrt{x}|\sqrt{{x} - 1}}}[/tex]
[tex]y = \sqrt{x} * sec^{-1}(-\sqrt{x}})[/tex]
Use the chain rule and multiplication rules to solve:
g(x) * f(x) = f'(x)g(x) + g'(x)f(x)
g(f(x)) = g'(f(x)) * 'f(x))
Thus:
f(x) = √x
g(x) = sec⁻¹ (√x)
[tex]\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \sqrt{x} * \frac{1}{\sqrt{x}\sqrt{\sqrt{x}^{2} - 1}} * \frac{1}{2\sqrt{x}}[/tex]
Simplify:
[tex]\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \sqrt{x} * \frac{1}{2|x|\sqrt{{x} - 1}}[/tex]
[tex]\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \frac{1}{2|\sqrt{x}|\sqrt{{x} - 1}}[/tex]
Answer:
[tex]\displaystyle y' = \frac{arcsec(\sqrt{x})}{2\sqrt{x}} + \frac{1}{2|\sqrt{x}|\sqrt{x - 1}}[/tex]
General Formulas and Concepts:
Algebra I
Exponential Rule [Rewrite]: [tex]\displaystyle b^{-m} = \frac{1}{b^m}[/tex]Exponential Rule [Root Rewrite]: [tex]\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}[/tex]Calculus
Derivatives
Derivative Notation
Basic Power Rule:
f(x) = cxⁿf’(x) = c·nxⁿ⁻¹Derivative Rule [Product Rule]: [tex]\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)[/tex]
Derivative Rule [Chain Rule]: [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]
Arctrig Derivative: [tex]\displaystyle \frac{d}{dx}[arcsec(u)] = \frac{u'}{|u|\sqrt{u^2 - 1}}[/tex]
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle y = \sqrt{x}sec^{-1}(\sqrt{x})[/tex]
Step 2: Differentiate
Rewrite: [tex]\displaystyle y = \sqrt{x}arcsec(\sqrt{x})[/tex]Product Rule: [tex]\displaystyle y' = \frac{d}{dx}[\sqrt{x}]arcsec(\sqrt{x}) + \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})][/tex]Chain Rule: [tex]\displaystyle y' = \frac{d}{dx}[\sqrt{x}]arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{d}{dx}[\sqrt{x}] \bigg][/tex]Rewrite [Exponential Rule - Root Rewrite]: [tex]\displaystyle y' = \frac{d}{dx}[x^\bigg{\frac{1}{2}}]arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{d}{dx}[x^\bigg{\frac{1}{2}}] \bigg][/tex]Basic Power Rule: [tex]\displaystyle y' = \frac{1}{2}x^\bigg{\frac{1}{2} - 1}arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{1}{2}x^\bigg{\frac{1}{2} - 1} \bigg][/tex]Simplify: [tex]\displaystyle y' = \frac{1}{2}x^\bigg{\frac{-1}{2}}arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{1}{2}x^\bigg{\frac{-1}{2}} \bigg][/tex]Rewrite [Exponential Rule - Rewrite]: [tex]\displaystyle y' = \frac{1}{2x^\bigg{\frac{1}{2}}}arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{1}{2x^\bigg{\frac{1}{2}}} \bigg][/tex]Rewrite [Exponential Rule - Root Rewrite]: [tex]\displaystyle y' = \frac{1}{2\sqrt{x}}arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{1}{2\sqrt{x}} \bigg][/tex]Arctrig Derivative: [tex]\displaystyle y' = \frac{1}{2\sqrt{x}}arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{1}{|\sqrt{x}|\sqrt{(\sqrt{x})^2 - 1}} \cdot \frac{1}{2\sqrt{x}} \bigg][/tex]Simplify: [tex]\displaystyle y' = \frac{arcsec(\sqrt{x})}{2\sqrt{x}} + \frac{1}{2|\sqrt{x}|\sqrt{x - 1}}[/tex]Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Derivatives
Book: College Calculus 10e
A merchant keeps marble in a cylindrical plastic container that has a diameter of 28cm and height of 35cm. A marble has a diameter of 25mm. Determine the number of marbles that can be stored in such a container if air space accounts for 20% between marbles.
Answer:
2107 marbles can be stored in the container.
Step-by-step explanation:
Since a merchant keeps marble in a cylindrical plastic container that has a diameter of 28cm and height of 35cm, and a marble has a diameter of 25mm, to determine the number of marbles that can be stored in such a container if air space accounts for 20 % between marbles, the following calculation must be performed, knowing that the volume of a cylinder is equal to height x π x radius²:
35 x 3.14 x (28/2) ² = X
109.9 x (14 x 14) = X
109.9 x 196 = X
21,540.4 = X
In turn, the volume of each 25mm diameter marble is equal to:
25mm = 2.5cm
4/3 x 3.14 x 1.25³ = X
4.18666 x 1.953125 = X
8.1770 = X
21,540.4 x 0.8 = 17,232.32
17,232.32 / 8,177 = 2,107.41
Therefore, 2107 marbles can be stored in the container.
-9(m + 2) + 406 - 7m)
Answer:
-9(m+2)+406-7m)
=-16+388
Thirty-six percent of customers who purchased products from an e-commerce site had orders exceeding 110. If 17% of customers have orders exceeding 110 and also pay with the e-commerce site's sponsored credit card, determine the probability that a customer whose order exceeds 110 will pay with the sponsored credit card.
Answer:
The right solution is "0.5".
Step-by-step explanation:
According to the question,
P(pay with the sponsored credit card | order exceeds $110)
= [tex]\frac{P(Pay \ with \ the \ sponsored \ credit\ card\ and\ order\ exceeds\ 110)}{P(order \ exceeds \ 110)}[/tex]
= [tex]\frac{P(A \ and \ B)}{P(A)}[/tex]
By putting the values, we get
= [tex]\frac{0.17}{0.34}[/tex]
= [tex]0.5[/tex]
Thus, the above is the right solution.
Find the missing term in the following pattern.
1984, 992, 496, blank space, 124, 62
Answer:
248
Step-by-step explanation:
common ratio for two consecutive terms is 2/1
for eg: 1984÷992 =2
992÷ 496 = 2
124÷ 62 = 2
that means 124 ×2 = 248 Answer
What are the solutions of the quadratic equation 49x2 = 9?
A. x = 1/9 and x = -1/9
B. x = 3/7 and x = -3/7
C. x = 3/4 and x = -3/4
D. x = 9/49 and x = -9/49
Brainliest if you explain how. got stumped on this one
Answer:
B
Step-by-step explanation:
49x^2=9
solve for x
x^2= 9/49
x=± [tex]\sqrt{9/49\\}[/tex]
which is x = ±3/7 (B)
Answer: b x=1/9 and x=-1/9
Step-by-step explanation:
The MD orders 50mg of an elixir to be given every 12 hours. Available is 125mg/5ml. How much should be administered every 12 hours?
Answer:
2ml
Step-by-step explanation:
50mg of some potent agent has to be given every 12 hours.
there is a solution that has a concentration of that agent of 125mg/5ml
we need to administer some part of this solution, which we cannot (or should not) change in its structure.
that means the ratio of agent to overall solution stays the same, no matter how much of the solution we administer.
all we need to do is to transit the ratio of 125/5 to represent 50/x (maintaining the said ratio).
in other words, we need to find how many ml we need to administer, so that 50mg of the agent enter the body.
so,
125/5 = 50/x
125x/5 = 50
25x = 50
x = 50/25 = 2
2ml of the solution needs to be administered every 12 hours.
Solve the system, or show that it has no solution. (If there is no solution, enter NO SOLUTION. If there are an infinite number of solutions, enter the general solution in terms of x, where x is any real number.)
20x − 80y = 100
−14x + 56y = −70
(x, y) =
Answer:
The system has an infinite set of solutions [tex](x,y) = (x, \frac{x-5}{4})[/tex]
Step-by-step explanation:
From the first equation:
[tex]20x - 80y = 100[/tex]
[tex]20x = 100 + 80y[/tex]
[tex]x = \frac{100 + 80y}{20}[/tex]
[tex]x = 5 + 4y[/tex]
Replacing on the second equation:
[tex]-14x + 56y = -70[/tex]
[tex]-14(5 + 4y) + 56y = -70[/tex]
[tex]-70 - 56y + 56y = -70[/tex]
[tex]0 = 0[/tex]
This means that the system has an infinite number of solutions, considering:
[tex]x = 5 + 4y[/tex]
[tex]4y = x - 5[/tex]
[tex]y = \frac{x - 5}{4}[/tex]
The system has an infinite set of solutions [tex](x,y) = (x, \frac{x-5}{4})[/tex]
The movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on your answer
Which sentence correctly compares the two numbers 5.395 and 5.385?
05.395 < 5.385
05.385 > 5.395
o 5.395 = 5.385
5.385 < 5.395
h
Submit
Pass
Don't know answer
Answer:
5.385 < 5.395
Step-by-step explanation:
Compare digits one by one starting form the left.
5.395 and 5.385
The 5s in the ones place are equal.
5.395 and 5.385
The 3s in the tenths place are equal.
5.395 and 5.385
The 9 in the hundredths place is greater than the 8 in the hundredths place, so the number with the 9 is grater than the number with the 8.
That makes the number with the 8 less than the number with the 9.
Answer: 5.385 < 5.395
Uma pizzaria oferece em seu cardápio 12 sabores de pizza. Se um cliente pretende pedir 3 pizzas, então o número de maneiras que ele pode realizar esse pedido é;
•364
•220
•440
•1320
Answer:
Step-by-step explanation:
Partindo do pressuposto de que você pode ter coberturas duplas e triplas do mesmo item, o cálculo é relativamente simples. Para calcular as combinações possíveis; deve-se multiplicar as coberturas disponíveis pelo número total de coberturas permitidas. Este cálculo é semelhante a como olhamos para diferentes sistemas de contagem de base. Normalmente contamos com decimais (base 10), portanto, o número de combinações, se usar 3 dígitos, seria calculado por 10 x 10 x 10.
10x10 = 100
100x10 = 1000 combinações (0 a 999)
Sua pergunta sobre coberturas de pizza é a mesma, mas assumindo um sistema de numeração de base 12, então 12x12x12 ou 12³
Portanto, 1.728 combinações incluindo 0 (sem coberturas?) E também incluindo 12 ocasiões em que todas as 3 coberturas seriam iguais. Se esses cenários de pessoas forem restritos de modo que você só possa ter coberturas duplas máximas, etc., então essas combinações devem ser removidas (subtraídas do total de combinações permitidas).
Espero ter ajudado você a entender os princípios, então você deve ser capaz de trabalhar a partir disso, de muitas outras soluções semelhantes
Solve (x - 5)2 = 3.
Answer:
x = 5±√3
Step-by-step explanation:
Equation: (x-5)² = 3
Step 1: Take the square root of both side of the equation
√(x-5)² = ±√3
x-5 = ±√3
Step 2: add 5 to both side of the equation
x-5+5 = 5±√3
x = 5±√3
Hence, from the options above, the right answer is
B. x = 5±√3
What is net cash flow
al of
10. A square field has four sprinklers that spray
in the areas represented by the circles below. If
the shaded portion represents area that is not
reached by the sprinklers, find the total area that
is not reached by the sprinklers.
Using the areas of the sqaure and of the circle, it is found that the total area that is not reached by the sprinklers is of 343.36 ft².
What is the area of a square?The area of a square of side length l is given by:
A = l²
In this problem, we have that l = 40 ft, hence:
A = (40 ft)² = 1600 ft².
What is the area of a circle?The area of a circle of radius r is given by:
[tex]A = \pi r^2[/tex]
In this problem, we have four circles of radius r = 10 ft, hence it's combined area, in square feet, is given by:
[tex]A_c = 4\pi (10)^2 = 400\pi = 1256.64 \text{ft}^2[/tex]
The area not reached by the sprinklers is the subtraction of the area of the square by the area of the circle, hence:
1600 - 1256.64 = 343.36 ft².
More can be learned about the area of a rectangle at https://brainly.com/question/10489198
NEED HELP
The average amount of money spent for lunch per person in the college cafeteria is $6.75 and the standard deviation is $2.28. Suppose that 18 randomly selected lunch patrons are observed. Assume the distribution of money spent is normal, and round all answers to 4 decimal places where possible.
C. For a single randomly selected lunch patron, find the probability that this
patron's lunch cost is between $7.0039 and $7.8026.
D. For the group of 18 patrons, find the probability that the average lunch cost is between $7.0039 and $7.8026.
Answer:
C.[tex]P(7.0039<x<7.8026)=0.1334[/tex]
D.[tex]P(7.0039<\bar{x}<7.8026)\approx 0.2942[/tex]
Step-by-step explanation:
We are given that
n=18
Mean, [tex]\mu=6.75[/tex]
Standard deviation, [tex]\sigma=2.28[/tex]
c.
[tex]P(7.0039<x<7.8026)=P(\frac{7.0039-6.75}{2.28}<\frac{x-\mu}{\sigma}<\frac{7.8026-6.75}{2.28})[/tex]
[tex]P(7.0039<x<7.8026)=P(0.11<Z<0.46)[/tex]
[tex]P(a<z<b)=P(z<b)-P(z<a)[/tex]
Using the formula
[tex]P(7.0039<x<7.8026)=P(Z<0.46)-P(Z<0.11)[/tex]
[tex]P(7.0039<x<7.8026)=0.67724-0.54380[/tex]
[tex]P(7.0039<x<7.8026)=0.1334[/tex]
D.[tex]P(7.0039<\bar{x}<7.8026)=P(\frac{7.0039-6.75}{2.28/\sqrt{18}}<\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}})<\frac{7.8026-6.75}{2.28/\sqrt{18}})[/tex]
[tex]P(7.0039<\bar{x}<7.8026)=P(0.47<Z<1.96)[/tex]
[tex]P(7.0039<\bar{x}<7.8026)=P(Z<1.96)-P(Z<0.47)[/tex]
[tex]P(7.0039<\bar{x}<7.8026)=0.97500-0.68082[/tex]
[tex]P(7.0039<\bar{x}<7.8026)=0.29418\approx 0.2942[/tex]
95, 86, 78, 71, 65, 60 _____
Answer:
hello there here is your answer
51 is your next term.
Step-by-step explanation:
you are subtracting 9 from each number
95-9= 86
86-9=78
78-9=65
65-9=60
60-9=51
so on and so on
Hope this help
have a good day
bye
Step-by-step explanation:
[tex]here \: is \: your \: solution: - \\ \\ given \: number \: = 95.86.78.71.65.60 \\ \\ = > 95 - 9 = 86 \\ \\ = > 86 - 8 = 78 \\ \\ = > 78 - 7 = 71 \\ \\ = > 71 - 6 = 65 \\ \\ = > 65 - 5 = 60 \\ \\ \: now \: follow \: the \: sequence \: \\ \\ subtract \: 4 \: from \: 60 \\ \\ = > 60 - 4 = 56 \\ \\ = > \: \: 56 \: \:( ANSWER✓✓✓)[/tex]
You bought a car that was $25500 and the value depreciates by 4.5% each year.
How much will the car be worth after 5 years?
How much after 8 years?
Answer:
(a) 20256.15625
(b) 17642.78546
Step-by-step explanation:
(a) There's a formula for this problem y = A(d)^t where, A is the initial value you are given, d is the growth or decay rate and t is the time period. So, in this case, as the car cost is decreasing it is a decay problem and we can write the formula as such; y = A(1-R)^t
So, in 5 years the car will be worth, 25500(1-4.5%)^5 or 20256.15625 dollars
(b) And after 8 years the car will be worth 25500(1-4.5%)^8 or 17642.78546 dollars.
Police Chase: A speeder traveling 40 miles per hour (in a 25 mph zone) passes a stopped police car which immediately takes off after the speeder. If the police car speeds up steadily to 55 miles/hour in 10 seconds and then travels at a steady 55 miles/hour, how long and how far before the police car catches the speeder who continued traveling at 40 miles/hour
Answer:
a. 18.34 s b. 327.92 m
Step-by-step explanation:
a. How long before the police car catches the speeder who continued traveling at 40 miles/hour
The acceleration of the car a in 10 s from 0 to 55 mi/h is a = (v - u)/t where u = initial velocity = 0 m/s, v = final velocity = 55 mi/h = 55 × 1609 m/3600 s = 24.58 m/s and t = time = 10 s.
So, a = (v - u)/t = (24.58 m/s - 0 m/s)/10 s = 24.58 m/s ÷ 10 s = 2.458 m/s².
The distance moved by the police car in 10 s is gotten from
s = ut + 1/2at² where u = initial velocity of police car = 0 m/s, a = acceleration = 2.458 m/s² and t = time = 10 s.
s = 0 m/s × 10 s + 1/2 × 2.458 m/s² (10)²
s = 0 m + 1/2 × 2.458 m/s² × 100 s²
s = 122.9 m
The distance moved when the police car is driving at 55 mi/h is s' = 24.58 t where t = driving time after attaining 55 mi/h
The total distance moved by the police car is thus S = s + s' = 122.9 + 24.58t
The total distance moved by the speeder is S' = 40t' mi = (40 × 1609 m/3600 s)t' = 17.88t' m where t' = time taken for police to catch up with speeder.
Since both distances are the same,
S' = S
17.88t' = 122.9 + 24.58t
Also, the time taken for the police car to catch up with the speeder, t' = time taken for car to accelerate to 55 mi/h + rest of time taken for police car to catch up with speed, t
t' = 10 + t
So, substituting t' into the equation, we have
17.88t' = 122.9 + 24.58t
17.88(10 + t) = 122.9 + 24.58t
178.8 + 17.88t = 122.9 + 24.58t
17.88t - 24.58t = 122.9 - 178.8
-6.7t = -55.9
t = -55.9/-6.7
t = 8.34 s
So, t' = 10 + t
t' = 10 + 8.34
t' = 18.34 s
So, it will take 18.34 s before the police car catches the speeder who continued traveling at 40 miles/hour
b. how far before the police car catches the speeder who continued traveling at 40 miles/hour
Since the distance moved by the police car also equals the distance moved by the speeder, how far the police car will move before he catches the speeder is given by S' = 17.88t' = 17.88 × 18.34 s = 327.92 m
Solve algebraically.
6(t-2) + 15t < 5(5 + 3t)
With work shown please!!
Step-by-step explanation:
6t-12+15t | 25+15t
21t-12 | 25+15t
21t-12 < 25+15t
hence proved..
Answer:
21t - 12 < 25 + 15t
Step-by-step explanation:
6( t - 2 ) + 15t < 5 ( 5 + 3t )
Distribute .6t - 12 + 15t < 25 + 15t
Combine like terms.21t - 12 < 25 + 15t.
Hence , Proved.
Assume that the breaking system of a train consists of two components connected in series with both of them following Weibull distributions. For the first component the shape parameter is 2.1 and the characteristic life is 100,000 breaking events. For the second component the shape parameter is 1.8 and characteristic life of 80,000. Find the reliability of the system after 2,000 breaking events:
Answer:
0.9984
Step-by-step explanation:
we have shape parameter for the first component as 2.1
characteristics life = 100000
for this component
we have
exp(-2000/100000)².¹
= e^-0.0002705
= 0.9997
for the second component
shape parameter = 1.8
characteristic life = 80000
= exp(-2000/80000)¹.⁸
= e^-0.001307
= 0.9987
the reliability oif the system after 2000 events
= 0.9987 * 0.9997
= 0.9984
