How do you solve the equation (X+3)^3=81

If a radius of a circle is 12 feet, what is the area and perimeter of the circle?

Vesna [10]

Area of a circle=πr²
area of this cirlce=π*(12 ft)²=144π ft²≈452.39 ft².

answer 1= the area of this circle is 452.39 ft².

Perimeter of a circle=2πr.
Perimeter of this cirlce=2π(12 ft)=24π ft≈75.4 ft

<span>answer 2= the perimeter of this cirlce is 75.4 ft.</span>

4 0

2 years ago

Read 2 more answers

I don’t get this one please HELPPP MEE

iren2701 [21]

Answer:

135

Step-by-step explanation:

The given expression is

$\frac{2 {x}^{2} }{x} +x(100 - 15x)$

From the question, x=5

So,by substitution,we obtain

$=\frac{2({5}^{2})}{5} +5(100 - 15(5))$

$=\frac{2(25)}{5} +5(100 - 75))$

$=2(5)+5(25)$

$= 10+125$

$= 135$

Hence the answer is 135

8 0

2 years ago

The rabbit population of Springfield, Ohio was 144,000 in 2016. It is expected to decrease by about 7.2% per year. Write an expo

xeze [42]

<h2>Hello!</h2>

The answer is:

In 2036 there will be a population of 32309 rabbits.

We can calculate the exponential decay using the following function:

$P(t)=StartAmount*(1-\frac{percent}{100})^{t}$

Where,

Start Amount, is the starting value or amount.

Percent, is the decay rate.

t, is the time elapsed.

We are given:

$StartAmount=144,000\\x=7.2(percent)\\t=2036-2016=20years$

Now, substituting it into the equation, we have:

$P(t)=StartAmount(1-\frac{percent}{100})^{t}$

$P(t)=144000*(1-\frac{7.2}{100})^{20}$

$P(t)=144000*(1-0.072)^{20}$

$P(t)=144000*(0.928)^{20}$

$P(t)=144000*0.861=32308.888=32309$

Hence, we have that in 2036 the population of rabbis will be 32309 rabbits.

Have a nice day!

5 0

2 years ago

What is the MEDIAN of the data - (12, 8, 6, 12, 16, 12) *

Lena [83]

Answer:

the median of this data is 12

6 0

2 years ago

In Exercises 5-7, find all the exact t-values for which the given statement is true,

bearhunter [10]

Answer: See Below

<u>Step-by-step explanation:</u>

NOTE: You need the Unit Circle to answer these (attached)

5) cos (t) = 1

Where on the Unit Circle does cos = 1?

Answer: at 0π (0°) and all rotations of 2π (360°)

In radians: t = 0π + 2πn

In degrees: t = 0° + 360n

******************************************************************************

$6)\quad sin (t) = \dfrac{1}{2}$

Where on the Unit Circle does $sin = \dfrac{1}{2}$

<em>Hint: sin is only positive in Quadrants I and II</em>

$\text{Answer: at}\ \dfrac{\pi}{6}\ (30^o)\ \text{and at}\ \dfrac{5\pi}{6}\ (150^o)\ \text{and all rotations of}\ 2\pi \ (360^o)$

$\text{In radians:}\ t = \dfrac{\pi}{6} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{6} + 2\pi n$

In degrees: t = 30° + 360n and 150° + 360n

******************************************************************************

$7)\quad tan (t) = -\sqrt3$

Where on the Unit Circle does $\dfrac{sin}{cos} = \dfrac{-\sqrt3}{1}\ or\ \dfrac{\sqrt3}{-1}\quad \rightarrow \quad (1,-\sqrt3)\ or\ (-1, \sqrt3)$

<em>Hint: sin and cos are only opposite signs in Quadrants II and IV</em>

$\text{Answer: at}\ \dfrac{2\pi}{3}\ (120^o)\ \text{and at}\ \dfrac{5\pi}{3}\ (300^o)\ \text{and all rotations of}\ 2\pi \ (360^o)$

$\text{In radians:}\ t = \dfrac{2\pi}{3} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{3} + 2\pi n$

In degrees: t = 120° + 360n and 300° + 360n

4 0

2 years ago