3 divided by 4 1/2 write in simplest form

Answer:

m=1/2

y-8=1/2(x-1)

y-8=1/2x-1/2

multiply through by 2

2y-16=x-1

2y-16+1-x=0

2y-15-x=0

2y-x-15=0

Answer:

354°15'38.99''

Step-by-step explanation:

Answer: C

Step-by-step explanation:

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

0.08 cm/min

Step-by-step explanation:

Given:

[tex]\dfrac{dV}{dt}=5\:\text{cm}^3\text{/min}[/tex]

Find [tex]\frac{dr}{dt}[/tex] when diameter D = 20 cm.

We know that the volume of a sphere is given by

[tex]V = \dfrac{4\pi}{3}r^3[/tex]

Taking the time derivative of V, we get

[tex]\dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt} = 4\pi\left(\dfrac{D}{2}\right)^2\dfrac{dr}{dt} = \pi D^2\dfrac{dr}{dt}[/tex]

Solving for [tex]\frac{dr}{dt}[/tex], we get

[tex]\dfrac{dr}{dt} = \left(\dfrac{1}{\pi D^2}\right)\dfrac{dV}{dt} = \dfrac{1}{\pi(20\:\text{cm}^2)}(5\:\text{cm}^3\text{/min})[/tex]

[tex]\:\:\:\:\:\:\:= 0.08\:\text{cm/min}[/tex]

Answer:

(3/2,10)

Step-by-step explanation:

Mid point is ((10-7)/2,(13+7)/2)=(1.5,10)

Answer:

y = -x + 5

Step-by-step explanation:

The point is in quadrant 2, so the line must pass through points that look like (a, 0) and (0, a) where a is a positive number.  The slope of such a line is -1.

If (x, y) is a point on the line, then the slope between points (x, y) and (-5, 10) is 1, and you can write

[tex]\frac{y-10}{x-(-5)}=-1\\y-10 = -1(x+5)\\y-10=-x-5\\y=-x+5[/tex]

Answer:

Step-by-step explanation:

In order to find out how long it takes for the rocket to hit the ground, we only need set that position equation equal to 0 (that's how high something is off the ground when it is sitting ON the ground) and factor to solve for t:

[tex]0=-4.9t^2+112t+395[/tex]

Factor that however you are factoring in class to get

t = -3.1 seconds and t = 25.9 seconds.

Since time can NEVER be negative, it takes the rocket approximately 26 seconds to hit the ground.

Answer:

34

Step-by-step explanation:

let the missing value is x

(25+x) /2 = 29.5

25+x = 29.5(2)

25+x = 59

x = 59-25

x = 34

Answer:

The third choice down

Step-by-step explanation:

|-2x| = 4

There are two solutions, one positive and one negative

-2x = 4  and -2x = -4

Divide by -2

-2x/-2 = 4/-2    -2x/-2 = -4/-2

x = -2   and x = 2

Answer:

2:3 is equal to 8:12

Step-by-step explanation:

2:3

To get the first number to 8

8/2 = 4

Multiply by all terms 4

2*3 : 3*4

8:12

2:3 is equal to 8:12

8:12 = 8/12

= 2/3

= 2:3

Therefore 2:3 and 8:12 are equalent to each other.

Answered by Gauthmath must click thanks and mark brainliest

Answer:

2025 cm

Step-by-step explanation:

Given the length of pieces - 45 cm, 75 cm and 81 cm

To find the length of the rope we have to find the L.C.M. of 45, 75 and 81 :

3  |  45, 75, 81

    | ________________      

3  |  15, 25, 27

    |________________

3  |    5, 25, 9

    |________________

3  |    5, 25, 3

    |________________

5  |    5, 25, 1

    |________________

5  |     1, 5, 1

    |________________

    |      1, 1, 1

L.C.M. = 3 × 3 × 3 × 3 × 5 × 5

= 2025 cm

So, the least length of the rope should be 2025 cm which can be cut into a whole number of pieces of length 45 cm, 75 cm and 81 cm.

Answer:

1:3

Step-by-step explanation:

27 : 81

Divide each side by 27

27/27 : 81/27

1:3

Answer:

0.475: Z = -0.062706778

0.525: Z = 0.062706778

Step-by-step explanation:

Answer:

"A"

Step-by-step explanation:

a+b >c

a+c>b

b+c>a

~~~~~~~~~~~~

A. T,T,T

B. T,T,F

C. T,F,T

It looks like the limit you want to find is

[tex]\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4}[/tex]

One way to compute this limit relies only on the definition of the constant e and some basic properties of limits. In particular,

[tex]e = \displaystyle\lim_{x\to\infty}\left(1+\frac1x\right)^x[/tex]

The idea is to recast the given limit to make it resemble this definition. The definition contains a fraction with x as its denominator. If we expand the fraction in the given limand, we have a denominator of x - 1. So we rewrite everything in terms of x - 1 :

[tex]\left(\dfrac{x+4}{x-1}\right)^{x+4} = \left(\dfrac{x-1+5}{x-1}\right)^{x-1+5} \\\\ = \left(1+\dfrac5{x-1}\right)^{x-1+5} \\\\ =\left(1+\dfrac5{x-1}\right)^{x-1} \times \left(1+\dfrac5{x-1}\right)^5[/tex]

Now in the first term of this product, we substitute y = (x - 1)/5 :

[tex]\left(\dfrac{x+4}{x-1}\right)^{x+4} = \left(1+\dfrac1y\right)^{5y} \times \left(1+\dfrac5{x-1}\right)^5[/tex]

Then use a property of exponentiation to write this as

[tex]\left(\dfrac{x+4}{x-1}\right)^{x+4} = \left(\left(1+\dfrac1y\right)^y\right)^5 \times \left(1+\dfrac5{x-1}\right)^5[/tex]

In terms of end behavior, (x - 1)/5 and x behave the same way because they both approach ∞ at a proportional rate, so we can essentially y with x. Then by applying some limit properties, we have

[tex]\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4} = \lim_{x\to\infty} \left(\left(1+\dfrac1x\right)^x\right)^5 \times \left(1+\dfrac5{x-1}\right)^5 \\\\ = \lim_{x\to\infty}\left(\left(1+\dfrac1x\right)^x\right)^5 \times \lim_{x\to\infty}\left(1+\dfrac5{x-1}\right)^5 \\\\ =\left(\lim_{x\to\infty}\left(1+\dfrac1x\right)^x\right)^5 \times \left(\lim_{x\to\infty}\left(1+\dfrac5{x-1}\right)\right)^5[/tex]

By definition, the first limit is e and the second limit is 1, so that

[tex]\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4} = e^5\times1^5 = \boxed{e^5}[/tex]

You can also use L'Hopital's rule to compute it. Evaluating the limit "directly" at infinity results in the indeterminate form [tex]1^\infty[/tex].

Rewrite

[tex]\left(\dfrac{x+4}{x-1}\right)^{x+4} = \exp\left((x+4)\ln\dfrac{x+4}{x-1}\right)[/tex]

so that

[tex]\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4} = \lim_{x\to\infty}\exp\left((x+4)\ln\dfrac{x+4}{x-1}\right) \\\\ = \exp\left(\lim_{x\to\infty}(x+4)\ln\dfrac{x+4}{x-1}\right) \\\\ =\exp\left(\lim_{x\to\infty}\frac{\ln\dfrac{x+4}{x-1}}{\dfrac1{x+4}}\right)[/tex]

and now evaluating "directly" at infinity gives the indeterminate form 0/0, making the limit ready for L'Hopital's rule.

We have

[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\ln\dfrac{x+4}{x-1}\right] = -\dfrac5{(x-1)^2}\times\dfrac{1}{\frac{x+4}{x-1}} = -\dfrac5{(x-1)(x+4)}[/tex]

[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1{x+4}\right]=-\dfrac1{(x+4)^2}[/tex]

and so

[tex]\displaystyle \exp\left(\lim_{x\to\infty}\frac{\ln\dfrac{x+4}{x-1}}{\dfrac1{x+4}}\right) = \exp\left(\lim_{x\to\infty}\frac{-\dfrac5{(x-1)(x+4)}}{-\dfrac1{(x+4)^2}}\right) \\\\ = \exp\left(5\lim_{x\to\infty}\frac{x+4}{x-1}\right) \\\\ = \exp(5) = \boxed{e^5}[/tex]

Answer:

0.2725

0.0431

Step-by-step explanation:

The distribution here is a poisson distribution :

λ = 1.3

The poisson distribution :

p(x) = [(e^-λ * λ^x)] ÷ x!

Expected probability of finding male with 0 accident ; x = 0

p(0) = [(e^-1.3 * 1.3^0)] ÷ 0!

p(0) = [0.2725317 * 1] ÷ 1

p(0) = 0.2725317

= 0.2725

2.)

P(x ≥ 4) = 1 - P(x < 4)

P(x < 4) = p(x = 0) + p(x. = 1) + p(x = 2) + p(x = 3)

p(x = 0) =  p(0) = [(e^-1.3 * 1.3^0)] ÷ 0! = 0.2725

p(x = 1) = p(1) = [(e^-1.3 * 1.3^1)] ÷ 1! = 0.35429

p(x = 2) = p(2) = [(e^-1.3 * 1.3^2)] ÷ 2! = 0.23029 p(x = 3) = p(3) = [(e^-1.3 * 1.3^3)] ÷ 0! = 0.09979

P(x < 4) = 0.2725 + 0.35429 + 0.23029 + 0.09979 = 0.95687

P(x ≥ 4) = 1 - 0.95687 = 0.0431

Step-by-step explanation:

before Tax

Coast = 888

in 2ND Item = $21

• 888/21

= $42.28

The Total mileage is "400" and the further solution can be defined as follows:

Let t become the time he spent commuting on the first day of his vacation.

It is then calculated as [tex]t + 2[/tex].

[tex]\to 40\times(t+2) = 60(t) + 20 \\\\\to 40t+80 = 60t + 20 \\\\\to 80-20 = 60t + 40t \\\\\to 60 = 20t \\\\\to t=\frac{60}{20} \\\\\to t=\frac{6}{2} \\\\\to t= 3\\\\[/tex]

It traveled [tex]40\times (3 + 2) + 20 = 40\times 5 + 20 = 200+20=220[/tex] miles on its first day of operation.

The car traveled [tex]180\ miles[/tex] on the second day, which was [tex]60 \ miles \times 3[/tex].

So,

Total mileage= first day traveled + second day traveled [tex]= 220+ 180= 400 \miles[/tex]

Learn more:

Total distance traveled: brainly.com/question/20670144

Answer:

9/25

Step-by-step explanation:

3 novels , 1 bio , 1 poetry = 5 books

P( novel) = novels / books

               = 3/5

Book is returned

3 novels , 1 bio , 1 poetry = 5 books

P( novel) = novels / books

               = 3/5

P(novel, return, novel) = 3/5 * 3/5 = 9/25

Answer:

rt3/2

Step-by-step explanation:

first off cosine is the x coordinate

now if you do't want to use a calculator, you can use use the unit circle.

360 - 330 = 30 (360 degrees is a whole circle)

a 30 60 90 triangle is made, then use the law for 30 60 90 triangles:

if the shortest leg is x, the other leg is x*rt3 and the hypotenuse is 2x.

Answer:

D

Step-by-step explanation:

cos 330 = cos (360-330)

= cos 30

= √3 /2

9514 1404 393

Answer:

Step-by-step explanation:

For continuous compounding the "rule of 69" applies. That is the doubling time can be found from ...

  t = 69.3147/r . . . . where r is the interest rate in percent.

Here, r=2, so ...

  t = 69.3147/2 = 34.6574 . . . years

That's 34 years and 240 days.

Answer:

8 units^2

Step-by-step explanation:

The area of a tringle is 1/2 bh. The base, LK, measures 4 while the height is also 4(you can get these values by counting the squares). This means the area is:

1/2 * (4)(4) = 1/2 * 16 = 8 units^2

Answer:

[tex]\sqrt{128x^8y^3} = 8 x^4 y \sqrt{2y}[/tex]

Step-by-step explanation:

Given

[tex]\sqrt{128x^8y^3}[/tex] --- the complete expression

Required

The equivalent expression

We have:

[tex]\sqrt{128x^8y^3}[/tex]

Expand

[tex]\sqrt{128x^8y^3} = \sqrt{128* x^8 * y^3}[/tex]

Further expand

[tex]\sqrt{128x^8y^3} = \sqrt{64 * 2* x^8 * y^2 * y}[/tex]

Rewrite as:

[tex]\sqrt{128x^8y^3} = \sqrt{64 * x^8 * y^2* 2 * y}[/tex]

Split

[tex]\sqrt{128x^8y^3} = \sqrt{64 * x^8 * y^2} * \sqrt{2 * y}[/tex]

Express as:

[tex]\sqrt{128x^8y^3} = (64 * x^8 * y^2)^\frac{1}{2} * \sqrt{2y}[/tex]

Remove bracket

[tex]\sqrt{128x^8y^3} = (64)^\frac{1}{2} * (x^8)^\frac{1}{2} * (y^2)^\frac{1}{2} * \sqrt{2y}[/tex]

[tex]\sqrt{128x^8y^3} = 8 * x^\frac{8}{2} * y^\frac{2}{2} * \sqrt{2y}[/tex]

[tex]\sqrt{128x^8y^3} = 8 * x^4 * y * \sqrt{2y}[/tex]

[tex]\sqrt{128x^8y^3} = 8 x^4 y \sqrt{2y}[/tex]

The answer is A and D

good luck

Answer:

(x-1)^2 + (y+4)^2 = 4

Step-by-step explanation:

The equation for a circle is given by

(x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius

(x-1)^2 + (y- -4)^2 = 2^2

(x-1)^2 + (y+4)^2 = 4

Answer:

m=[tex]7\sqrt3[/tex]

n=7

Step-by-step explanation:

Hi there!

We are given a right triangle (notice the 90°) angle, the measure of one of the acute angles as 60°, and the measure of the hypotenuse (the side OPPOSITE from the 90 degree angle) as 14

We need to find the lengths of m and n

Firstly, let's find the measure of the other acute angle

The acute angles in a right triangle are complementary, meaning they add up to 90 degrees

Let's make the measure of the unknown acute angle x

So x+60°=90°

Subtract 60 from both sides

x=30°

So the measure of the other acute angle is 30 degrees

This makes the right triangle a special kind of right triangle, a 30°-60°-90°  triangle

In a 30°-60°-90° triangle, if the length of the hypotenuse is a, then the length of the leg (the side that makes up the right angle) opposite from the 30 degree angle is [tex]\frac{a}{2}[/tex], and the leg opposite from the 60 degree angle is [tex]\frac{a\sqrt3}{2}[/tex]

In this case, a=14, n=[tex]\frac{a}{2}[/tex], and m=[tex]\frac{a\sqrt3}{2}[/tex]

Now substitute the value of a into the formulas to find n and m to find the lengths of those sides

So that means that n=[tex]\frac{14}{2}[/tex], which is equal to 7

And m=[tex]\frac{14\sqrt3}{2}[/tex], which simplified, is equal to [tex]7\sqrt3[/tex]

Hope this helps!

Answer:

rectangle

Step-by-step explanation:

Perimeter of 20 feet

rectangle (square is technically a rectangle):

sides 5 and 5

5*5 = 25ft²

Circle:

20/(2π) = 3.18309...

3.1809...²π = 31.831ft²

Max area of rectangle (i.e. square) has a smaller area than a circle.

Step-by-step explanation:

f(x)=log(x)

     =d(log(x)/dx)

=>y=1/x

Answer:

The number of hits would follow a binomial distribution with [tex]n =10,\!000[/tex] and [tex]p \approx 4.59 \times 10^{-6}[/tex].

The probability of finding [tex]0[/tex] hits is approximately [tex]0.955[/tex] (or equivalently, approximately [tex]95.5\%[/tex].)

The mean of the number of hits is approximately [tex]0.0459[/tex]. The variance of the number of hits is approximately [tex]0.0459\![/tex] (not the same number as the mean.)

Step-by-step explanation:

There are [tex](26 + 10)^{6} \approx 2.18 \times 10^{9}[/tex] possible passwords in this set. (Approximately two billion possible passwords.)

Each one of the [tex]10^{9}[/tex] randomly-selected passwords would have an approximately [tex]\displaystyle \frac{10,\!000}{2.18 \times 10^{9}}[/tex] chance of matching one of the users' password.

Denote that probability as [tex]p[/tex]:

[tex]p := \displaystyle \frac{10,\!000}{2.18 \times 10^{9}} \approx 4.59 \times 10^{-6}[/tex].

For any one of the [tex]10^{9}[/tex] randomly-selected passwords, let [tex]1[/tex] denote a hit and [tex]0[/tex] denote no hits. Using that notation, whether a selected password hits would follow a bernoulli distribution with [tex]p \approx 4.59 \times 10^{-6}[/tex] as the likelihood of success.

Sum these [tex]0[/tex]'s and [tex]1[/tex]'s over the set of the [tex]10^{9}[/tex] randomly-selected passwords, and the result would represent the total number of hits.

Assume that these [tex]10^{9}[/tex] randomly-selected passwords are sampled independently with repetition. Whether each selected password hits would be independent from one another.

Hence, the total number of hits would follow a binomial distribution with [tex]n = 10^{9}[/tex] trials (a billion trials) and [tex]p \approx 4.59 \times 10^{-6}[/tex] as the chance of success on any given trial.

The probability of getting no hit would be:

[tex](1 - p)^{n} \approx 7 \times 10^{-1996} \approx 0[/tex].

(Since [tex](1 - p)[/tex] is between [tex]0[/tex] and [tex]1[/tex], the value of [tex](1 - p)^{n}[/tex] would approach [tex]0\![/tex] as the value of [tex]n[/tex] approaches infinity.)

The mean of this binomial distribution would be:[tex]n\cdot p \approx (10^{9}) \times (4.59 \times 10^{-6}) \approx 0.0459[/tex].

The variance of this binomial distribution would be:

[tex]\begin{aligned}& n \cdot p \cdot (1 - p)\\ & \approx(10^{9}) \times (4.59 \times 10^{-6}) \times (1- 4.59 \times 10^{-6})\\ &\approx 4.59 \times 10^{-6}\end{aligned}[/tex].