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3 years ago

What is the exact volume of the cone?

Mathematics

2 answers:

lions [1.4K]3 years ago

4 0

Answer:

THE ANSWER IS D <3

Step-by-step explanation:

soldier1979 [14.2K]3 years ago

3 0

The Answer I got is D

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a hat contains 2 red apples and 3 green apples. a bucket contains 7 red apples and 3 green apples. a container is selected at ra

Leokris [45]

Answer:

Step-by-step explanation:

8 0

3 years ago

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio

JulijaS [17]

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

$Initial =1\ million$

$6\ years\ later = 500,000$

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

$\frac{dA}{dt} = kA$

Where k represents the constant of proportionality

$\frac{dA}{dt} = kA$

Multiply both sides by dt/A

$\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}$

$\frac{dA}{A} = k\ dt$

Integrate both sides

$\int\ {\frac{dA}{A} = \int\ {k\ dt}$

$ln\ A = kt + lnC$

Make A, the subject

$A = Ce^{kt}$

$t = 0\ when\ A =1\ million$ i.e. At initial

So, we have:

$A = Ce^{kt}$

$1000000 = Ce^{k*0}$

$1000000 = Ce^{0}$

$1000000 = C*1$

$1000000 = C$

$C =1000000$

Substitute $C =1000000$ in $A = Ce^{kt}$

$A = 1000000e^{kt}$

To solve for k;

$6\ years\ later = 500,000$

i.e.

$t = 6\ A = 500000$

So:

$500000= 1000000e^{k*6}$

Divide both sides by 1000000

$0.5= e^{k*6}$

Take natural logarithm (ln) of both sides

$ln(0.5) = ln(e^{k*6})$

$ln(0.5) = k*6$

Solve for k

$k = \frac{ln(0.5)}{6}$

$k = \frac{-0.693}{6}$

$k = -0.1155$

Recall that:

$\frac{dA}{dt} = kA$

Where

$\frac{dA}{dt}$ = Rate

So, when

$A = 600000$

The rate is:

$\frac{dA}{dt} = -0.1155 * 600000$

$\frac{dA}{dt} = -69300$

<em>Hence, the amount of oil was decreasing at 69300 barrels, yearly</em>

7 0

2 years ago

In the 1990s the demand for personal computers in the home went up with household income. For a given community in the 1990s, th

WITCHER [35]

Answer:

a) 0.5198 computers per household

b) 0.01153 computers

Step-by-step explanation:

Given:

number of computers in a home,

q = 0.3458 ln x - 3.045 ; 10,000 ≤ x ≤ 125,000

here x is mean household income

mean income = $30,000

increasing rate, $\frac{dx}{dt}$ = $1,000

Now,

a) computers per household are

since,

mean income of $30,000 lies in the range of 10,000 ≤ x ≤ 125,000

thus,

q = 0.3458 ln(30,000) - 3.045

q = 0.5198 computers per household

b) Rate of increase in computers i.e $\frac{dq}{dt}$

$\frac{dq}{dt}$ = $\frac{d(0.3458 ln x - 3.045)}{dt}$

$\frac{dq}{dt}=0.3458\times(\frac{1}{x})\frac{dx}{dt} - 0$

on substituting the values, we get

$\frac{dq}{dt}=0.3458\times(\frac{1}{30,000})\times1,000$

= 0.01153 computers

6 0

3 years ago

What is the greatest common factor of 3062, 8062, and 2013?

Zanzabum

One
the gcf all the numbers is 1

6 0

3 years ago

Lithium has 3 protons. how many neutrons are in the isotope lithium-8?

bulgar [2K]

5 neutrons in lithium-8

7 0

3 years ago

Read 2 more answers