All categories

3 years ago

How many plants spaced every 6" are needed to surround a circular walkway with a 25' radius

Mathematics

1 answer:

Neporo4naja [7]3 years ago

6 0

Answer:= 2 × 3.14 (25)

= 157.

circumference of circle feet is 157 × 12 inches.

Step-by-step explanation:

consider the following formula to find the circumference of circle

C= 2^_r

here r is radius of circle

the radius of circular is 25 ft.

Therefore circumference is as follows

C = 2^_r

You might be interested in

I will mark you if you help me

Paraphin [41]

Slope intercept form: y=-0.25x+2.5

Graph shown below

7 0

3 years ago

If x greater than 7,then the absolute value of x is greater than 7 <br> X=10

r-ruslan [8.4K]

The absolute value is 10/7

4 0

3 years ago

Can anyone figure this out

kipiarov [429]

To figure this out simply, find where the first number is on the x axis and then move over to the y axis, the Horizontal line is the x and the vertical is the y.

The systolic pressure is 130 mmHg

3 0

3 years ago

ann has 5 brand new CD’s, which is 3 less than twice the that bob (x) has. translate the situation into an algebraic equation

Nataliya [291]

Ann = 3 CD's

Ann = 2x-3

5 = 2x-3

x= CD's bob has

5 = 2x-3 is the equation

to solve

add 3 to each side

8 = 2x

divide by 2

x =4

bob has 4 CD's, ann has 5

6 0

3 years ago

A 250-gal tank is initially filled with brine (salt-water mixture) containing 100 lb of salt. Brine containing 3 lb of salt per

gulaghasi [49]

Answer:

x(t) = 581.66 lb

Step-by-step explanation:

From the given information:

Consider the salt quantity in the tank at time t be x(t) lb.

∴ the rate of change of salt content in the tank is $\dfrac{dx}{dt}= Rate \ of \ \ inflow - rate \ of \ \ outflow$

where:

the rate of inflow = salt conc. × flow rate = 3 lb/gallon × 7 gallons

=21 lb/s

rate of outflow = salt conc. in the tank × flow rate

= x/250 × 9

= 9x/250 lb/s

∴

$\dfrac{dx}{dt} = 21 - \dfrac{9x}{250}$

$\dfrac{dx}{dt} + \dfrac{9x}{250}= 21$

This is a 1st order linear differentiation,

The integrating factor if $e^{^{ \int \dfrac{9}{250}\ dt}} = e^{^{ \dfrac{9t}{250} }}$

∴

$e^{^{ \dfrac{9t }{250}}} \ \ x(t) = \int 21 e^{\dfrac{9t}{250}} \ dt + C$

$e^{^{ \dfrac{9t }{250}}} \ \ x(t) = 21 \times \dfrac{250}{9}e^{\dfrac{9t}{250}} + C$

$x(t) = \dfrac{1750}{3}+Ce^{^{\dfrac{-9t}{250}}}$ at t = (0) and x(0) = 100 lb

Hence;

$100 = \dfrac{1750}{3}+C$

$C = \dfrac{1750}{3} -100$

$C = - \dfrac{1450}{3}$

∴ $x(t) = \dfrac{1750}{3}-\dfrac{1450}{100}e^{^{\dfrac{-9t}{250}}}$

after time t = 1 minute i.e 60 s

$x(t) = \dfrac{1750}{3}-\dfrac{1450}{100}e^{^{\dfrac{-9 \times 60}{250}}}$

$x(t) = \dfrac{1750}{3}-\dfrac{1450}{100}e^{^{\dfrac{-540}{250}}}$

x(t) = 581.66 lb

5 0

3 years ago