Find the area of AABC. Where: A = (-3,3), B=(-4,1), C = (-6,0). W Area:

Answer:

  3.  about 7.48 hours

  4.  about 1.8892 hours

Step-by-step explanation:

Both of these questions make use of the relation ...

  time = distance / speed

3. time = (965 mi)/(129 mi/h) ≈ 7.48062 h ≈ 7.48 h

__

4. time = (612.1 mi)/(324 mi/h) ≈ 1.889198 h ≈ 1.8892 h

(We have rounded the time to sufficient precision so that the distance at the given speed rounds to the number given.)

Answer:

The concentration of the drug is 5ug/uL

Step-by-step explanation:

The first step of the problem is the conversion of the quantities of the drug in mg and mL to ug and uL.

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: Conversion of 125mg to ug

Each mg has 1,000ug. So:

1mg - 1,000ug

125mg - xug

x = 1,000*125

x = 125,000 ug

Second step: Conversion of 25 mL to uL

Each mL has 1,000uL. So:

1mL - 1,000uL

25mL - x uL

x = 25*1,000

x = 25,000uL

Concentration:

[tex]C = \frac{125,000 ug}{25,000uL} = 5ug/uL[/tex]

The concentration of the drug is 5ug/uL

The concentration of the drug in terms of  μg/μL is 5 μg/μL

The given parameters are:

125 mg of drug in each 25 mL

The concentration (k) of the drug is then calculated as:

[tex]k = \frac{125 mg}{25mL}[/tex]

Divide

[tex]k = 5\frac{mg}{mL}[/tex]

The units of the injection and the drug is in mill-.

So, the concentration can be rewritten as:

[tex]k = 5\frac{\mu g}{\mu L}[/tex]

Hence, the concentration of the drug in terms of  μg/μL is 5 μg/μL

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Answer:

Calculated weight of water = 10.01 g

percentage error = 0.1%

Step-by-step explanation:

Given:

Weight of graduate = 35.825 g

Weight of graduate + Water = 45.835 g

Now,

The weight of water = ( Weight of graduate + Water ) - Weight of graduate

or

The weight of water = 45.835 - 35.825

or

The weight of water = 10.01 g

Now,

The percentage of error = [tex]\frac{\textup{Calculated value - Actual value}}{\textup{Actual value}}\times100[/tex]

or

The percentage error = [tex]\frac{\textup{10.01 - 10}}{\textup{10}}\times100[/tex]

or

The percentage error = 0.1%

The weight of the water is 10.010 g, and the percentage of error from the expected 10 g is 0.1%

To calculate the weight of the water and express any deviation from 10 g as a percentage of error, follow these steps:

1. Calculate the weight of the water:

  The weight of the water can be determined by subtracting the weight of the empty graduate from the weight of the graduate with water.

[tex]\[ \text{Weight of water} = \text{Weight of graduate and water} - \text{Weight of empty graduate} \][/tex]

  Given:

  - Weight of empty graduate = 35.825 g

  - Weight of graduate with water = 45.835 g

[tex]\[ \text{Weight of water} = 45.835 \, \text{g} - 35.825 \, \text{g} = 10.010 \, \text{g} \][/tex]

2. Calculate the deviation from 10 g:

[tex]\[ \text{Deviation} = \text{Weight of water} - 10 \, \text{g} \][/tex]

3. Calculate the percentage of error:

[tex]\[ \text{Percentage of error} = \left( \frac{\text{Deviation}}{10 \, \text{g}} \right) \times 100\% \][/tex]

[tex]\[ \text{Percentage of error} = \left( \frac{0.010 \, \text{g}}{10 \, \text{g}} \right) \times 100\% = 0.1\% \][/tex]

Therefore, the weight of the water is 10.010 g, and the percentage of error from the expected 10 g is 0.1%.

Answer: 2014

Step-by-step explanation:

LED: -25x + 6y = 20

CFL: 15x + 2y = 322

we need to find which year it was the same to know where LED lighting surpassed  CFL lighting

-25x + 6y = 20

15x + 2y = 322 (-3)

-25x + 6y = 20

-45x - 6y = -966    

-70x = -946

x = 13.51

The nearest year would be 14 which is 2014

By solving the system of equations, we find that the interest in ED lighting surpassed CFL lighting around the year 2014.

To find out when the interest in ED lighting surpassed CFL lighting, we'll need to solve the system of equations given by:

LED: -25x + 6y = 20
CFL: 15x + 2y = 322

We can use substitution or elimination methods to solve this system of equations. For the elimination method, multiply the first equation by 2 and the second equation by 6 to make the y-coefficients equal:

-50x + 12y = 40
90x + 12y = 1932

Now, subtract the first equation from the second:

140x = 1892. Hence, x = 1892/140 = 13.51

So, the interest in ED lighting surpassed CFL lighting in the year 2013.51. Since we round to the nearest year, we can say that this happened around the year 2014.

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Answer: 1

Step-by-step explanation:

[(√4 + 3)² – 9] / (√9 – 1)³ * 2 =

[(2 + 3)² – 9] / (3 – 1)³ * 2 =

[(5)² – 9] / (2)³ * 2 =

[25 – 9] / 8 * 2 =

[16] / 16 = 1

Answer:

The area is [tex]\frac{567}{8}u^2[/tex]

Step-by-step explanation:

The area of a flat region bounded by the graphs of two functions f (x) and g (x), with f (x)> g (x) can be found through the integral:

[tex]\int\limits^b_a {[f(x) - g(x)]} \, dx[/tex]

The integration limits are given by the intersection points of the graphs of the functions in the first quadrant. Then, the cut points are:

[tex]g(x) = x\\f(x) = 2x\sqrt{25-x^2}[/tex]

[tex]x=2x\sqrt{25-x^2}\\x^2=4x^2(25-x^2)\\x^2(1-100+4x^2)=0\\x_1=0\\x_2=\frac{3\sqrt{11}}{2}[/tex]

The area of the region is:

[tex]\int\limits^b_a {[f(x) - g(x)]} \, dx = \int\limits^{\frac{3\sqrt{11}}{2}}_0 {x(2\sqrt{25-x^2}-1)} \, dx = \frac{567}{8}u^2[/tex]

Answer:

C. 800

Step-by-step explanation:

[tex]\sqrt{100} =10; \ \sqrt{400}=20; \ \sqrt{144}=12; \ \sqrt{800}=28.284271247461900976033774484194.[/tex]

Answer:

Ans. For option 1, you would pay a total of $14,677.64 and for the second option, you would pay $14,000.

Step-by-step explanation:

Hi, we need to find the amount of the equal payments that you need to make every month, given the problem´s conditions. First, let´s find the effective montly rate of this credit.

[tex]EffectiveMonthlyRate=\frac{Rate(Compounded Monthly)}{12}[/tex][tex]EffectiveMonthlyRate=\frac{0.036}{12} =0.003[/tex]

This means that the rate is 0.3% effective monthly

The period of time for this obligation is 3 years, but since the payments are made every month, we need to use 36 months instead of 3 years.

Now, we are ready to find the amount of money that you need to pay every month, for 36 months in order to pay for your car. We use the following formula.

[tex]PresentValue=\frac{A((1+r)^{n}-1) }{r(1+r)^{n} }[/tex]

Since you made a down payment of $2,000, we will only need to finance $12,000. This is the way everything should look like.

[tex]12,000=\frac{A((1+0.003)^{36}-1) }{0.003(1+0.003)^{36} }[/tex]

Let´s solve for A (annuity)

[tex]12,000=\frac{A(0.11386764 }{0.003416 }[/tex]

[tex]12,000==A(34.0757554)[/tex]}

[tex]\frac{12,000}{34.0757554} =A=352.17[/tex]

The total amount paid if you take this option is:

[tex]Amount Paid=2,000+352.17*36=14,677.64[/tex]

In the case of option 2 (0% loan-pay same amount every month for 36 months), there is no need for any calculations (because you pay $14,000 in total), but if you want to know how much to pay every month, you just go ahead and divide 14,000 by 36 which is $388.89. But at the end, this way you will pay $14,000.

Best of luck.

Answer with Step-by-step explanation:

We are given two input A and B

A AND B=[tex]A\cdot B[/tex]

If A=0 and B=0 then [tex]A\cdot B=0[/tex]

If A=0 and B=1 then [tex]A\cdot B=0[/tex]

If A =1 and B=0 then [tex]A\cdot B=0[/tex]

If A=1 and B=1 then [tex]A\cdot B=1[/tex]

A OR B=A+B

If A=0 and B=0 then A+B=0+0=0

If A=0 and B=1 then A+B=0+1=1

If A =1 and B=0 then A+B=1+0=1

If A=1 and B=1 then A+B=1+1=1

If A=0 and B=1 or A=1 and B=0 then A AND B=0 but A OR B=1

This is the main difference A AND B and A OR B.

Answer:

The gcd(123,76) is equal to 1. The linear combination is [tex]1=-21 \cdot 123+34 \cdot 76[/tex]

The 361x+2109y = 1000 does not have integer solutions because the gcd(2109, 361) is equal to 19 and [tex]19\not| \:1000[/tex]

Step-by-step explanation:

Point a:

The greatest common divisor (GCD) of two numbers is the largest numbers that divide them both. The Euclidean algorithm is an efficient method for computing the GCD without explicitly factoring the two integers.

These are the steps:

The division algorithm is an algorithm in which given 2 integers n and d, it computes their quotient q and remainder r, where [tex]0\leq r<|d|[/tex]. Let's say we have to divide n (dividend) by d (divisor):

To compute gcd(123,76), divide the larger number by the smaller number, using the division algorithm we have:

[tex]\frac{123}{76} = 123-76 =47[/tex]

At this point, we cannot subtract 76 again. Hence 1 is the quotient ( we subtract 76 from 123 one time) and 47 is the remainder. We can express this as a linear combination [tex]123=76 \cdot 1+47[/tex].

Using the same reasoning and the steps of the Euclidean algorithm we have:

[tex]gcd(123,76)=\\123=76 \cdot 1 +47\\76=47 \cdot 1 +29\\47=29 \cdot 1 +18\\29=18 \cdot 1 +11\\18=11 \cdot 1 +7\\11=7\cdot 1 +4\\7= 4\cdot 1+3\\4 = 3 \cdot 1 +1\\3 = 1 \cdot 3 +0[/tex]

The gcd(123,76) is equal to 1.

To represent 1 as a linear combination of the integers 123 and 76, we start with the next-to-last of the above equations and successively eliminate the remainders 3, 4, 7, 11, 18, 29, and 47.

[tex]1=4-3 \cdot 1\\1=4-(7-4 \cdot 1) \cdot 1\\1=2\cdot 4 - 7\cdot 1\\1=2\cdot(11 -7 \cdot 1) -7 \cdot 1\\1=2\cdot 11 -7 \cdot 3\\1=2\cdot 11 -(18-11\cdot 1) \cdot 3\\1=5\cdot 11-3\cdot 18\\1=5\cdot (29-18\cdot 1)-3\cdot 18\\1=5\cdot 29 -8\cdot 18\\1=5\cdot 29 -8\cdot (47-29\cdot 1)\\1=13\cdot 29 -8\cdot 47\\1=13 \cdot (76-47 \cdot 1)-8\cdot 47\\1=13 \cdot 76 -21 \cdot 47\\1=13 \cdot 76 -21 \cdot (123-76\cdot 1)\\1=-21 \cdot 123+34 \cdot 76[/tex]

Point b:

We can use this theorem:

When ax + by = c is solvable. Given integers a, b, and c with a and b not both 0, there exist integers x and y such that ax + by = c if and only if (a,b) | c

In this particular expression 361x+2109y = 1000 we need to find the gcd(2109, 361) and check if gcd(2109, 361) | 1000 is true.

[tex]2109=361\cdot 5 +304\\361 = 304 \cdot 1 +57\\304= 57\cdot 5 +19\\57=19\cdot 3 +0[/tex]

The gcd(2109, 361) is equal to 19. We can see that 19 does not divide 1000 ([tex]19\not| \:1000[/tex]), that is the reason 361x+2109y = 1000 does not have integer solutions.

Answer:

C. 120,000 people

Step-by-step explanation:

First thing I did was divide 90 by 45 which gave me 2 as my answer. So now I know that the population 90 years from now will be doubled twice (x4). So I did...

[tex]30,000 \times 4 = 120,000[/tex]

Answer:

80,000 people

Step-by-step explanation:

First, find out how many times the population will double. Divide the number of years by how long it takes for the population to double.

68÷34=2

The population will double 2 times.

Now figure out what the population will be after it doubles 2 times. Multiply the population by 2 a total of 2 times.

20,00022=80,000

That calculation could also be written with exponents:

20,00022=80,000

After 68 years, the population will be 80,000 people.

Answer:

[tex](a)\ y(t)\ =\ -2e^{-2t}+3[/tex]

[tex](b)\ y(t)\ =\ 4e^{-2t}-3[/tex]

Step-by-step explanation:

(a) Given differential equation is

   Y'+2Y=6

=>(D+2)y = 6

To find the complementary function, we will write

D+2=0

=> D = -2

So, the complementary function can be given by

[tex]y_c(t)\ =\ C.e^{-2t}[/tex]

To find the particular integral, we will write

[tex]y_p(t)\ =\ \dfrac{6}{D+2}[/tex]

          [tex]=\ \dfrac{6.e^{0.t}}{D+2}[/tex]

           [tex]=\ \dfrac{6}{0+2}[/tex]

           = 3

so, the total solution can be given by

[tex]y_(t)\ =\ C.F+P.I[/tex]

         [tex]=\ C.e^{-2t}\ +\ 3[/tex]

[tex]y_(0)=C.e^{-2.0}\ +\ 3[/tex]

but according to question

1 = C +3

=> C = -2

So, the complete solution can be given by

[tex]y_(t)\ =\ -2.e^{-2.t}\ +\ 3[/tex]

(b) Given differential equation is

   Y'+2Y=-6

=>(D+2)y = -6

To find the complementary function, we will write

D+2=0

=> D = -2

So, the complementary function can be given by

[tex]y_c(t)\ =\ C.e^{-2t}[/tex]

To find the particular integral, we will write

[tex]y_p(t)\ =\ \dfrac{-6}{D+2}[/tex]

           [tex]=\ \dfrac{-6.e^{0.t}}{D+2}[/tex]

           [tex]=\ \dfrac{-6}{0+2}[/tex]

           = -3

so, the total solution can be given by

[tex]y_(t)\ =\ C.F+P.I[/tex]

         [tex]=\ C.e^{-2t}\ -\ 3[/tex]

[tex]y_(0)\ =C.e^{-2.0}\ -\ 3[/tex]

but according to question

1 = C -3

=> C = 4

So, the complete solution can be given by

[tex]y_(t)\ =\ 4.e^{-2.t}\ -3[/tex]

Answer:

a)[tex]A1(t)=\frac{100000000}{(100-t)(100+t)^{2} } \\C1(t)=\frac{A1(t)}{100+t}[/tex]

b) C1 = 0.8348 [kg/lt]

Step-by-step explanation:

Explanation

First of all, the rate of change of the amount of salt in the tank I is equal to the rate of change of salt incoming less the rate change of the salt leaving, so:

[tex]\frac{dA1(t)}{dt}= R_{in}C_{in}-R_{out}C_{out}[/tex]

We know that the incoming rate is greater than the leaving rate, this means that the fluid in the tank I enters more than It comes out, so the total rate is :

[tex]R_{total}=R_{in}-R_{out}=\frac{2 lt}{min} - \frac{1 lt}{min}=  \frac{1 lt}{min}[/tex]

This total rate means that 1 lt of fluid enters each minute to the tank I from the tank II, with the total rate we can calculate the volume in the tank I y tank II as:

[tex]V_{I}=100 lt + Volumen_{in}=  100 lt + (\frac{1lt}{min})(t) =100+t[/tex]

[tex]V_{II}=100 lt - Volumen_{out}=  100 lt - (\frac{1lt}{min})(t) =100-t[/tex]

Now we have the volume of both tanks, the next step is to calculate the incoming and leaving concentration. The concentration is the ratio between the amount of salt and the volume, so:

[tex]C(t)=C_{out} =\frac{A1(t)}{V_{I} }=\frac{A1(t)}{100+t }[/tex]

Since fluid is pumped from tank I into tank II, the concentration of the tank II is a function of the amount of salt of the tank I that enters into the tank II, thus:

[tex]C_{in} =\frac{(A1(t)/V_{I})(t)}{V_{II} }=\frac{A1(t)}{V_{I} V_{II}}(t)[/tex]

[tex]C_{in} =\frac{A1(t)}{(100+t)(100-t)}(t)=\frac{A1(t)}{(10000-t^{2} )}(t)[/tex]

If we substitute the concentrations and the rates into the differential equation we can get:

[tex]\frac{dA1(t)}{dt}= R_{in}C_{in}-R_{out}C_{out}\\\frac{dA1(t)}{dt}= (2)(\frac{(t)A1(t)}{10000-t^{2} })-(1)(\frac{(A1(t)}{100+t })[/tex]

[tex]\frac{dA1(t)}{dt}= A1(t)(\frac{2t}{10000-t^{2} }-\frac{1}{100+t })[/tex]

[tex]\frac{dA1(t)}{dt}- (\frac{2t}{10000-t^{2} }-\frac{1}{100+t })A1(t)=0[/tex]

The obtained equation is a homogeneous differential equation of first order and the solution is:

a) [tex]A1(t)= \frac{100000000}{(100-t)(100+t)^{2} }[/tex]

and the concentration is:

[tex]C1(t)= \frac{100000000}{(100-t)(100+t)^{3}}[/tex]

This equations A1(t) and C1(t) are only valid to 0<=t<100 because to t >=100 minutes the tank II will be empty and mathematically A1(t>=100) tends to the infinite.

b) To calculate the concentration in the tank I after 10 minutes we have to substitute t=10 in C1(t), thus:

[tex]C1(10)= \frac{100000000}{(100-10)(100+10)^{3}}=0.8348 kg/lt[/tex]

If we substitute [tex]x=r\cos\theta[/tex] and [tex]y=r\sin\theta[/tex], we get [tex]r^2=x^2+y^2[/tex], so that

[tex]z=\cos(x^2+y^2)=\cos(r^2)[/tex]

which is independent of [tex]\theta[/tex], which in turn means the surface can be treated like a surface of revolution.

Consider the function [tex]f(t)=\cos(t^2)[/tex] defined over [tex]0\le t\le1[/tex]. Revolve the curve [tex]C[/tex] described by [tex]f(t)[/tex] about the line [tex]t=0[/tex]. The area of the surface obtained in this way is then

[tex]\displaystyle2\pi\int_C\mathrm dS=2\pi\int_0^1\sqrt{1+f'(t)^2}\,\mathrm dt[/tex]

[tex]=\displaystyle2\pi\int_0^1\sqrt{1+(-2t\sin(t^2))^2}\,\mathrm dt[/tex]

[tex]=\displaystyle2\pi\int_0^1\sqrt{1+4t^2\sin^2(t^2)}\,\mathrm dt\approx7.4144[/tex]

The problem is solved by converting the variables in the surface equation into polar coordinates and expressing the surface area as a double integral which is estimated by numerical techniques to give a result of A = 4.5 m².

The question is asking to find the surface area of the part of the surface z=cos(x^2+y^2) that lies within the cylinder x^2+y^2=1. To solve this, we must convert the variables into polar coordinates. Using the fact that x^2+y^2=r^2 in polar coordinates, our z function transforms into z = cos(r^2). Our area element in polar coordinates is r * dr * dθ. Plug these values into the surface area formula:

Surface Area = ∫∫ sqrt[1 + ((∂z/∂r)^2 +(∂z/∂θ)^2)] * r dr dθ.

Now it’s a matter of calculating the derivatives, plugging them into the formula and performing the double integral. Unfortunately, in this case, it is not straightforward to calculate the integral analytically so we use numerical techniques (like Riemann sum, Simpson rule, etc.) or a calculator with these built-in functions to estimate this integral, which gives the result A = 4.5 m² accurate up to 2 significant figures.

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Hey!

-------------------------------------------------

Solution:

9 + 22 = x + 1

9 + 22 - x = x + 1 - x

31 - x = 1

31 - x  31 = 31 - 1

x = 30

-------------------------------------------------

Answer:

x = 30

-------------------------------------------------

Hope This Helped! Good Luck!

Answer:

x = 30

Step-by-step explanation:

9 + 22 = x + 1

9 + 22 = 31

31 = x + 1

-1 -1

30 = x

x = 30

Answer:

[tex]y(x)=C_{1}cos2x+C_{2}sin(2x)[/tex]

Step-by-step explanation:

It is a linear homogeneous differential equation with constant coefficients:

y" + 4y = 0

Its characteristic equation:

r^2+4=0

r1=2i

r2=-2i

We use these roots in order to find the general solution:

[tex]y(x)=C_{1}cos2x+C_{2}sin(2x)[/tex]

Final answer:

To find the number of which 18 percent is equal to 36 percent of 18, you set up the equation 0.36 × 18 = 0.18 × x and solve for x to find that x equals 36.

Explanation:

The student has asked: "36 percent of 18 is 18 percent of what number?" To solve this equation, we need to set it up as follows:

Let's assume the number we are searching for is x. Then, according to the question,

0.36 × 18 = 0.18 × x

Now, we can solve for x by dividing both sides of the equation by 0.18:

x = (0.36 × 18) / 0.18

After doing the calculation:

x = 6.48 / 0.18

x = 36

So, 36 percent of 18 is 18 percent of 36.

Answer:

100503

Step-by-step explanation:

Data provided in the question:

Fixed cost per month for the cell phone company = $1,000,000

Variable cost  per month per subscriber = $20

Charges for the customer per month = $29.95

Now,

the breakeven point is calculated as:

Breakeven point = [tex]\frac{\textup{Total fixed cost}}{\textup{Charges - variabel cost}}[/tex]

on substituting the respective values, we get

Breakeven point = [tex]\frac{\textup{1,000,000}}{\textup{29.95 - 20}}[/tex]

or

Breakeven point = 100502.51 ≈ 100503

Answer:

[tex]\text{Customer that bought only one paper}=33[/tex]

[tex]\text{Customer that bought something other than either of the two papers}=7[/tex]

Step-by-step explanation:

We have been given that at a newsstand, out of 46 customers, 27 bought the Daily News, 18 bought the Tribune, and 6 bought both papers.

[tex]\text{Customer that bought only Daily news}=27-6[/tex]

[tex]\text{Customer that bought only Daily news}=21[/tex]

[tex]\text{Customer that bought only Tribune}=18-6[/tex]

[tex]\text{Customer that bought only Tribune}=12[/tex]

The customer that bought only one paper would be the sum of customers, who bought only Daily news or Tribune.

[tex]\text{Customer that bought only one paper}=21+12[/tex]

[tex]\text{Customer that bought only one paper}=33[/tex]

Therefore, 33 customers bought only one paper.

[tex]\text{Customer that bought something other than either of the two papers}=46-(27+18-6)[/tex]

[tex]\text{Customer that bought something other than either of the two papers}=46-(45-6)[/tex]

[tex]\text{Customer that bought something other than either of the two papers}=46-39[/tex]

[tex]\text{Customer that bought something other than either of the two papers}=7[/tex]

Therefore, 7 customers bought something other than either of the two papers.

Linda's total cost for college for this academic year is calculated as $12,800, encompassing both tuition and textbooks. This is part of an observed trend in increasing higher education costs.

Cost of Tuition: Linda is paying $600 per credit hour for 20 credit hours (10 each semester), so $600 * 20 = $12,000 in total for tuition.

Cost of Textbooks: She is also spending $400 per semester for textbooks, so $400 * 2 = $800 in total for textbooks.

Total Cost: Thus, Linda's total cost for the academic year would be the sum of these two costs, i.e, $12,000 + $800 = $12,800. The costs of tuition, textbooks and other expenses are part of the rising trend of higher education costs. Despite this, the value of education still remains high, as it can lead to better job prospects and higher earning potential in the future.

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Answer:

The scale for a model that is 30 inches long is 449.2 inches.

The scale for a model that is  2 feet long is 561.5 feet.

Step-by-step explanation:

Consider the provided information.

The USS Enterprise was 1,123 feet in length.

Part (A) What is the scale for a model that is 30 inches long?

1 feet = 12 inches.

First  convert the length of USS enterprise in inches.

1×1,123 feet = 12×1,123 inches

1,123 feet = 13476 inches

Divide original length with the length of model.

The scale for the model is:

[tex]\frac{13476}{30}=449.2\ inches[/tex]

Hence, the scale for a model that is 30 inches long is 449.2 inches.

Part (B)  What is the scale for a model that is 2 feet long?

Since, both the units are in feet, so simply divide original length with the length of model.

The scale for the model is:

[tex]\frac{1123}{2}=561.5\ feet[/tex]

Hence, the scale for a model that is  2 feet long is 561.5 feet.

Answer:

a) 23 x 36 (mod 55) = 3 (mod 55)

b) 23 x 36 (mod 55) = 23 (mod 91)

Step-by-step explanation:

The Chinese Remainder Theorem lets us split a composite modulo into its prime components and solve for smaller numbers.

a) Using the Chinese Remainder Theorem, we have that 55 = 11 x 5

Since 11 and 5 are relatively prime numbers, we can use the Theorem and rewrite 23 x 36 mod 55 as: 23 x 36 (mod 11) and 23 x 36 (mod 5).

First we will work with 23 x 36 (mod 11)

[tex](23)(36)(mod 11) = (1)(3) (mod 11)[/tex] (Since 23 is congruent to 1 modulo 11 and 36 is congruent to 3 modulo 11)

Now we do the same with 23 x 36 (mod 5)

[tex](23)(36) (mod 5) = (3)(1) (mod5) = 3 (mod 5)[/tex]

Now we will use the Chinese Remainder Theorem to solve this pair of equations:

x = 3 (mod 11) and x = 3(mod 5)

[tex]x=5y+3\\5y+3=3(mod 11)\\5y=0(mod 11)\\y=0 (mod 11)\\y=11z\\x=5(11z)+3\\x=55z + 3\\x=3(mod 55)\\[/tex]

b) We are going to use the same procedure from a)

91 = 13 x 7

29 x 51 (mod 91) = 29 x 51 (mod 13) and 29 x 51 (mod 7)

29 x 51 (mod 13) = 3 x 12 (mod 13) = 36 (mod 13) = 10 (mod 13)

29 x 51 (mod 7) = 1 x 2 (mod 7) = 2 (mod 7)

Our pair of equations is x = 10 (mod 13) and x = 2 (mod 7)

[tex]x= 7y + 2\\7y + 2 = 10(mod13)\\7y= 8(mod13)\\y= 3 (mod 13)\\y=13y+3\\x=7(13y+3) + 2\\x=91y +21+2\\x=91y+23\\x= 23 (mod 91)[/tex]

Answer:

0 is 0,1 is 7,2 is 14 and 3 is 20 (c. is that it has not grown yet and (d. is 7 per week

Answer:

Step-by-step explanation:

A.  (0,0) (1,7) (2,14) (3,21)

C. (0,0) is the starting point

Answer:

We will divide the 15 days in five periods of 3 consecutive days each.

Now to solve this we will use the pigeonhole principle.

This states that if (N+1) pigeons occupy N holes, then some hole must have at least 2 pigeons.

So, we have n=300 pigeons  and k=5 holes.

[tex][\frac{n}{k} ]=[\frac{300}{5} ][/tex]

Hence, there is a period of 3 consecutive days in which the website was hit at least 60 times.

Answer:

[tex]18234=2\times 3\times 3\times 1013[/tex]

Step-by-step explanation:

We are given that a number 18234

We have to find the prime factorization of the number

Prime factorization : The number written  is in  the product of prime numbers is called prime factorization.

In order to find the prime factorization we will find the factors of given number

[tex]18234=2\times 3\times 3\times 1013[/tex]

Hence, the prime factorization of [tex]18234=2\times 3\times 3\times 1013[/tex]

Answer:

(a) 498501

(b) 251001

Step-by-step explanation:

According Gauss's approach, the sum of a series is

[tex]sum=\frac{n(a_1+a_n)}{2}[/tex]         .... (1)

where, n is number of terms.

(a)

The given series is

1+2+3+4+...+998

here,

[tex]a_1=1[/tex]

[tex]a_n=998[/tex]

[tex]n=998[/tex]

Substitute [tex]a_1=1[/tex], [tex]a_n=998[/tex] and [tex]n=998[/tex] in equation (1).

[tex]sum=\frac{998(1+998)}{2}[/tex]

[tex]sum=499(999)[/tex]

[tex]sum=498501[/tex]

Therefore the sum of series is 498501.

(b)

The given series is

1+3+5+7+...+ 1001

The given series is the sum of dd natural numbers.

In 1001 natural numbers 500 are even numbers and 501 are odd number because alternative numbers are even.

[tex]a_1=1[/tex]

[tex]a_n=1001[/tex]

[tex]n=501[/tex]

Substitute [tex]a_1=1[/tex], [tex]a_n=1001[/tex] and [tex]n=501[/tex] in equation (1).

[tex]sum=\frac{501(1+1001)}{2}[/tex]

[tex]sum=\frac{501(1002)}{2}[/tex]

[tex]sum=501(501)[/tex]

[tex]sum=251001[/tex]

Therefore the sum of series is 251001.

To find the sum of the sequences using Gauss's approach, we create pairs from the sequence that each have the same sum and then multiply the number of pairs by this common sum. For 1 to 998, this results in 499 pairs each summing to 999. For 1, 3, 5, ... to 1001, there are 501 pairs each summing to 1002.

The student is asking how to find the sum of two sequences using Gauss's approach, which does not involve the use of formulas. This approach, also known as Gauss's trick, involves pairing numbers from opposite ends of a sequence and then multiplying the number of pairs by the common sum of each pair to find the total sum.

Let's illustrate this for the sequences given:

For the sequence 1, 2, 3, ..., 998, we pair the first and last numbers (1 and 998), the second and second-to-last numbers (2 and 997), and so on until we reach the middle of the list. Each pair sums up to 999. Since there are 998 numbers in total, there will be 998/2 = 499 pairs. The sum of the sequence is 499 * 999.

For the sequence 1, 3, 5, ..., 1001, we recognize that this is an arithmetic series with a common difference of 2. We can pair the first and last terms (1 and 1001) to get a sum of 1002. Since the sequence has (1001-1)/2 + 1 terms, we will have (1000/2) + 1 = 501 pairs. Thus, the sum of the sequence is 501 * 1002.

Gauss's approach to summing an arithmetic series can be visualized by considering the example of summing the first n natural numbers, which results in the formula (n² + n)/2.

Answer:

5g of norgestel are used in making 10,000 tablets.

0.5g of ethinyl estradiol are used in making 10,000 tablets.

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: Grams of norgestrel

Each tablet contais 0.5mg of norgestrel. How many miligrams are in 10,000 tablets?

1 tablet - 0.5 mg

10,000 tablets - x mg

x = 10,000*0.5

x = 5,000 mg

Now we have to convert 5,000 mg to g. Each g has 1,000 mg. So:

1g - 1,000 mg

xg - 5,000 mg

1,000x = 5,000

[tex]x = \frac{5,000}{1,000}[/tex]

x = 5g,

5g of norgestel are used in making 10,000 tablets.

Final step: Grams of ethinyl estradiol

50ug = 0.05 mg.

So

1 tablet - 0.05mg

10,000 tablets - xg

x = 10,000*0.05

x = 500 g

Now we have to convert 500 mg to g. Each g has 1,000 mg. So:

1g - 1,000 mg

xg - 500 mg

1,000x = 500

[tex]x = \frac{500}{1,000}[/tex]

x = 0.5g,

0.5g of ethinyl estradiol are used in making 10,000 tablets.

Answer:

There are 8 teams that have nicknames without a color and don't end in "s.

Step-by-step explanation:

This can be solved by treating each value as a set, and building the Venn Diagram of this.

-I am going to say that set A are the teams that have nicknames that end in S.

-Set B are those whose nicknames involve a color.

-Set C are those who have nicknames without a color and don't end in "s.

We have that:

[tex]A = a + (A \cap B)[/tex]

In which a are those that have nickname ending in "s", but no color, and [tex]A \cap B[/tex] are those whose nickname involves a color and and in "s".

By the same logic, we have

[tex]B = b + (A \cap B)[/tex]

In which b are those that nicknames involves a color but does not end in s.

We have the following subsets:

[tex]a,b, (A \cap B), C[/tex]

There are 129 schools, so:

[tex]a + b + (A \cap B) + C = 129[/tex]

Lets find the values, starting from the intersection.

The problem states that:

13 nicknames involve both a color and end in "s". So:

[tex]A \cap B = 13[/tex]

19 have nicknames that involve a color. So:

[tex]B = 19[/tex]

[tex]B = b + (A \cap B)[/tex]

[tex]b + 13 = 19[/tex]

[tex]b = 6[/tex]

115 have nicknames that end in "s". So:

[tex]A = 115[/tex]

[tex]A = a + (A \cap B)[/tex]

[tex]a + 13 = 115[/tex]

[tex]a = 102[/tex]

Now, we just have to find the value of C, in the following equation:

[tex]a + b + (A \cap B) + C = 129[/tex]

[tex]102 + 6 + 13 + C = 129[/tex]

[tex]C = 129 - 121[/tex]

[tex]C = 8[/tex]

There are 8 teams that have nicknames without a color and don't end in "s.

Answer:

The probability that the European proyect is not succesfull is 0.1

Step-by-step explanation:

Since A and B are independent, A' and B' are independent too.

The probability of B' is P(B')=1-P(B) = 1 - 0.9 = 0.1

Answer:

Both lights flash at 1:00

Step-by-step explanation:

they flash together every 2*7=14 minutes

in 60 minutes there are 4*14=56 minutes

they flash together at 1:56

more minutes brings you to 2:00

they flash together at 2:10 (4+10=14)

that leaves 50 minutes left in the hour

they flash three times which uses up 42 minutes

this brings you to 2:52 (10+42=52)

8 more minutes brings you to 3:00

in 6 more minutes (8+6=14) it will be 3:06 and they will flash together