Anybody help me out??????

Find the area and the circumference of a circle with the radius 3 ft.Use 3.14 for pi

Lilit [14]

ANSWER

• Circumference: 18.84 ft

,

• Area: 28.26 ft²

EXPLANATION

The circumference of a circle of radius r is,

$C=2\pi r$

In this case, the radius is r = 3 ft and we have to use 3.14 for π,

$C=2\cdot3.14\cdot3ft=18.84ft$

Hence, the circumference of this circle is 18.84 feet.

The area of a circle of radius r is,

$A=\pi r^2$

In this case, r = 3 ft and π = 3.14,

$A=3.14\cdot3^2ft^2=28.26ft^2$

Hence, the area of this circle is 28.26 square feet.

6 0

1 year ago

The two-way table shows the distribution of book style to genre. Miguel claims that given that the book is paperback (PB) does n

Grace [21]

Answer:

Yes, the two events are independent because P(NF|PB)=P(NF).

Step-by-step explanation:

5 0

3 years ago

a) Suppose that a large mixing tank initially holds 300 gallons of water in which 50 pounds ofsalt have been dissolved. Pure wat

Kay [80]

Answer:

x = 50*e∧ -t/100

Step-by-step explanation:

We assume:

1.-That the volume of mixing is always constant 300 gallons

2.-The mixing is instantaneous

Δ(x)t = Amount in - Amount out

Amount = rate * concentration*Δt

Amount in = 3 gallons/ min * 0 = 0

Amount out = 3 gallons/min * x/ 300*Δt

Then

Δ(x)t/Δt = - 3*x/300 Δt⇒0 lim Δ(x)t/Δt = dx/dt

dx/dt = - x/100

dx/ x = - dt/100

A linear first degree differential equation

∫ dx/x = ∫ - dt/100

Ln x = - t/100 + C

initial conditions to determine C

t= 0 x = 50 pounds

Ln (50) = 0/100 * C

C = ln (50)

Then final solution is:

Ln x = - t/100 + Ln(50) or

e∧ Lnx = e ∧ ( -t/100 + Ln(50))

x = e∧ ( -t/100) * e∧Ln(50)

x = e∧ ( -t/100) * 50

x = 50*e∧ -t/100

4 0

3 years ago

What is the value of (7 – 3i)(6 + i)?

Dimas [21]

45 - 11i is the answer! And make sure you double check.

8 0

3 years ago

Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves

Vadim26 [7]

The expression on the left side describes a parabola. Factorize it to determine where it crosses the y-axis (i.e. the line x = 0) :

-3y² + 9y - 6 = -3 (y² - 3y + 2)

… = -3 (y - 1) (y - 2) = 0

⇒ y = 1 or y = 2

Also, complete the square to determine the vertex of the parabola:

-3y² + 9y - 6 = -3 (y² - 3y) - 6

… = -3 (y² - 3y + 9/4 - 9/4) - 6

… = -3 (y² - 2•3/2 y + (3/2)²) + 27/4 - 6

… = -3 (y - 3/2)² + 3/4

⇒ vertex at (x, y) = (3/4, 3/2)

I've attached a sketch of the curve along with one of the shells that make up the solid. For some value of x in the interval 0 ≤ x ≤ 3/4, each cylindrical shell has

radius = x

height = y⁺ - y⁻

where y⁺ refers to the half of the parabola above the line y = 3/2, and y⁻ is the lower half. These halves are functions of x that we obtain from its equation by solving for y :

x = -3y² + 9y - 6

x = -3 (y - 3/2)² + 3/4

x - 3/4 = -3 (y - 3/2)²

-x/3 + 1/4 = (y - 3/2)²

± √(1/4 - x/3) = y - 3/2

y = 3/2 ± √(1/4 - x/3)

y⁺ and y⁻ are the solutions with the positive and negative square roots, respectively, so each shell has height

(3/2 + √(1/4 - x/3)) - (3/2 - √(1/4 - x/3)) = 2 √(1/4 - x/3)

Now set up the integral and compute the volume.

$\displaystyle 2\pi \int_{x=0}^{x=3/4} 2x \sqrt{\frac14 - \frac x3} \, dx$

Substitute u = 1/4 - x/3, so x = 3/4 - 3u and dx = -3 du.

$\displaystyle 2\pi \int_{u=1/4-0/3}^{u=1/4-(3/4)/3} 2\left(\frac34 - 3u\right) \sqrt{u} \left(-3 \, du\right)$

$\displaystyle -12\pi \int_{u=1/4}^{u=0} \left(\frac34 - 3u\right) \sqrt{u} \, du$

$\displaystyle 12\pi \int_{u=0}^{u=1/4} \left(\frac34 u^{1/2} - 3u^{3/2}\right) \, du$

$\displaystyle 12\pi \left(\frac34\cdot\frac23 u^{3/2} - 3\cdot\frac25u^{5/2}\right) \bigg|_{u=0}^{u=1/4}$

$\displaystyle 12\pi \left(\frac12 u^{3/2} - \frac65u^{5/2}\right) \bigg|_{u=0}^{u=1/4}$

$\displaystyle 12\pi \left(\frac12 \left(\frac14\right)^{3/2} - \frac65\left(\frac14\right)^{5/2}\right) - 12\pi (0 - 0)$

$\displaystyle 12\pi \left(\frac1{16} - \frac3{80}\right) = \frac{12\pi}{40} = \boxed{\frac{3\pi}{10}}$

6 0

2 years ago