A city of 6688 residents has an area of 16 square miles. What I the population density of the city

nika2105 [10]

(2^8 x 3^-5 x 6^0) x [(3^-2/2^3) x 2^28] = x
(256 x .004 x 1) x [(.111/8) x 268,435,456] = x
(1.024 x 1) x .(0138 x 268,435,456) = x
1.024 x (.0138 x 268,435,456) = x
1.024 x 3,704,409.2928 = x
3,793,315.1158272 = x
Correct me if I am wrong. :)

7 0

2 years ago

(c). It is well known that the rate of flow can be found by measuring the volume of blood that flows past a point in a given tim

aleksklad [387]

(i) Given that

$V(R) = \displaystyle \int_0^R 2\pi K(R^2r-r^3) \, dr$

when R = 0.30 cm and v = (0.30 - 3.33r²) cm/s (which additionally tells us to take K = 1), then

$V(0.30) = \displaystyle \int_0^{0.30} 2\pi \left(0.30-3.33r^2\right)r \, dr \approx \boxed{0.0425}$

and this is a volume so it must be reported with units of cm³.

In Mathematica, you can first define the velocity function with

v[r_] := 0.30 - 3.33r^2

and additionally define the volume function with

V[R_] := Integrate[2 Pi v[r] r, {r, 0, R}]

Then get the desired volume by running V[0.30].

(ii) In full, the volume function is

$\displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

Compute the integral:

$V(R) = \displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

$V(R) = \displaystyle 2\pi K \int_0^R (R^2r-r^3) \, dr$

$V(R) = \displaystyle 2\pi K \left(\frac12 R^2r^2 - \frac14 r^4\right)\bigg_0^R$

$V(R) = \displaystyle 2\pi K \left(\frac{R^4}2- \frac{R^4}4\right)$

$V(R) = \displaystyle \boxed{\frac{\pi KR^4}2}$

In M, redefine the velocity function as

v[r_] := k*(R^2 - r^2)

(you can't use capital K because it's reserved for a built-in function)

Then run

Integrate[2 Pi v[r] r, {r, 0, R}]

This may take a little longer to compute than expected because M tries to generate a result to cover all cases (it doesn't automatically know that R is a real number, for instance). You can make it run faster by including the Assumptions option, as with

Integrate[2 Pi v[r] r, {r, 0, R}, Assumptions -> R > 0]

which ensures that R is positive, and moreover a real number.

5 0

2 years ago

In the morning, the temperature was 15∘F. During the day, the temperature fell 37 degrees. Which expression shows how to find th

tino4ka555 [31]

Answer:

- 22°F

Step-by-step explanation:

Given that :

Morning temperature = 15°F

Change in temperature = fall by 37 degree

The temperature of the end of the day:

Morning temperature + (change in temperature)

Fall by 37° = decrease in temperature by 37° = - 37°

15°F + (-37°F)

15°F - 37°F

= - 22°F

6 0

2 years ago

Line ef is tangent to circle g at point h. segment gh is a radius of circle g. what can be concluded about triangle fhg?

solong [7]

We know that

A line tangent to a circle is perpendicular to the radius to the point of tangency.
so
<span>Line ef is </span>perpendicular to the segment gh

hence
Triangle FHG is a right triangle

3 0

3 years ago

Read 2 more answers

How to find upper and lower bound of a definite integral given an unknown equation?

worty [1.4K]

Finding the upper and lower bounds for a definite integral without an equation is pretty hard because how can we find the upper and lower bounds of definite integral if there is no equation given. But I will teach you how to find the lower and upper bounds of a definite integral, when the equation is like this

$\int\limits^6_1 t^{2} - 6t + 11dt$

So, i integrate this,
$( \frac{t^{3} }{3} - 3t^{2} + 11t) \int\limits^6_1$

I know I have a minimum at x=3 because;
f(t )= t^2 − 6t + 11
f′(t) = 2
t−6 = 0
2(t−3) = 0
t = 3
f(5) = 4
f(1) = −4

4 0

2 years ago