Consider the function defined by f(x) lnx/x^4 for x > 0 and its graph y = f(x).
The graph of f has a horizontal tangent at point P. Find the coordinates of P.

53x95 = 5035!

hopes this helps! ^^

Answer:

5605

Step-by-step explanation:

Hey so I have a question do u need to explain?

#1

First the rectangle

Area

For semicircle

Area

Total area

#2

Area:-

Answer:

its 54

Step-by-step explanation:

1+2=3 3x5=15 15x2=30 30+16=46 46+8=54

Answer: 0.27

Population density of the species in water:

Population density of a species in the water is given by the expression,

       Density =Given in the question,

Number of goldfish in the tank = 12

Volume of the tank = 45 cubic feet

Substitute these values in the expression,

Density =

            = 0.267

            ≈ 0.27 goldfish per cubic feet

      Therefore, density of the tank will be 0.27 goldfish per cubic feet.

I know you're looking for an answer but i'm doing a challenge where i have to answer 10 questions by the end of today, I'm sorry! :(

Answer:

50.29in

Step-by-step instructions:

First find the semi-circular part:- 1/2×2πr

=1/2×2×22/7×9

=28.29

Find whole perimeter:- 28.29+5+5+12 = 50.29 in

Answer:

She translated each vertex of Figure 1 seven units to the right.

Step-by-step explanation:

Answer:

9.09

Step-by-step explanation:

[tex]\frac{0.81}{0.09} + \frac{0.81}{9}[/tex]

= 9 + [tex]\frac{0.81}{9}[/tex]

= 9+ 0.09

= 9.09

I hope this helped!! If so please rate and mark brainliest <3!!

The required equation f(10) = 13.52 mg remains.

We have given that ,

m is the initial mass and h is the half-life in years. cobalt-60 has a half-life of about 5.3 years. which equation gives the mass of a 50 mg cobalt-60

The amount of a radioactive substance remaining after t years is given by the function  

[tex]f(t)=m(0.5)^{t/h}[/tex]............ (1),

where m = initial mass and h= half-life in years.

Now, for Cobalt-60, h = 5.3 years, m = 50 mg and t = 10 years,

then from equation (1) we get,

[tex]f(10)=50(0.5)^{10/5.3}[/tex]

Therefore  the required equation  f(10) = 13.52 mg .

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Answer:

The required equation f(10) = 13.52 mg remains.

We have given that ,

m is the initial mass and h is the half-life in years. cobalt-60 has a half-life of about 5.3 years. which equation gives the mass of a 50 mg cobalt-60

What is the fromula for he amount of a radioactive substance remaining after t years?

The amount of a radioactive substance remaining after t years is given by the function  

............ (1),

where m = initial mass and h= half-life in years.

Now, for Cobalt-60, h = 5.3 years, m = 50 mg and t = 10 years,

then from equation (1) we get,

Therefore  the required equation  f(10) = 13.52 mg .

To learn more about the mass of radioactive substance visit:

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Answer:

Step-by-step explanation:

Notation for Cereal Set Roster

The roster form is one of the simplest ways to represent the set. The elements of a set are represented in a row and separated by a comma in roster form. For example, the first five positive even numbers are denoted as A = 2, 4, 6, 8, 10.

A = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100 is an example of roster notation: A = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100.

Hope that helps you understand :)

Answer:

6.0 ft^2

Step-by-step explanation:

hope this helps

Step-by-step explanation:

Only 3 numbers are possible at units place (2, 4 & 6) as we need even number.

But at tens & hundreds place all 6 are possible

Number of 3 digit even numbers

= 3x6x6 = 108

36

[tex]Solution:[/tex]

[tex]6\times6=36[/tex]

[tex]such\ as:[/tex]

[tex]11\ \ \ 22\ \ \ 33\ \ \ 44\ \ \ 55\ \ \ 66\\ 12\ \ \ 21\ \ \ 31\ \ \ 41\ \ \ 51\ \ \ 61\\ 13\ \ \ 23\ \ \ 32\ \ \ 42\ \ \ 52\ \ \ 62\\ 14\ \ \ 24\ \ \ 34\ \ \ 43\ \ \ 53\ \ \ 63\\ 15\ \ \ 25\ \ \ 35\ \ \ 45\ \ \ 54\ \ \ 64\\ 16+263646+56\\ 6+6+6+6+6+6\\ =36.[/tex]

I hope this helps you

:)

Answer: 44 degrees

Step-by-step explanation:

Answer:

200/7

Step-by-step explanation:

(2 meters)/(7 centimeters) or (2m/7cm)

1 meter = 100 cm

(2m/7cm)*(100cm/1m) = 200/7 (or   28.57) scale factor.

The function is given by y = 550(0.76)ˣ, after 10 passes, the number of dirts is about 35.36 ml

Let y represent the amount of dirt on the floor after x passes.

The floor initially has 550 ml of dirt, hence a = 550.

During each pass, 24% of the dirt is removed. Hence b = 100% - 24% = 76% = 0.76

The equation is given by:

y = 550(0.76)ˣ

After 10 passes:

y = 550(0.76)¹⁰ = 35.36 ml

The function is given by y = 550(0.76)ˣ, after 10 passes, the number of dirts is about 35.36 ml

Find out more on exponential function at: https://brainly.com/question/12940982

As per linear equation, there are 372 more girls in school than boys.

"A linear equation is an equation in which the highest power of the variable is always 1. It is also known as a one-degree equation. The standard form of a linear equation in one variable is of the form Ax + B = 0. Here, x is a variable, A is a coefficient and B is constant."

Given, there are 1970 pupils.

There are 799 boys in the school.

Therefore, the number of girls in the school is

[tex]= (1970-799)\\= 1171[/tex]

Therefore, the number of girls more than boys in the school is

[tex]= (1171-799)\\= 372[/tex]

In that school , there are 372 more girls than total number of the boys.

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Answer:

Third option

Step-by-step explanation:

Given the following question:

[tex]7x-4=5x+15[/tex]

To find the answer we need to use inverse of operations to simplify the equation in till we isolate the variable and find the value of x.

[tex]7x-4=5x+15[/tex]
[tex]-4+4=0[/tex]
[tex]15+4=19[/tex]
[tex]7x=5x+19[/tex]
[tex]5x-5x=0[/tex]
[tex]7x-5x=2x[/tex]
[tex]2x=19[/tex]
[tex]2x\div2=x[/tex]
[tex]19\div2=\frac{19}{2}[/tex]
[tex]=\frac{19}{2}[/tex]

Your answer is the third option or "19/2."

Hope this helps.

Answer:

t = 0, and t = 4/5

Step-by-step explanation:

1. Write the equation in standard form:

-75t^2+60t=0

2. Solve with the Quadratic formula:

[tex]t_{1,2} = \frac{-60 +- \sqrt{60^2 - 4(-75)}0}{2(-75)} \\\\t_{1,2} = \frac{-60+-60}{2(-75)} \\\\3. Separate the fractions:\\t_{1} = \frac{-60+60}{2(-75)} , t_{2} = \frac{-60-60}{2(-75)} \\\\4. Solve for each:\\t_{1} = \frac{0}{-150} = 0\\t_{2} = \frac{-120}{-150} = 4/5[/tex]

hope this helps!

first add the ratio numbers which is 12

[tex] \frac{2}{12} \times 6000 = 1000[/tex]

then divide 1000 with 175

=5.71

5.71×2.25

=12.85 ruro

Answer: X-axis

Step-by-step explanation: The arrow is pointing to the horizontal line on the on the graph. That horizontal line is the x-axis.

Using the midline property;

[tex]2wy = vz[/tex]

[tex]2(p - 21) = p - 7[/tex]

[tex]2p - 42 = p - 7[/tex]

[tex]2p - p = - 7 + 42[/tex]

[tex]p = 35[/tex]

Answer:

2 semi circles put together

Step-by-step explanation:

just cut the circle in half

Answer:

1200

Step-by-step explanation:

since the first week has 75 doubling it is 150 doubling that is 300 doubling that is 600 doubling that for the final week is 1200.

1200 more zombies will there be after an additional 4 weeks.

A sequence is an enumerated group of items in mathematics where repetitions are permitted and order is important. It has members (also known as elements or phrases), just like a set. The length of the series is the number of elements (potentially infinite).

At initial the zombies are= 75

Now, the zombies doubles at every week.

So, After 1 st week the population of zombies are

= 75 x 2

= 150

Then, After 2 nd week the population of zombies are

= 150 x 2

= 300

Then, After 3 rd week the population of zombies are

= 300 x 2

= 600

Then, After 4th week the population of zombies are

= 300 x 2

= 1200

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Answer:

the answer to your question is B

Answer:

2x8=16 so there for y=x^18

Step-by-step explanation:

Answer:

Let a = side length of a cube

Let S = surface area of a cube

Area of a square = a²

Since a cube has 6 square sides:  S = 6a²

To make a the subject:

S = 6a²

Divide both sides by 6:

[tex]\sf \implies \dfrac{S}{6}=a^2[/tex]

Square root both sides:

[tex]\sf \implies a=\sqrt{\dfrac{S}{6}}[/tex]

(positive square root only as distance is positive)

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

[tex]\sf x=-3-\sqrt{2} \implies (x+[3+\sqrt{2}])=0[/tex]

[tex]\sf x=-3+\sqrt{2} \implies (x+[3-\sqrt{2}])=0[/tex]

Therefore,

[tex]\sf y=a(x+[3+\sqrt{2}]) (x+[3-\sqrt{2}])[/tex]  for some constant a

Given the y-intercept is at (0, -5)

[tex]\sf \implies a(0+3+\sqrt{2}) (0+3-\sqrt{2})=-5[/tex]

[tex]\sf \implies a(3+\sqrt{2}) (3-\sqrt{2})=-5[/tex]

[tex]\sf \implies a(9-2)=-5[/tex]

[tex]\sf \implies 7a=-5[/tex]

[tex]\sf \implies a=-\dfrac57[/tex]

Substituting found value of a into the equation and simplifying:

[tex]\sf y=-\dfrac57(x+[3+\sqrt{2}]) (x+[3-\sqrt{2}])[/tex]

[tex]\sf \implies y=-\dfrac57(x^2+6x+7)[/tex]

[tex]\sf \implies y=-\dfrac57x^2-\dfrac{30}{7}x-5[/tex]

Answer:

20% of 30

= 20/100 * 30

= 6

hence 6 students recieved A grade