Find the equation of a line perpendicular to 2x-4y=1 that contains the point (-4, -2).

Answer:

  G(x) = x^2 -10x +25

Step-by-step explanation:

To translate F(x) 5 units to the right, replace x with (x-5).

  G(x) = F(x-5) = (x -5)^2

  G(x) = x^2 -10x +25

Answer:

D

Step-by-step explanation:

Answer:

66%

Step-by-step explanation:

15% of homes have both features.

The percentage of homes that have a pool and no garage is:

Pool only = 19% - 15% = 4%

The percentage of homes that have a garage and no pool is:

Garage only = 62% - 15% = 47%

Therefore, the percentage of homes that have a pool, a garage or both is:

[tex]P = 4\%+47\%+15\%\\P=66\%[/tex]

Answer:

-1, 1

13, 15

Step-by-step explanation:

x and x+2 are the integers

Roots of the quadratic equation are: -1 and 13.

So the integers pairs are: -1, 1 and 13, 15

Answer:

D. (2, 3, 4)

Step-by-step explanation:

The range is the y values. The y values, in numerical order, range from 2 to 4. The 2s do not need to be repeated.

Answer:

Since there are 2 possibilities for each bat (hit or out), the amount of total possibilities is 2 * 2 * 2 = 8. There is only one possibility out of those eight that gives us three hits, therefore the probability is 1 / 8 or 0.125.

1) sorry i don't know =(

2)- 5/12 is the simplified fraction for 60/144.

3)A=54cm²

Answer:

15-6x= -12-4x

15-2x= -12

-2x= -27

x= -13.5

Step-by-step explanation:

The answer would be -3 13/81 (simplified)

Answer:

Class width = 20

Approximate lower class limit of the first class = 110

Approximate Upper class limit of the first class = 119

Step-by-step Explanation:

The class width of the histogram attached below can be gotten by finding the difference between successive lower class limits.

Thus, class width = 130 - 110 = 20

The approximate lower class limit of the first class is the lowest score we have in the first class = 110

The approximate upper class limit of the first class is the closest highest score that fall within the first class and is below the lower limit of the second class. Thus approximate upper class limit of the first class = 129

Answer:

2 & 8

-4 & -4

Step-by-step explanation:

x and y are the numbers, as per question:

x= 4+2y and xy= 16

Answer:

Step-by-step explanation:

The original equation is [tex]121y''+110y'-24y=0[/tex]. We propose that the solution of this equations is of the form [tex] y = Ae^{rt}[/tex]. Then, by replacing the derivatives we get the following

[tex]121r^2Ae^{rt}+110rAe^{rt}-24Ae^{rt}=0= Ae^{rt}(121r^2+110r-24)[/tex]

Since we want a non trival solution, it must happen that A is different from zero. Also, the exponential function is always positive, then it must happen that

[tex]121r^2+110r-24=0[/tex]

Recall that the roots of a polynomial of the form [tex]ax^2+bx+c[/tex] are given by the formula

[tex] x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}[/tex]

In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions

[tex]r_1 = -\frac{12}{11}[/tex]

[tex]r_2 = \frac{2}{11}[/tex]

So, in this case, the general solution is [tex]y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}[/tex]

a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations

[tex]c_1 + c_2 = 1[/tex]

[tex]c_1\frac{-12}{11} + c_2\frac{2}{11} = 0[/tex](or equivalently [tex]c_2 = 6c_1[/tex]

By replacing the second equation in the first one, we get [tex]7c_1 = 1 [/tex] which implies that [tex] c_1 = \frac{1}{7}, c_2 = \frac{6}{7}[/tex].

So [tex]y_1 = \frac{1}{7}e^{\frac{-12t}{11}} + \frac{6}{7}e^{\frac{2t}{11}}[/tex]

b) By using y(0) =0 and y'(0)=1 we get the equations

[tex] c_1+c_2 =0[/tex]

[tex]c_1\frac{-12}{11} + c_2\frac{2}{11} = 1[/tex](or equivalently [tex]-12c_1+2c_2 = 11[/tex]

By solving this system, the solution is [tex]c_1 = \frac{-11}{14}, c_2 = \frac{11}{14}[/tex]

Then [tex]y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}[/tex]

c)

The Wronskian of the solutions is calculated as the determinant of the following matrix

[tex]\left| \begin{matrix}y_1 & y_2 \\ y_1' & y_2'\end{matrix}\right|= W(t) = y_1\cdot y_2'-y_1'y_2[/tex]

By plugging the values of [tex]y_1[/tex] and

We can check this by using Abel's theorem. Given a second degree differential equation of the form y''+p(x)y'+q(x)y the wronskian is given by

[tex]e^{\int -p(x) dx}[/tex]

In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is

[tex]e^{\int -\frac{10}{11} dx} = e^{\frac{-10x}{11}}[/tex]

Note that this function is always positive, and thus, never zero. So [tex]y_1, y_2[/tex] is a fundamental set of solutions.

Answer:

The mean number of bags that arrive on time to its intended destination is 1260 with a standard deviation of 21.59.

Step-by-step explanation:

For each bag, there are only two possible outcomes. Either it arrives on time to it's intended destination, or it does not. The probability of a bag arriving on time is independent of other bags. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Luggage checked-in at Airline A arrives on time to its intended destination with a probability of 0.63.

This means that [tex]p = 0.63[/tex]

In a random sample of 2000 bags

This means that [tex]n = 2000[/tex]

Mean and standard deviation of the number of bags that arrive on time to its intended destination:

[tex]E(X) = np = 2000*0.63 = 1260[/tex]

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{2000*0.63*0.37} = 21.59[/tex]

The mean number of bags that arrive on time to its intended destination is 1260 with a standard deviation of 21.59.

Answer:

[tex](\frac{3}{8}\,in )^2=\frac{9}{64} \,in^2=0.140625\,\,i^2[/tex]

Step-by-step explanation:

Recall that the formula for the area of a square of side L is: [tex]Area=L^2[/tex]

Therefore, for this case:

[tex]Area=L^2\\Area = (\frac{3}{8} \,in)^2\\Area=\frac{9}{64} \,\,in^2\\Area=0.140625\,\,in^2[/tex]

Answer:

40°

Step-by-step explanation:

2x = 3x - 20 add like terms

x = 20 and angle l is equal to 3x minus 20 so 3 × 20 - 20 = 40°

Answer:

Step-by-step explanation:

Hello!

The statistical model predicts the appraised value of houses in a section of the county East Meadow (Y) in relationship with the number of rooms of the house (X)

For a sample of n=64 houses the simple linear regression was estimated:

^Y= 74.80 + 24.93X

Range of X: 5 - 11

Range of Y: 160 - 300 ($ thousands of dollars)

Interpretation of the estimates of the y-intercept and the slope

y-intercept:

74.80 thousand dollars is the estimated average value of a house in a section of the county East Meadow when the house has zero rooms.

Slope:

24.93 [tex]\frac{thousand dollars}{rooms}[/tex] is the modification of the estimated average value of a house in a section of the county East Meadow when the number of rooms increases on one.

I hope this helps!

Answer:

Peter gets 18£

Gordon and Gavin each get 9£

Answer:

peter = 18 Gordon = 9 Gavin = 9

Step-by-step explanation:

2+1+1 = 4

36 div 4 = 9

2 times 9 = 18

1 times 9 = 9

Answer:

5/6

Step-by-step explanation:

Find the factor that divides both numbers...

25/5=5

30/5=6

5/6 is the simplified ratio

P.S. Please give me brainliest, i have only have two!

Answer:

1/3:1/4

Step-by-step explanation:

203/259

write in simplest form

2 hours to 45 seconds

Express ratio

15:1

simplest form

1/3:1/4

Answer:

Option (B)

Step-by-step explanation:

Coordinates of the vertices of the rectangle were A(0, 0), B(a, 0), C(a, b) and D(0, b)

Formula to determine the distance between two points with the vertices (x, y) and (x', y') is,

d = [tex]\sqrt{(x-x')^2+(y-y')^2}[/tex]

For the length of AD,

AD = [tex]\sqrt{(0-0)^2+(b-0)^2}[/tex]

     = [tex]\sqrt{b^2}[/tex]

     = b

Therefore, Option (B) will be the answer.

Answer:

Set the height up to 4

Step-by-step explanation:

Since there are 4 numbers between 1-5, set the height up to 4

Answer:

4 temperature recordings.

Step-by-step explanation:

2, 2, 4, 5

There are 4 recordings in the range of 1-5°C.

Answer:

X^6

Step-by-step explanation:

Answer:

-4

Step-by-step explanation:

Answer:

bh

6, 17

102

51

Step-by-step explanation:

Answer:

1/2 (bh)

1/2(17)(6)

51

Answer:

B

Step-by-step explanation:

Answer:

20.58

Step-by-step explanation:

Force A = 16 pounds

direction = N75E = 75° in the first quadrant

Force B = 20 pounds

direction = S20E = 180° - 20° = 160° in the fourth quadrants

we resolve the forces into their x and y resultant forces.

For the y axis forces,

-(16 x sin 75°) + (20 x sin 160°) = Fy

-15.45 + 6.84 = Fy

Fy = -8.61 N

For the x axis forces'

(16 x cos 75°) + (20 x cos 160°) = Fx

-4.14 - 18.79 = Fx

Fx = -22.93 N

Resultant force = [tex]\sqrt{(Fx)^{2} + (Fy)^{2} }[/tex] =  [tex]\sqrt{(-8.61)^{2} + (-22.93)^{2} }[/tex]

Resultant force = 24.49 N

angle made by the resultant force is,

∅ = [tex]tan^{-1}[/tex] [tex]\frac{Fy}{Fx}}[/tex]

∅  =  [tex]tan^{-1}[/tex] [tex]\frac{-8.61}{-22.93}}[/tex]

∅ =  [tex]tan^{-1}[/tex] 0.3755

∅ = 20.58

Answer:

f(2) = f(3) = 102 ft

Step-by-step explanation:

The height f at t = 2 seconds is given by:

[tex]f(t) = -16t^2 + 80t + 6\\f(2) = -16*2^2 + 80*2 + 6\\f(2)=-64+160+6\\f(2)=102\ ft[/tex]

The height f at t = 3 seconds is given by:

[tex]f(t) = -16t^2 + 80t + 6\\f(3) = -16*3^2 + 80*3 + 6\\f(3)=-144+240+6\\f(3)=102\ ft[/tex]

For both t =2 and t =3, the expression results in a height of 102 ft, therefore f(2) = f(3) = 102 ft.

Answer:

B and G

Step-by-step explanation:

Square and rectangle

Answer:

4cos=2X

X=3-4COS

X=-1

Answer:

C:

Step-by-step explanation:

In a circle, the measure of an inscribed angle is half the measure of the central angle with the same intercepted arc

So

AQD = ARC AD/2

<AQD = 78/2

<AQD = 39°